{"id":2102,"date":"2015-04-22T20:50:16","date_gmt":"2015-04-22T20:50:16","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=2102"},"modified":"2016-08-09T18:24:43","modified_gmt":"2016-08-09T18:24:43","slug":"stoichiometry-of-gaseous-substances-mixtures-and-reactions-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/stoichiometry-of-gaseous-substances-mixtures-and-reactions-formulas\/","title":{"raw":"Stoichiometry of Gaseous Substances, Mixtures, and Reactions","rendered":"Stoichiometry of Gaseous Substances, Mixtures, and Reactions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>LEARNING OBJECTIVES<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Use the ideal gas law to compute gas densities and molar masses<\/li>\r\n \t<li>Perform stoichiometric calculations involving gaseous substances<\/li>\r\n \t<li>State Dalton\u2019s law of partial pressures and use it in calculations involving gaseous mixtures<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp201619456\">The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine <span class=\"no-emphasis\" data-type=\"term\">Lavoisier<\/span>, widely regarded as the \u201cfather of modern chemistry,\u201d changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, \u201cIt took the mob only a moment to remove his head; a century will not suffice to reproduce it.\u201d<\/p>\r\n<p id=\"fs-idm31863392\">As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask \u201cHow much?\u201d We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.<\/p>\r\n\r\n<section id=\"fs-idp78729184\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Density of a Gas<\/h2>\r\n<p id=\"fs-idm26134320\">Recall that the density of a gas is its mass to volume ratio, [latex]\\rho =\\frac{m}{V}.[\/latex] Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can used to determine its molar mass and thereby assist in its identification. The ideal gas law, <em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em>, provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in Example 1.<\/p>\r\n\r\n<div id=\"fs-idp128586304\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1<\/h3>\r\n<h4 id=\"fs-idm63453648\"><strong><span data-type=\"title\">Derivation of a Density Formula from the Ideal Gas Law<\/span><\/strong><\/h4>\r\nUse <em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em> to derive a formula for the density of gas in g\/L\r\n<h4 id=\"fs-idp146685296\"><span data-type=\"title\">Solution<\/span><\/h4>\r\n<ol id=\"fs-idp85675504\" class=\"stepwise\" data-number-style=\"arabic\">\r\n \t<li><em data-effect=\"italics\">PV = nRT<\/em><\/li>\r\n \t<li><em data-effect=\"italics\">Rearrange to get (mol\/L)<\/em>: [latex]\\frac{n}{v}=\\frac{P}{RT}[\/latex]<\/li>\r\n \t<li><em data-effect=\"italics\">Multiply each side of the equation by the molar mass, \u2133.<\/em> When moles are multiplied by \u2133 in g\/mol, g are obtained:\r\n<div data-type=\"newline\"><\/div>\r\n[latex]\\left(\\mathcal{M}\\right)\\left(\\frac{n}{V}\\right)=\\left(\\frac{P}{RT}\\right)\\left(\\mathcal{M}\\right)[\/latex]<\/li>\r\n \t<li>[latex]g\\text{\/L}=\\rho =\\frac{P\\mathcal{M}}{RT}[\/latex]<\/li>\r\n<\/ol>\r\n<h4 id=\"fs-idm8187184\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA gas was found to have a density of 0.0847 g\/L at 17.0 \u00b0C and a pressure of 760 torr. What is its molar mass? What is the gas?\r\n<div id=\"fs-idp1319536\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0[latex]\\rho =\\frac{P\\mathcal{M}}{RT}[\/latex]<\/div>\r\n<div style=\"text-align: right;\" data-type=\"newline\">[latex]0.0847\\text{g\/L}=760\\cancel{\\text{torr}}\\times \\frac{1\\cancel{\\text{atm}}}{760\\cancel{\\text{torr}}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\mathcal{M}}{\\text{0.0821 L}\\cancel{\\text{atm}}\\text{\/mol K}}\\times \\text{290 K}[\/latex]<\/div>\r\n<div style=\"text-align: right;\" data-type=\"newline\">\u2133 = 2.02 g\/mol; therefore, the gas must be hydrogen (H<sub>2<\/sub>, 2.02 g\/mol)<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp140587168\">We must specify both the temperature and the pressure of a gas when calculating its density, because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or pressure. Gas densities are often reported at STP.<\/p>\r\n\r\n<div id=\"fs-idp74274064\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2<\/h3>\r\n<h4 id=\"fs-idm20853312\"><strong><span data-type=\"title\">Empirical\/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas<\/span><\/strong><\/h4>\r\nCyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 \u00b0C, what is the molecular formula for cyclopropane?\r\n<h4 id=\"fs-idp234218112\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nStrategy: First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:\r\n<div id=\"fs-idm18588864\" data-type=\"equation\">[latex]\\text{85.7 g C}\\times \\frac{\\text{1 mol C}}{\\text{12.01 g C}}=\\text{7.136 mol C}\\frac{7.136}{7.136}=\\text{1.00 mol C}[\/latex]<\/div>\r\n<div id=\"fs-idm150258896\" data-type=\"equation\">[latex]\\text{14.3 g H}\\times \\frac{\\text{1 mol H}}{\\text{1.01 g H}}=\\text{14.158 mol H}\\frac{14.158}{7.136}=\\text{1.98 mol H}[\/latex]<\/div>\r\n<p id=\"fs-idp144237600\">Empirical formula is CH<sub>2<\/sub> [empirical mass (EM) of 14.03 g\/empirical unit].<\/p>\r\n<p id=\"fs-idp40237504\">Next, use the density equation related to the ideal gas law to determine the molar mass:<\/p>\r\n\r\n<div id=\"fs-idp110330000\" data-type=\"equation\">[latex]\\text{d}=\\frac{\\text{P}\\mathcal{M}}{\\text{RT}}\\frac{\\text{1.56 g}}{\\text{1.00 L}}=\\text{0.984 atm}\\times \\frac{\\mathcal{M}}{\\text{0.0821 L atm\/mol K}}\\times \\text{323 K}[\/latex]<\/div>\r\n<p id=\"fs-idp221616976\">\u2133 = 42.0 g\/mol, [latex]\\frac{\\mathcal{M}}{\\text{E}\\mathcal{M}}=\\frac{42.0}{14.03}=2.99,[\/latex] so (3)(CH<sub>2<\/sub>) = C<sub>3<\/sub>H<sub>6<\/sub> (molecular formula)<\/p>\r\n\r\n<h4 id=\"fs-idp259910672\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nAcetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 \u00b0C, what is the molecular formula for acetylene?\r\n<div id=\"fs-idp130463408\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0Empirical formula, CH; Molecular formula, C<sub>2<\/sub>H<sub>2<\/sub><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-idp6528928\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Molar Mass of a Gas<\/h2>\r\n<p id=\"fs-idp56968688\">Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, <em data-effect=\"italics\">m<\/em>, to its amount in moles, <em data-effect=\"italics\">n<\/em>:<\/p>\r\n\r\n<div id=\"fs-idp45609344\" data-type=\"equation\">[latex]\\mathcal{M}=\\frac{\\text{grams of substance}}{\\text{moles of substance}}=\\frac{m}{n}[\/latex]<\/div>\r\n<p id=\"fs-idp44359360\">The ideal gas equation can be rearranged to isolate <em data-effect=\"italics\">n<\/em>:<\/p>\r\n\r\n<div id=\"fs-idp78574320\" data-type=\"equation\">[latex]n=\\frac{PV}{RT}[\/latex]<\/div>\r\n<p id=\"fs-idp18382336\">and then combined with the molar mass equation to yield:<\/p>\r\n\r\n<div id=\"fs-idm6613536\" data-type=\"equation\">[latex]\\mathcal{M}=\\frac{mRT}{PV}[\/latex]<\/div>\r\n<p id=\"fs-idm17646432\">This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.<\/p>\r\n\r\n<div id=\"fs-idp207136496\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3<\/h3>\r\n<h4 id=\"fs-idp30589168\"><strong><span data-type=\"title\">Determining the Molar Mass of a Volatile Liquid<\/span><\/strong><\/h4>\r\nThe approximate molar mass of a volatile liquid can be determined by:\r\n<ol id=\"fs-idp104043696\" data-number-style=\"arabic\">\r\n \t<li>Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole<\/li>\r\n \t<li>Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure<\/li>\r\n \t<li>Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample\u2019s mass (see\u00a0Figure 1)<\/li>\r\n<\/ol>\r\n<figure id=\"CNX_Chem_09_03_liquidgas\">[caption id=\"\" align=\"alignnone\" width=\"839\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212041\/CNX_Chem_09_03_liquidgas1.jpg\" alt=\"This figure shows four photos each connected by a right-facing arrow. The first photo shows a glass flask with aluminum foil covering the top sitting on a scale. The scale reads 89.516. The second photo shows a syringe being inserted into the flask through the aluminum foil covering. The third photo shows the glass flask being inserted into a beaker of water. The water appears to be heated at 100. The fourth photo shows the glass flask being weighed again. This time the scale reads 89.512.\" width=\"839\" height=\"203\" data-media-type=\"image\/jpeg\" \/> Figure 1. When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At [latex]{t}_{l\\rightarrow g},[\/latex] the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)[\/caption]<\/figure>\r\n<p id=\"fs-idm10831984\">Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm<sup>3<\/sup> at 99.6 \u00b0C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?<\/p>\r\n\r\n<h4 id=\"fs-idp60723232\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nSince [latex]\\mathcal{M}=\\frac{m}{n}[\/latex] and [latex]n=\\frac{PV}{RT},[\/latex] substituting and rearranging gives [latex]\\mathcal{M}=\\frac{mRT}{PV},[\/latex]\r\n<p id=\"fs-idp43964944\">then<\/p>\r\n\r\n<div id=\"fs-idm11022240\" data-type=\"equation\">[latex]\\mathcal{M}=\\frac{mRT}{PV}=\\frac{\\left(\\text{0.494 g}\\right)\\times \\text{0.08206 L\\cdot atm\/mol K}\\times \\text{372.8 K}}{\\text{0.976 atm}\\times \\text{0.129 L}}=120\\text{g\/mol}.[\/latex]<\/div>\r\n<h4 id=\"fs-idp146723968\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA sample of phosphorus that weighs 3.243 \u00d7 10<sup>-2<\/sup> g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 \u00b0C. What are the molar mass and molecular formula of phosphorus vapor?\r\n<div id=\"fs-idp108389616\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0124 g\/mol P<sub>4<\/sub><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-idp148167024\" data-depth=\"2\">\r\n<h2 data-type=\"title\">The Pressure of a Mixture of Gases: Dalton\u2019s Law<\/h2>\r\n<p id=\"fs-idp70155072\">Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other\u2019s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it present alone in the container (Figure 2). The pressure exerted by each individual gas in a mixture is called its <strong><span data-type=\"term\">partial pressure<\/span><\/strong>. This observation is summarized by <strong><span data-type=\"term\">Dalton\u2019s law of partial pressures<\/span><\/strong>: <em data-effect=\"italics\">The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases<\/em>:<\/p>\r\n\r\n<div id=\"fs-idp65742144\" data-type=\"equation\">[latex]{P}_{Total}={P}_{A}+{P}_{B}+{P}_{C}+\\ldots ={\\text{\\Sigma}}_{\\text{i}}{P}_{\\text{i}}[\/latex]<\/div>\r\n<p id=\"fs-idp97232880\">In the equation <em data-effect=\"italics\">P<sub>Total<\/sub><\/em> is the total pressure of a mixture of gases, <em data-effect=\"italics\">P<sub>A<\/sub><\/em> is the partial pressure of gas A; <em data-effect=\"italics\">P<sub>B<\/sub><\/em> is the partial pressure of gas B; <em data-effect=\"italics\">P<sub>C<\/sub><\/em> is the partial pressure of gas C; and so on.<\/p>\r\n\r\n<figure id=\"CNX_Chem_09_03_DaltonLaw1\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"881\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212042\/CNX_Chem_09_03_DaltonLaw11.jpg\" alt=\"This figure includes images of four gas-filled cylinders or tanks. Each has a valve at the top. The interior of the first cylinder is shaded blue. This region contains 5 small blue circles that are evenly distributed. The label \u201c300 k P a\u201d is on the cylinder. The second cylinder is shaded lavender. This region contains 8 small purple circles that are evenly distributed. The label \u201c600 k P a\u201d is on the cylinder. To the right of these cylinders is a third cylinder. Its interior is shaded pale yellow. This region contains 12 small yellow circles that are evenly distributed. The label \u201c450 k P a\u201d is on this region of the cylinder. An arrow labeled \u201cTotal pressure combined\u201d appears to the right of these three cylinders. This arrow points to a fourth cylinder. The interior of this cylinder is shaded a pale green. It contains evenly distributed small circles in the following quantities and colors; 5 blue, 8 purple, and 12 yellow. This cylinder is labeled \u201c1350 k P a.\u201d\" width=\"881\" height=\"395\" data-media-type=\"image\/jpeg\" \/> Figure 2. If equal-volume cylinders containing gas A at a pressure of 300 kPa, gas B at a pressure of 600 kPa, and gas C at a pressure of 450 kPa are all combined in the same-size cylinder, the total pressure of the mixture is 1350 kPa.[\/caption]\r\n\r\n<\/figure>\r\n<p id=\"fs-idp110759216\">The partial pressure of gas A is related to the total pressure of the gas mixture via its <strong><span data-type=\"term\">mole fraction<\/span><\/strong>, a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components):<\/p>\r\n\r\n<div id=\"fs-idp18188304\" data-type=\"equation\">[latex]{P}_{A}={X}_{A}\\times {P}_{Total}\\text{where}{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex]<\/div>\r\n<p id=\"fs-idp75739968\">where <em data-effect=\"italics\">P<sub>A<\/sub><\/em>, <em data-effect=\"italics\">X<sub>A<\/sub><\/em>, and <em data-effect=\"italics\">n<sub>A<\/sub><\/em> are the partial pressure, mole fraction, and number of moles of gas A, respectively, and <em data-effect=\"italics\">n<sub>Total<\/sub><\/em> is the number of moles of all components in the mixture.<\/p>\r\n\r\n<div id=\"fs-idp143452448\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4<\/h3>\r\n<h4 id=\"fs-idp147731856\"><strong><span data-type=\"title\">The Pressure of a Mixture of Gases<\/span><\/strong><\/h4>\r\nA 10.0-L vessel contains 2.50 \u00d7 10<sup>-3<\/sup> mol of H<sub>2<\/sub>, 1.00 \u00d7 10<sup>-3<\/sup> mol of He, and 3.00 \u00d7 10<sup>-4<\/sup> mol of Ne at 35 \u00b0C.\r\n<p id=\"fs-idp100506960\">(a) What are the partial pressures of each of the gases?<\/p>\r\n<p id=\"fs-idp147676624\">(b) What is the total pressure in atmospheres?<\/p>\r\n\r\n<h4 id=\"fs-idp17315648\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nThe gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using [latex]P=\\frac{nRT}{V}:[\/latex]\r\n<div id=\"fs-idp201684064\" data-type=\"equation\">[latex]{P}_{{\\text{H}}_{2}}=\\frac{\\left(2.50\\times {10}^{\\text{-3}}\\cancel{\\text{mol}}\\right)\\left(0.08206\\cancel{\\text{L}}\\text{atm}\\cancel{{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}}\\right)\\left(\\text{308 K}\\right)}{10.0\\cancel{\\text{L}}}=6.32\\times {10}^{\\text{-3}}\\text{atm}[\/latex]<\/div>\r\n<div id=\"fs-idp100280736\" data-type=\"equation\">[latex]{P}_{\\text{He}}=\\frac{\\left(1.00\\times {10}^{\\text{-3}}\\cancel{\\text{mol}}\\right)\\left(\\text{0.08206 L atm}\\cancel{{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}}\\right)\\left(\\text{308 K}\\right)}{10.0\\cancel{\\text{L}}}=2.53\\times {10}^{\\text{-3}}\\text{atm}[\/latex]<\/div>\r\n<div id=\"fs-idp36264816\" data-type=\"equation\">[latex]{P}_{\\text{Ne}}=\\frac{\\left(3.00\\times {10}^{\\text{-4}}\\cancel{\\text{mol}}\\right)\\left(0.08206\\cancel{\\text{L}}\\text{atm}\\cancel{{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}}\\right)\\left(\\text{308 K}\\right)}{10.0\\cancel{\\text{L}}}=7.58\\times {10}^{\\text{-4}}\\text{atm}[\/latex]<\/div>\r\n<p id=\"fs-idp49915536\">The total pressure is given by the sum of the partial pressures:<\/p>\r\n\r\n<div id=\"fs-idp207663808\" data-type=\"equation\">[latex]{P}_{\\text{T}}={P}_{{\\text{H}}_{2}}+{P}_{\\text{He}}+{P}_{\\text{Ne}}=\\left(0.00632+0.00253+0.00076\\right)\\text{atm}=9.61\\times {10}^{\\text{-3}}\\text{atm}[\/latex]<\/div>\r\n<h4 id=\"fs-idp1986144\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA 5.73-L flask at 25 \u00b0C contains 0.0388 mol of N<sub>2<\/sub>, 0.147 mol of CO, and 0.0803 mol of H<sub>2<\/sub>. What is the total pressure in the flask in atmospheres?\r\n<div id=\"fs-idm10367824\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a01.137 atm<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp107652512\">Here is another example of this concept, but dealing with mole fraction calculations.<\/p>\r\n\r\n<div id=\"fs-idp107854880\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5<\/h3>\r\n<h4 id=\"fs-idp279668416\"><strong><span data-type=\"title\">The Pressure of a Mixture of Gases<\/span><\/strong><\/h4>\r\nA gas mixture used for anesthesia contains 2.83 mol oxygen, O<sub>2<\/sub>, and 8.41 mol nitrous oxide, N<sub>2<\/sub>O. The total pressure of the mixture is 192 kPa.\r\n<p id=\"fs-idp41662496\">(a) What are the mole fractions of O<sub>2<\/sub> and N<sub>2<\/sub>O?<\/p>\r\n<p id=\"fs-idp132439904\">(b) What are the partial pressures of O<sub>2<\/sub> and N<sub>2<\/sub>O?<\/p>\r\n\r\n<h4 id=\"fs-idp86077792\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nThe mole fraction is given by [latex]{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex] and the partial pressure is <em data-effect=\"italics\">P<sub>A<\/sub><\/em> = <em data-effect=\"italics\">X<sub>A<\/sub><\/em> \u00d7 <em data-effect=\"italics\">P<sub>Total<\/sub><\/em>.\r\n<p id=\"fs-idm32258688\">For O<sub>2<\/sub>,<\/p>\r\n\r\n<div id=\"fs-idm52855552\" data-type=\"equation\">[latex]{X}_{{O}_{2}}=\\frac{{n}_{{O}_{2}}}{{n}_{Total}}=\\frac{\\text{2.83 mol}}{\\left(2.83+8.41\\right)\\text{mol}}=0.252[\/latex]<\/div>\r\n<p id=\"fs-idp192169392\">and [latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=0.252\\times \\text{192 kPa}=\\text{48.4 kPa}[\/latex]<\/p>\r\n<p id=\"fs-idp263870464\">For N<sub>2<\/sub>O,<\/p>\r\n\r\n<div id=\"fs-idp202334992\" data-type=\"equation\">[latex]{X}_{{O}_{2}}=\\frac{{n}_{{O}_{2}}}{{n}_{Total}}=\\frac{\\text{2.83 mol}}{\\left(2.83+8.41\\right)\\text{mol}}=0.252[\/latex]<\/div>\r\n<p id=\"fs-idp259944368\">and<\/p>\r\n[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=\\left(0.252\\right)\\times \\text{192 kPa}[\/latex]\r\n<p id=\"fs-idp66867744\">[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=0.252\\times \\text{192 kPa}=\\text{48.4 kPa}[\/latex]<\/p>\r\n\r\n<h4 id=\"fs-idm31767024\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nWhat is the pressure of a mixture of 0.200 g of H<sub>2<\/sub>, 1.00 g of N<sub>2<\/sub>, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 \u00b0C?\r\n<div id=\"fs-idp69846704\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a01.87 atm<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-idp221977104\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Collection of Gases over Water<\/h2>\r\n<figure id=\"CNX_Chem_09_03_WaterVapor\">\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"300\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212044\/CNX_Chem_09_03_WaterVapor1.jpg\" alt=\"This figure shows a diagram of equipment used for collecting a gas over water. To the left is an Erlenmeyer flask. It is approximately two thirds full of a lavender colored liquid. Bubbles are evident in the liquid. The label \u201cReaction Producing Gas\u201d appears below the flask. A line segment connects this label to the liquid in the flask. The flask has a stopper in it through which a single glass tube extends from the open region above the liquid in the flask up, through the stopper, to the right, then angles down into a pan that is nearly full of light blue water. This tube again extends right once it is well beneath the water\u2019s surface. It then bends up into an inverted flask which is labeled \u201cCollection Flask.\u201d This collection flask is positioned with its mouth beneath the surface of the light blue water and appears approximately half full. Bubbles are evident in the water in the inverted flask. The open space above the water in the inverted flask is labeled \u201ccollected gas.\u201d\" width=\"300\" height=\"257\" data-media-type=\"image\/jpeg\" \/> Figure 3. When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).[\/caption]\r\n\r\n<\/figure>A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 3), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.\r\n\r\nHowever, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor\u2014this is referred to as the \u201cdry\u201d gas pressure, that is, the pressure of the gas only, without water vapor. The <strong><span data-type=\"term\">vapor pressure of water<\/span><\/strong>, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 4); more detailed information on the temperature dependence of water vapor can be found in Table 1, and vapor pressure will be discussed in more detail in the next chapter on liquids.<span data-type=\"media\" data-alt=\"A graph is shown. The horizontal axis is labeled \u201cTemperature ( degrees C )\u201d with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled \u201cVapor pressure ( torr )\u201d with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, \u201cVapor pressure at ( 100 degrees C ).\u201d\"><span data-type=\"media\" data-alt=\"A graph is shown. The horizontal axis is labeled \u201cTemperature ( degrees C )\u201d with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled \u201cVapor pressure ( torr )\u201d with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, \u201cVapor pressure at ( 100 degrees C ).\u201d\">\r\n<\/span><\/span>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"650\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212045\/CNX_Chem_09_03_WaterVapor21.jpg\" alt=\"A graph is shown. The horizontal axis is labeled \u201cTemperature ( degrees C )\u201d with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled \u201cVapor pressure ( torr )\u201d with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, \u201cVapor pressure at ( 100 degrees C ).\u201d\" width=\"650\" height=\"506\" data-media-type=\"image\/jpeg\" \/> Figure 4. This graph shows the vapor pressure of water at sea level as a function of temperature.[\/caption]\r\n<table id=\"fs-idm68841392\" summary=\"This table has six columns and 13 rows. The first row is a header and it labels each column, \u201cTemperature (degree sign C),\u201d \u201cPressure (torr),\u201d \u201cTemperature (degree sign C),\u201d \u201cPressure (torr),\u201d \u201cTemperature (degree sign C),\u201d and \u201cPressure (torr).\u201d Under the first column are the following: negative 10, negative 5, negative 2, 0, 2, 4, 6, 8, 10, 12, 14, and 16. Under the second column are the following: 1.95, 3.0, 3.9, 4.6, 5.3, 6.1, 7.0, 8.0, 9.2, 10.5, 12.0, and 13.6. Under the third column are the following: 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, and 29. Under the fourth column are the following: 15.5, 16.5, 17.5, 18.7, 19.8, 21.1, 22.4, 23.8, 25.2, 26.7, 28.3, and 30.0. Under the fifth column are the following: 30, 35, 40, 50, 60, 70, 80, 90, 95, 99, 100.0, and 101.0. Under the sixth column are the following: 31.8, 42.2, 55.3, 92.5, 149.4, 233.7, 355.1, 525.8, 633.9, 733.2, 760.0, and 787.6.\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"6\">Table 1. Vapor Pressure of Ice and Water in Various Temperatures at Sea Level<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<th>Temperature (\u00b0C)<\/th>\r\n<th>Pressure (torr)<\/th>\r\n<th>Temperature (\u00b0C)<\/th>\r\n<th>Pressure (torr)<\/th>\r\n<th>Temperature (\u00b0C)<\/th>\r\n<th>Pressure (torr)<\/th>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>\u201310<\/td>\r\n<td>1.95<\/td>\r\n<td>18<\/td>\r\n<td>15.5<\/td>\r\n<td>30<\/td>\r\n<td>31.8<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>\u20135<\/td>\r\n<td>3.0<\/td>\r\n<td>19<\/td>\r\n<td>16.5<\/td>\r\n<td>35<\/td>\r\n<td>42.2<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>\u20132<\/td>\r\n<td>3.9<\/td>\r\n<td>20<\/td>\r\n<td>17.5<\/td>\r\n<td>40<\/td>\r\n<td>55.3<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>0<\/td>\r\n<td>4.6<\/td>\r\n<td>21<\/td>\r\n<td>18.7<\/td>\r\n<td>50<\/td>\r\n<td>92.5<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>5.3<\/td>\r\n<td>22<\/td>\r\n<td>19.8<\/td>\r\n<td>60<\/td>\r\n<td>149.4<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4<\/td>\r\n<td>6.1<\/td>\r\n<td>23<\/td>\r\n<td>21.1<\/td>\r\n<td>70<\/td>\r\n<td>233.7<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>6<\/td>\r\n<td>7.0<\/td>\r\n<td>24<\/td>\r\n<td>22.4<\/td>\r\n<td>80<\/td>\r\n<td>355.1<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>8<\/td>\r\n<td>8.0<\/td>\r\n<td>25<\/td>\r\n<td>23.8<\/td>\r\n<td>90<\/td>\r\n<td>525.8<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>10<\/td>\r\n<td>9.2<\/td>\r\n<td>26<\/td>\r\n<td>25.2<\/td>\r\n<td>95<\/td>\r\n<td>633.9<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>12<\/td>\r\n<td>10.5<\/td>\r\n<td>27<\/td>\r\n<td>26.7<\/td>\r\n<td>99<\/td>\r\n<td>733.2<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>14<\/td>\r\n<td>12.0<\/td>\r\n<td>28<\/td>\r\n<td>28.3<\/td>\r\n<td>100.0<\/td>\r\n<td>760.0<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>16<\/td>\r\n<td>13.6<\/td>\r\n<td>29<\/td>\r\n<td>30.0<\/td>\r\n<td>101.0<\/td>\r\n<td>787.6<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-idp46681200\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example\u00a06<\/h3>\r\n<h4 id=\"fs-idp35760544\"><strong><span data-type=\"title\">Pressure of a Gas Collected Over Water<\/span><\/strong><\/h4>\r\nIf 0.200 L of argon is collected over water at a temperature of 26 \u00b0C and a pressure of 750 torr in a system like that shown in Figure 3, what is the partial pressure of argon?\r\n<h4 id=\"fs-idp18515888\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nAccording to Dalton\u2019s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:\r\n<div id=\"fs-idp297283264\" data-type=\"equation\">[latex]{P}_{\\text{T}}={P}_{\\text{Ar}}+{P}_{{\\text{H}}_{2}\\text{O}}[\/latex]<\/div>\r\n<p id=\"fs-idp135677552\">Rearranging this equation to solve for the pressure of argon gives:<\/p>\r\n\r\n<div id=\"fs-idp33156768\" data-type=\"equation\">[latex]{P}_{\\text{Ar}}={P}_{\\text{T}}-{P}_{{\\text{H}}_{2}\\text{O}}[\/latex]<\/div>\r\n<p id=\"fs-idp144314112\">The pressure of water vapor above a sample of liquid water at 26 \u00b0C is 25.2 torr (<a class=\"target-chapter\" title=\"Water Properties\" href=\".\/chapter\/water-properties-2\/\" target=\"_blank\">Water Properties<\/a>), so:<\/p>\r\n\r\n<div id=\"fs-idm40258608\" data-type=\"equation\">[latex]{P}_{\\text{Ar}}=750\\text{torr}-25.2\\text{torr}=725\\text{torr}[\/latex]<\/div>\r\n<h4 id=\"fs-idp98186880\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA sample of oxygen collected over water at a temperature of 29.0 \u00b0C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?\r\n<div id=\"fs-idp98213776\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0734 torr<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/section><section id=\"fs-idp132626656\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Chemical Stoichiometry and Gases<\/h2>\r\n<p id=\"fs-idp137887648\">Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.<\/p>\r\n<p id=\"fs-idp221629776\">We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.<\/p>\r\n\r\n<\/section><section id=\"fs-idp45488016\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Avogadro\u2019s Law Revisited<\/h2>\r\n<p id=\"fs-idm7424624\">Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.<\/p>\r\n<p id=\"fs-idp21344160\">We can extend Avogadro\u2019s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to [latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\phantom{\\rule{0.4em}{0ex}}\\rightarrow\\phantom{\\rule{0.4em}{0ex}}2{\\text{NH}}_{3}\\left(g\\right),[\/latex] a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.<\/p>\r\n<p id=\"fs-idp16937680\">The explanation for this is illustrated in Figure 5. According to Avogadro\u2019s law, equal volumes of gaseous N<sub>2<\/sub>, H<sub>2<\/sub>, and NH<sub>3<\/sub>, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N<sub>2<\/sub> reacts with three molecules of H<sub>2<\/sub> to produce two molecules of NH<sub>3<\/sub>, the volume of H<sub>2<\/sub> required is three times the volume of N<sub>2<\/sub>, and the volume of NH<sub>3<\/sub> produced is two times the volume of N<sub>2<\/sub>.<span data-type=\"media\" data-alt=\"This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled \u201cN subscript 2\u201d. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled \u201cH subscript 2.\u201d Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled \u201cN H subscript 3.\u201d Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.\"><span data-type=\"media\" data-alt=\"This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled \u201cN subscript 2\u201d. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled \u201cH subscript 2.\u201d Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled \u201cN H subscript 3.\u201d Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.\">\r\n<\/span><\/span><\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"880\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212046\/CNX_Chem_09_03_Ammonia1.jpg\" alt=\"This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled \u201cN subscript 2\u201d. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled \u201cH subscript 2.\u201d Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled \u201cN H subscript 3.\u201d Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.\" width=\"880\" height=\"438\" data-media-type=\"image\/jpeg\" \/> Figure 5. One volume of N<sub>2<\/sub> combines with three volumes of H<sub>2<\/sub> to form two volumes of NH<sub>3<\/sub>.[\/caption]\r\n\r\n<div id=\"fs-idp89809920\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7<\/h3>\r\n<h4 id=\"fs-idp31720640\"><strong><span data-type=\"title\">Reaction of Gases<\/span><\/strong><\/h4>\r\nPropane, C<sub>3<\/sub>H<sub>8<\/sub>(<em data-effect=\"italics\">g<\/em>), is used in gas grills to provide the heat for cooking. What volume of O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) measured at 25 \u00b0C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.\r\n<h4 id=\"fs-idp137922560\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nThe ratio of the volumes of C<sub>3<\/sub>H<sub>8<\/sub> and O<sub>2<\/sub> will be equal to the ratio of their coefficients in the balanced equation for the reaction:\r\n<div id=\"fs-idp57291280\" data-type=\"equation\">[latex]\\begin{array}{l}{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\text{ }\\text{ }\\rightarrow\\text{ }\\text{ }3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(l\\right)\\\\ \\text{1 volume}+\\text{5 volumes}\\phantom{\\rule{3.5em}{0ex}}\\text{3 volumes}+\\text{4 volumes}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idp90728656\">From the equation, we see that one volume of C<sub>3<\/sub>H<sub>8<\/sub> will react with five volumes of O<sub>2<\/sub>:<\/p>\r\n\r\n<div id=\"fs-idp8505312\" data-type=\"equation\">[latex]2.7\\cancel{\\text{L}{\\text{C}}_{3}{\\text{H}}_{8}}\\times \\frac{\\text{5 L}{\\text{O}}_{2}}{1\\cancel{\\text{L}{\\text{C}}_{3}{\\text{H}}_{8}}}=\\text{13.5 L}{\\text{O}}_{2}[\/latex]<\/div>\r\n<p id=\"fs-idp78474976\">A volume of 13.5 L of O<sub>2<\/sub> will be required to react with 2.7 L of C<sub>3<\/sub>H<sub>8<\/sub>.<\/p>\r\n\r\n<h4 id=\"fs-idp109294464\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nAn acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C<sub>2<\/sub>H<sub>2<\/sub>, at 0 \u00b0C and 1 atm. How many tanks of oxygen, each providing 7.00 \u00d7 10<sup>3<\/sup> L of O<sub>2<\/sub> at 0 \u00b0C and 1 atm, will be required to burn the acetylene?\r\n<div id=\"fs-idp172649872\" data-type=\"equation\">[latex]2{\\text{C}}_{2}{\\text{H}}_{2}+5{\\text{O}}_{2}\\rightarrow 4{\\text{CO}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]<\/div>\r\n<div id=\"fs-idm19557184\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a03.34 tanks (2.34 \u00d7 10<sup>4<\/sup> L)<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8<\/h3>\r\n<h4 id=\"fs-idp19707328\"><strong><span data-type=\"title\">Volumes of Reacting Gases<\/span><\/strong><\/h4>\r\nAmmonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 \u00b0C and 1 atm, was manufactured. What volume of H<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N<sub>2<\/sub>?\r\n<div id=\"fs-idp109603280\" data-type=\"equation\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightarrow 2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/div>\r\n<h4 id=\"fs-idp212669488\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nBecause equal volumes of H<sub>2<\/sub> and NH<sub>3<\/sub> contain equal numbers of molecules and each three molecules of H<sub>2<\/sub> that react produce two molecules of NH<sub>3<\/sub>, the ratio of the volumes of H<sub>2<\/sub> and NH<sub>3<\/sub> will be equal to 3:2. Two volumes of NH<sub>3<\/sub>, in this case in units of billion ft<sup>3<\/sup>, will be formed from three volumes of H<sub>2<\/sub>:\r\n<div id=\"fs-idp171418912\" data-type=\"equation\">[latex]683\\cancel{\\text{billion}{\\text{ft}}^{3}{\\text{NH}}_{3}}\\times \\frac{\\text{3 billion}{\\text{ft}}^{3}{\\text{H}}_{2}}{2\\cancel{\\text{billion}{\\text{ft}}^{3}{\\text{NH}}_{3}}}=1.02\\times {10}^{3}\\text{billion}{\\text{ft}}^{3}{\\text{H}}_{2}[\/latex]<\/div>\r\n<p id=\"fs-idp138098944\">The manufacture of 683 billion ft<sup>3<\/sup> of NH<sub>3<\/sub> required 1020 billion ft<sup>3<\/sup> of H<sub>2<\/sub>. (At 25 \u00b0C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)<\/p>\r\n\r\n<h4 id=\"fs-idp37301360\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nWhat volume of O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) measured at 25 \u00b0C and 760 torr is required to react with 17.0 L of ethylene, C<sub>2<\/sub>H<sub>4<\/sub>(<em data-effect=\"italics\">g<\/em>), measured under the same conditions of temperature and pressure? The products are CO<sub>2<\/sub> and water vapor.\r\n<div id=\"fs-idp105257584\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a051.0 L<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9<\/h3>\r\n<div id=\"fs-idp13049024\" data-type=\"example\">\r\n<h4 id=\"fs-idm2007840\"><strong><span data-type=\"title\">Volume of Gaseous Product<\/span><\/strong><\/h4>\r\nWhat volume of hydrogen at 27 \u00b0C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?\r\n<div id=\"fs-idm20890912\" data-type=\"equation\">[latex]2\\text{Ga}\\left(s\\right)+6\\text{HCl}\\left(aq\\right)\\rightarrow 2{\\text{GaCl}}_{3}\\left(aq\\right)+3{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/div>\r\n<h4 id=\"fs-idp98177696\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nTo convert from the mass of gallium to the volume of H<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), we need to do something like this:<span id=\"fs-idp205351920\" data-type=\"media\" data-alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled \u201cMass of G a.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled \u201cMoles of G a.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled \u201cMoles of H subscript 2 ( g ).\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded lavender and is labeled \u201cVolume of H subscript 2 ( g ).\u201d\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212048\/CNX_Chem_09_03_Example3_img1.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled \u201cMass of G a.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled \u201cMoles of G a.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled \u201cMoles of H subscript 2 ( g ).\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded lavender and is labeled \u201cVolume of H subscript 2 ( g ).\u201d\" width=\"881\" height=\"104\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idp47678000\">The first two conversions are:<\/p>\r\n\r\n<div id=\"fs-idm46693760\" data-type=\"equation\">[latex]8.88\\cancel{\\text{g Ga}}\\times \\frac{1\\cancel{\\text{mol Ga}}}{69.723\\cancel{\\text{g Ga}}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{3 mol}{\\text{H}}_{2}}{2\\cancel{\\text{mol Ga}}}=0.191{\\text{mol H}}_{2}[\/latex]<\/div>\r\n<p id=\"fs-idp35037744\">Finally, we can use the ideal gas law:<\/p>\r\n\r\n<div id=\"fs-idp48121168\" data-type=\"equation\">[latex]{V}_{{\\text{H}}_{2}}={\\left(\\frac{nRT}{P}\\right)}_{{\\text{H}}_{2}}=\\frac{0.191\\cancel{\\text{mol}}\\times \\text{0.08206 L}\\cancel{\\text{atm}{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}}\\times \\text{300 K}}{0.951\\cancel{\\text{atm}}}=\\text{4.94 L}[\/latex]<\/div>\r\n<h4 id=\"fs-idm17626704\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nSulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO<sub>2<\/sub> at 343 \u00b0C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen?\r\n<div id=\"fs-idp30333040\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a01.30 \u00d7 10<sup>3<\/sup> L<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm13395728\" class=\"chemistry sciences-interconnect\" data-type=\"note\"><\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"title\">\r\n<div id=\"fs-idm152886256\" class=\"chemistry sciences-interconnect\" data-type=\"note\">\r\n<div id=\"fs-idp13049024\" data-type=\"example\">\r\n<h3 id=\"fs-idm2007840\">Greenhouse Gases and Climate Change<\/h3>\r\n<\/div>\r\n<div id=\"fs-idm13395728\" class=\"chemistry sciences-interconnect\" data-type=\"note\">\r\n<p id=\"fs-idp143406944\">The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost [latex]\\frac{1}{3}[\/latex] is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions\u2014without atmosphere, the average global average temperature of 14 \u00b0C (57 \u00b0F) would be about \u201319 \u00b0C (\u20132 \u00b0F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth\u2019s climate (Figure 6).<\/p>\r\n\r\n<figure id=\"CNX_Chem_09_03_GlobalWarming\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"650\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212049\/CNX_Chem_09_03_GlobalWarming1.jpg\" alt=\"This diagram shows half of a two dimensional view of the earth in blue and green at the left of the image. A slight distance outside the hemisphere is a grey arc. A line segment connects the label \u201cAtmosphere\u201d to the region between the hemisphere and the grey arc. In this region, near the surface of the earth the chemical formulas C O subscript 2, C H subscript 3, and N subscript 2 O appear. Five red arrows formed from wavy lines extend from green regions on the earth out into and just beyond the region labeled \u201cAtmosphere.\u201d The label \u201cInfrared radiation\u201d points to one of these red arrows. At a fair distance outside of the grey arc appears a yellow circle with a jagged boundary. This circle is labeled \u201cSun.\u201d From it extend yellow arrows with wavy lines which extend toward the earth. Three of the arrows extend to the green region on the earth. One of the arrows appears to be reflected off the grey arc, causing its path to turn away from the earth.\" width=\"650\" height=\"436\" data-media-type=\"image\/jpeg\" \/> Figure 6. Greenhouse gases trap enough of the sun\u2019s energy to make the planet habitable\u2014this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.[\/caption]\r\n\r\n<\/figure>\r\n<p id=\"fs-idm12984240\">There is strong evidence from multiple sources that higher atmospheric levels of CO<sub>2<\/sub> are caused by human activity, with fossil fuel burning accounting for about [latex]\\frac{3}{4}[\/latex] of the recent increase in CO<sub>2<\/sub>. Reliable data from ice cores reveals that CO<sub>2<\/sub> concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO<sub>2<\/sub> concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure 7).<span data-type=\"media\" data-alt=\"This figure has the heading \u201cCarbon Dioxide in the Atmosphere.\u201d The first graph has a horizontal axis label \u201cYear ( B C )\u201d and a vertical axis label \u201cCarbon dioxide concentration ( p p m ).\u201d The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.\"><span data-type=\"media\" data-alt=\"This figure has the heading \u201cCarbon Dioxide in the Atmosphere.\u201d The first graph has a horizontal axis label \u201cYear ( B C )\u201d and a vertical axis label \u201cCarbon dioxide concentration ( p p m ).\u201d The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.\">\r\n<\/span><\/span><\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"879\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212050\/CNX_Chem_09_03_GlobalWarming21.jpg\" alt=\"This figure has the heading \u201cCarbon Dioxide in the Atmosphere.\u201d The first graph has a horizontal axis label \u201cYear ( B C )\u201d and a vertical axis label \u201cCarbon dioxide concentration ( p p m ).\u201d The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.\" width=\"879\" height=\"329\" data-media-type=\"image\/jpeg\" \/> Figure 7. CO<sub>2<\/sub> levels over the past 700,000 years were typically from 200\u2013300 ppm, with a steep, unprecedented increase over the past 50 years.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\n<a href=\"http:\/\/www.epa.gov\/climatestudents\/basics\/today\/greenhouse-effect.html\" target=\"_blank\">Click here to see a 2-minute video from the Environmental Protection Agency <\/a>explaining greenhouse gases and global warming.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"title\">\r\n<div id=\"fs-idm152886256\" class=\"chemistry sciences-interconnect\" data-type=\"note\">\r\n<h3 data-type=\"title\">Portrait of a Chemist: Susan Solomon<\/h3>\r\n[caption id=\"\" align=\"alignright\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212054\/CNX_Chem_09_03_SusanSolom1.jpg\" alt=\"A photograph is shown of Susan Solomon sitting next to a globe.\" width=\"325\" height=\"274\" data-media-type=\"image\/jpeg\" \/> Figure 8. Susan Solomon\u2019s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)[\/caption]\r\n<p id=\"CNX_Chem_09_03_SusanSolom\">Atmospheric and climate scientist Susan <span class=\"no-emphasis\" data-type=\"term\">Solomon<\/span> (Figure 8) is the author of one of <em data-effect=\"italics\">The New York Times<\/em> books of the year (<em data-effect=\"italics\">The Coldest March<\/em>, 2001), one of Time magazine\u2019s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.<\/p>\r\nFor more information, watch this <a href=\"http:\/\/techtv.mit.edu\/videos\/17040-introducing-atmospheric-chemist-prof-susan-solomon\" target=\"_blank\">video\u00a0from MIT about Susan Solomon<\/a>.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<section>\r\n<div data-type=\"note\">\r\n<p id=\"fs-idp236281408\">The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton\u2019s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro\u2019s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Equations<\/h3>\r\n<section>\r\n<div data-type=\"note\">\r\n<ul>\r\n \t<li><em data-effect=\"italics\">P<sub>Total<\/sub><\/em> = <em data-effect=\"italics\">P<sub>A<\/sub><\/em> + <em data-effect=\"italics\">P<sub>B<\/sub><\/em> + <em data-effect=\"italics\">P<sub>C<\/sub><\/em> + \u2026 = \u01a9<sub>i<\/sub><em data-effect=\"italics\">P<\/em><sub>i<\/sub><\/li>\r\n \t<li><em data-effect=\"italics\">P<sub>A<\/sub><\/em> = <em data-effect=\"italics\">X<sub>A<\/sub> P<sub>Total<\/sub><\/em><\/li>\r\n \t<li>[latex]{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section><\/div>\r\n<section id=\"fs-idm40478192\" class=\"summary\" data-depth=\"1\">\r\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\r\n<div class=\"entry-content\">\r\n<div class=\"im_section\">\r\n<div class=\"im_section\">\r\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<section id=\"fs-idp118960256\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idp231956592\" data-type=\"exercise\">\r\n<div id=\"fs-idp8716432\" data-type=\"problem\">\r\n<ol>\r\n \t<li id=\"fs-idp5305888\">What is the density of laughing gas, dinitrogen monoxide, N<sub>2<\/sub>O, at a temperature of 325 K and a pressure of 113.0 kPa?<\/li>\r\n \t<li>Calculate the density of Freon 12, CF<sub>2<\/sub>Cl<sub>2<\/sub>, at 30.0 \u00b0C and 0.954 atm.<\/li>\r\n \t<li>Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.<\/li>\r\n \t<li>A cylinder of O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) used in breathing by emphysema patients has a volume of 3.00 L at a pressure of 10.0 atm. If the temperature of the cylinder is 28.0 \u00b0C, what mass of oxygen is in the cylinder?<\/li>\r\n \t<li>What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 \u00b0C and a pressure of 307 torr?<\/li>\r\n \t<li>What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 \u00b0C and a pressure of 777 torr?<\/li>\r\n \t<li>How could you show experimentally that the molecular formula of propene is C<sub>3<\/sub>H<sub>6<\/sub>, not CH<sub>2<\/sub>?<\/li>\r\n \t<li>The density of a certain gaseous fluoride of phosphorus is 3.93 g\/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.<\/li>\r\n \t<li>Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies l25 mL with a pressure of 99.5 kPa at 22 \u00b0C?\r\n<ol>\r\n \t<li>Outline the steps necessary to answer the question.<\/li>\r\n \t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A 36.0\u2013L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO<sub>2<\/sub>, 805 g O<sub>2<\/sub>, and 4,880 g N<sub>2<\/sub>. What is the pressure in the flask in atmospheres, in torr, and in kilopascals?<\/li>\r\n \t<li>A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO<sub>2<\/sub>, 12.0% O<sub>2<\/sub>, and the remainder N<sub>2<\/sub> at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)<\/li>\r\n \t<li>A sample of gas isolated from unrefined petroleum contains 90.0% CH<sub>4<\/sub>, 8.9% C<sub>2<\/sub>H<sub>6<\/sub>, and 1.1% C<sub>3<\/sub>H<sub>8<\/sub> at a total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)<\/li>\r\n \t<li>A mixture of 0.200 g of H<sub>2<\/sub>, 1.00 g of N<sub>2<\/sub>, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.<\/li>\r\n \t<li>Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than 3.0 % O<sub>2<\/sub> is not. If enough O<sub>2<\/sub> is added to a cylinder of H<sub>2<\/sub> at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive?<\/li>\r\n \t<li>A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 \u00d7 10<sup>-6<\/sup> mg\/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 \u00b0C?<\/li>\r\n \t<li>A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of 18 \u00b0C. What is the pressure of the carbon monoxide? (See Table 9.2\u00a0for the vapor pressure of water.)<\/li>\r\n \t<li>In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 \u00b0C. The mass of the gas was 0.472 g. What was the molar mass of the gas?<\/li>\r\n \t<li>Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO:[latex]2\\text{HgO}\\left(s\\right)\\rightarrow 2\\text{Hg}\\left(l\\right)+{\\text{O}}_{2}\\left(g\\right)[\/latex]\r\n<ol>\r\n \t<li>Outline the steps necessary to answer the following question: What volume of O<sub>2<\/sub> at 23 \u00b0C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?<\/li>\r\n \t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:[latex]4{\\text{H}}_{2}\\text{O}\\left(g\\right)+3\\text{Fe}\\left(s\\right)\\rightarrow{\\text{Fe}}_{3}{\\text{O}}_{4}\\left(s\\right)+4{\\text{H}}_{2}\\left(g\\right)[\/latex]\r\n<ol>\r\n \t<li>Outline the steps necessary to answer the following question: What volume of H<sub>2<\/sub> at a pressure of 745 torr and a temperature of 20 \u00b0C can be prepared from the reaction of 15.O g of H<sub>2<\/sub>O?<\/li>\r\n \t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The chlorofluorocarbon CCl<sub>2<\/sub>F<sub>2<\/sub> can be recycled into a different compound by reaction with hydrogen to produce CH<sub>2<\/sub>F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), a compound useful in chemical manufacturing: [latex]{\\text{CCl}}_{2}{\\text{F}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{CH}}_{2}{\\text{F}}_{2}\\left(g\\right)+2\\text{HCl}\\left(g\\right).[\/latex]\r\n<ol>\r\n \t<li>Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and 35.5 \u00b0C would be required to react with 1 ton (1.000 \u00d7 10<sup>3<\/sup> kg) of CCl<sub>2<\/sub>F<sub>2<\/sub>?<\/li>\r\n \t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN<sub>3<\/sub>). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 \u00b0C and 756 torr formed by the decomposition of 125 g of sodium azide.<\/li>\r\n \t<li>Lime, CaO, is produced by heating calcium carbonate, CaCO<sub>3<\/sub>; carbon dioxide is the other product.\r\n<ol>\r\n \t<li>Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875\u00b0 and 0.966 atm is produced by the decomposition of 1 ton (1.000 \u00d7 10<sup>3<\/sup> kg) of calcium carbonate?<\/li>\r\n \t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C<sub>2<\/sub>H<sub>2<\/sub>, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC<sub>2<\/sub>, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.\r\n<ol>\r\n \t<li>Outline the steps necessary to answer the following question: What volume of C<sub>2<\/sub>H<sub>2<\/sub> at 1.005 atm and 12.2 \u00b0C is formed by the reaction of 15.48 g of CaC<sub>2<\/sub> with water?<\/li>\r\n \t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C<sub>2<\/sub>H<sub>6<\/sub>, to produce carbon dioxide and water, if the volumes of C<sub>2<\/sub>H<sub>6<\/sub> and O<sub>2<\/sub> are measured under the same conditions of temperature and pressure.<\/li>\r\n \t<li>What volume of O<sub>2<\/sub> at STP is required to oxidize 8.0 L of NO at STP to NO<sub>2<\/sub>? What volume of NO<sub>2<\/sub> is produced at STP?<\/li>\r\n \t<li>Consider the following questions:\r\n<ol>\r\n \t<li>What is the total volume of the CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) and H<sub>2<\/sub>O(<em data-effect=\"italics\">g<\/em>) at 600 \u00b0C and 0.888 atm produced by the combustion of 1.00 L of C<sub>2<\/sub>H<sub>6<\/sub>(<em data-effect=\"italics\">g<\/em>) measured at STP?<\/li>\r\n \t<li>What is the partial pressure of H<sub>2<\/sub>O in the product gases?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Methanol, CH<sub>3<\/sub>OH, is produced industrially by the following reaction:[latex]\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)\\stackrel{\\text{copper catalyst 300 \\textdegree C, 300 atm}}{\\to }{\\text{CH}}_{3}\\text{OH}\\left(g\\right)[\/latex]\r\nAssuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.<\/li>\r\n \t<li>What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO<sub>2<\/sub> to BaO and O<sub>2<\/sub>?<\/li>\r\n \t<li>A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N<sub>2<\/sub> and 1.25 L of O<sub>2<\/sub> at STP. What is the colorless gas?<\/li>\r\n \t<li>Ethanol, C<sub>2<\/sub>H<sub>5<\/sub>OH, is produced industrially from ethylene, C<sub>2<\/sub>H<sub>4<\/sub>, by the following sequence of reactions: [latex]3{\\text{C}}_{2}{\\text{H}}_{4}+2{\\text{H}}_{2}{\\text{SO}}_{4}\\rightarrow{\\text{C}}_{2}{\\text{H}}_{5}{\\text{HSO}}_{4}+{\\left({\\text{C}}_{2}{\\text{H}}_{5}\\right)}_{2}{\\text{SO}}_{4}[\/latex] [latex]{\\text{C}}_{2}{\\text{H}}_{5}{\\text{HSO}}_{4}+{\\left({\\text{C}}_{2}{\\text{H}}_{5}\\right)}_{2}{\\text{SO}}_{4}+3{\\text{H}}_{2}\\text{O}\\rightarrow 3{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}+2{\\text{H}}_{2}{\\text{SO}}_{4}[\/latex] What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?<\/li>\r\n \t<li>One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature (37 \u00b0C) and a pressure of 743 torr, what is the molar mass of hemoglobin?<\/li>\r\n \t<li>A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)<\/li>\r\n \t<li>One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (-NH<sub>2<\/sub>) in protein material are allowed to react with nitrous acid, HNO<sub>2<\/sub>, to form N<sub>2<\/sub> gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH<sub>2<\/sub>(NH<sub>2<\/sub>)COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N<sub>2<\/sub> collected over water at a pressure of 735 torr and 29 \u00b0C. What was the percentage of glycine in the sample? [latex]{\\text{CH}}_{2}\\left({\\text{NH}}_{2}\\right){\\text{CO}}_{2}\\text{H}+{\\text{HNO}}_{2}\\rightarrow{\\text{CH}}_{2}\\left(\\text{OH}\\right){\\text{CO}}_{2}\\text{H}+{\\text{H}}_{2}\\text{O}+{\\text{N}}_{2}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\r\n<div class=\"entry-content\">\r\n<div class=\"im_section\">\r\n<div class=\"im_section\">\r\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h4>Selected Answers<\/h4>\r\n2.\u00a0[latex]\\rho =\\frac{P\\mathcal{M}}{RT}=\\frac{0.954\\cancel{\\text{atm}}\\left[12.011+2\\left(18.9954\\right)+2\\left(35.453\\right)\\right]\\text{g}\\cancel{{\\text{mol}}^{\\text{-1}}}}{\\text{0.08206 L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 303.15\\cancel{\\text{K}}}=\\text{4.64 g}{\\text{L}}^{\\text{-1}}[\/latex]\r\n\r\n4.\u00a0[latex]\\text{mass}{\\text{O}}_{2}=\\frac{\\left(\\text{31.9988 g}\\cancel{{\\text{mol}}^{\\text{-1}}}\\right)\\left(10.0\\cancel{\\text{atm}}\\right)\\left(3.00\\cancel{\\text{L}}\\right)}{\\left(0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(301.15\\cancel{\\text{K}}\\right)}=\\text{38.8 g}[\/latex]\r\n\r\n&nbsp;\r\n\r\n6. From the ideal gas law, <em data-effect=\"italics\">PV = nRT<\/em>, set [latex]n=\\frac{\\text{mass}}{\\text{molar mass}}[\/latex] and solve the molar mass.\r\n\r\n[latex]\\text{molar mass}=\\frac{\\left(\\text{0.281 g}\\right)\\left(0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(399.15\\cancel{\\text{K}}\\right)}{\\left(\\frac{777\\cancel{\\text{torr}}}{760\\cancel{\\text{torr}}\\cancel{{\\text{atm}}^{\\text{-1}}}}\\right)\\left(0.125\\cancel{\\text{L}}\\right)}=\\text{72.0 g}{\\text{mol}}^{\\text{-1}}[\/latex]\r\n\r\n8. [latex]\\mathcal{M}=\\frac{mRT}{PV}D=\\frac{m}{V}\\mathcal{M}=\\frac{DRT}{P}[\/latex]\r\n<p style=\"text-align: left;\">[latex]\\mathcal{M}=\\frac{\\text{3.93 g}{\\text{L}}^{\\text{-1}}\\left(\\text{0.08206 L}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(273.15\\cancel{\\text{K}}\\right)}{1.00\\cancel{\\text{atm}}}=\\text{88.1 g}{\\text{mol}}^{\\text{-1}}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">\u2133<sub>phosphorous<\/sub> = 30.97376 g\/mol<\/p>\r\n\r\n<div style=\"text-align: left;\" data-type=\"newline\">\u2133<sub>fluorine<\/sub> = 18.998403 g\/mol<\/div>\r\n<div style=\"text-align: left;\" data-type=\"newline\"><\/div>\r\n<div style=\"text-align: left;\" data-type=\"newline\">[latex]\\begin{array}{l}\\text{molecular formula:}\\text{phosphorous:}30.97376\\\\ \\\\ \\phantom{\\rule{7.9em}{0ex}}\\text{flourine:}\\phantom{\\rule{1.3em}{0ex}}\\underline{3\\left(18.998403\\right)}\\\\ \\phantom{\\rule{13.5em}{0ex}}87.968969\\end{array}[\/latex]<\/div>\r\n<div style=\"text-align: left;\" data-type=\"newline\"><\/div>\r\n<div style=\"text-align: left;\" data-type=\"newline\">The molecular formula is PF<sub>3<\/sub>.<\/div>\r\n<div style=\"text-align: left;\" data-type=\"newline\"><\/div>\r\n<div style=\"text-align: left;\" data-type=\"newline\">To find this answer you can either use trial and error, or you can realize that since phosphorus is in group 5, it can fill its valence shell by forming three bonds. Fluorine, being in group 7, needs to form only one bond to fill its shell. Thus it makes sense to start with PF<sub>3<\/sub> as a probable formula.<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n10. Calculate the moles of each gas present and from that, calculate the pressure from the ideal gas law. Assume 25 \u00b0C. The calibration gas contains:\r\n<div data-type=\"newline\">[latex]\\frac{350\\cancel{\\text{g}}{\\text{CO}}_{2}}{44.0098\\cancel{\\text{g}}{\\text{mol}}^{\\text{-1}}{\\text{CO}}_{2}}=\\text{7.953 mol}{\\text{CO}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\frac{805\\cancel{\\text{g}}{\\text{O}}_{2}}{31.9988\\cancel{\\text{g}}{\\text{mol}}^{\\text{-1}}{\\text{O}}_{2}}=\\text{25.157 mol}{\\text{O}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\frac{4880\\cancel{\\text{g}}{\\text{N}}_{2}}{28.01348\\cancel{\\text{g}}{\\text{mol}}^{\\text{-1}}{\\text{N}}_{2}}=\\text{174.202 mol}{\\text{N}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">Total moles = 7.953 + 25.157 + 174.202 = 207.312 mol<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]P=\\frac{nRT}{V}=\\frac{207.312\\cancel{\\text{mol}}\\times 0.08206\\cancel{\\text{L}}\\text{atm}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 298.15\\cancel{\\text{K}}}{36.0\\cancel{\\text{L}}}=\\text{141 atm}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n12. Since these are percentages of the total pressure, the partial pressure can be calculated as follows:\r\n<div id=\"fs-idp207682832\" data-type=\"exercise\">\r\n<div id=\"fs-idm99835968\" data-type=\"solution\">\r\n<div data-type=\"newline\">CH<sup>4<\/sup>: 90% of 307.2 kPa = 0.900 \u00d7 307.2 = 276 kPa<\/div>\r\n<div data-type=\"newline\">C<sup>2<\/sup> H<sup>6<\/sup>: 8.9% of 307.2 kPa = 0.089 \u00d7 307.2 = 27 kPa<\/div>\r\n<div data-type=\"newline\">C<sup>3<\/sup> H<sup>8<\/sup>: 1.1% of 307.2 kPa = 0.011 \u00d7 307.2 = 3.4 kPa<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">14.\u00a0The oxygen increases the pressure within the tank to (34.5 atm \u2013 33.2 atm =) 1.3 atm. The percentage O<sub>2<\/sub> on a mole basis is [latex]\\frac{1.3}{34.5}\\times 100\\%=3.77\\%.[\/latex] The mixture is explosive. However, the percentage is given as a weight percent. Converting to a mass basis increases the percentage of oxygen even more, so the mixture is still explosive.<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n16. The vapor pressure of water at 18 \u00b0C is 15.5 torr. Subtract the vapor pressure of water from the total pressure to find the pressure of the carbon monoxide:\r\n<div data-type=\"newline\"><em data-effect=\"italics\">P<\/em><sub>T<\/sub> = <em data-effect=\"italics\">P<\/em><sub>gas<\/sub> + <em data-effect=\"italics\">P<\/em><sub>water<\/sub><\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">Rearrangement gives:<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\"><em data-effect=\"italics\">P<\/em><sub>T<\/sub> \u2013<em data-effect=\"italics\"> P<\/em><sub>water =<\/sub><em data-effect=\"italics\">P<\/em><sub>gas<\/sub><\/div>\r\n<div data-type=\"newline\">756 torr \u2013 15.5 torr = 740 torr<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n18. (a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O<sub>2<\/sub> produced by decomposition of this amount of HgO; and determine the volume of O<sub>2<\/sub> from the moles of O<sub>2<\/sub>, temperature, and pressure.\r\n<div data-type=\"newline\">(b) [latex]\\begin{array}{l}\\\\ \\\\ 5.36\\cancel{\\text{g HgO}}\\times \\frac{\\text{1 mol HgO}}{\\left(200.59+15.9994\\right)\\cancel{\\text{g HgO}}}=\\text{0.0247 mol HgO}\\\\ 0.0247\\cancel{\\text{mol HgO}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{2\\cancel{\\text{mol HgO}}}=\\text{0.01235 mol}{\\text{O}}_{2}\\end{array}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\"><em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em><\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\"><em data-effect=\"italics\">P<\/em> = 0.975 atm<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\"><em data-effect=\"italics\">T<\/em> = (23.0 + 273.15) K<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]V=\\frac{nRT}{P}=\\frac{0.01235\\cancel{\\text{mol}}\\left(\\text{0.08206 L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(296.15\\cancel{\\text{K}}\\right)}{0.975\\cancel{\\text{atm}}}\\phantom{\\rule{0.4em}{0ex}}=\\phantom{\\rule{0.4em}{0ex}}\\text{0.308 L}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div id=\"fs-idp46179152\" data-type=\"exercise\">\r\n<div id=\"fs-idp138325296\" data-type=\"solution\">\r\n<div data-type=\"newline\">\r\n\r\n20. (a) Determine the molar mass of CCl<sub>2<\/sub>F<sub>2<\/sub>. From the balanced equation, calculate the moles of H<sub>2<\/sub> needed for the complete reaction. From the ideal gas law, convert moles of H<sub>2<\/sub> into volume.\r\n<div id=\"fs-idp39473616\" data-type=\"exercise\">\r\n<div id=\"fs-idp75359296\" data-type=\"solution\">\r\n<div data-type=\"newline\">(b) Molar mass of CCl<sub>2<\/sub> F<sub>2<\/sub> = 12.011 + 2 \u00d7 18.9984 + 2 \u00d7 35.4527 = 120.913 g\/mol<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\text{mol}{\\text{H}}_{2}=1.000\\times {10}^{6}\\text{g}\\times \\frac{\\text{1 mol}{\\text{CCL}}_{2}{\\text{F}}_{2}}{\\text{120.913 g}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{4 mol}{\\text{H}}_{2}}{\\text{1 mol}{\\text{CCl}}_{2}{\\text{F}}_{2}}=3.308\\times {10}^{4}\\text{mol}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]V=\\frac{nRT}{P}=\\frac{\\left(3.308\\times {10}^{4}\\cancel{\\text{mol}}\\right)\\left(\\text{0.08206 L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}{\\cancel{\\text{K}}}^{\\text{-1}}\\right)\\left(308.65\\cancel{\\text{K}}\\right)}{225\\cancel{\\text{atm}}}=3.72\\times {10}^{3}\\text{L}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n22. (a) Balance the equation. Determine the grams of CO<sub>2<\/sub> produced and the number of moles. From the ideal gas law, determine the volume of gas.\r\n<div data-type=\"newline\">(b) [latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightarrow\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\text{mass}{\\text{CO}}_{2}=1.00\\times {10}^{6}\\text{g}\\times \\frac{\\text{1 mol}{\\text{CaCO}}_{2}}{\\text{100.087 g}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{44.01 g}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{CO}}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{1 mol}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{CaCO}}_{2}}=4.397\\times {10}^{5}\\text{g}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\text{mol}{\\text{CO}}_{2}=\\frac{4.397\\times {10}^{5}\\text{g}}{\\text{44.01 g}{\\text{mol}}^{\\text{-1}}}=\\text{9991 mol}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]V=\\frac{nRT}{P}=\\frac{\\left(\\text{9991 mol}\\right)\\left(\\text{0.08206 L atm}{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}\\right)\\left(\\text{875 K}\\right)}{\\text{0.966 atm}}=7.43\\times {10}^{5}\\text{L}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n24. [latex]2{\\text{C}}_{2}{\\text{H}}_{6}\\left(\\text{g}\\right)+7{\\text{O}}_{2}\\left(\\text{g}\\right)\\rightarrow 4{\\text{CO}}_{2}\\left(\\text{g}\\right)+6{\\text{H}}_{2}\\text{O}\\left(\\text{g}\\right)[\/latex]\r\n<div data-type=\"newline\">From the balanced equation, we see that 2 mol of C<sub>2<\/sub>H<sub>6<\/sub> requires 7 mol of O<sub>2<\/sub> to burn completely. Gay-Lussac\u2019s law states that gases react in simple proportions by volume. As the number of liters is proportional to the number of moles,<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\frac{\\text{12.00 L}}{\\text{2 mol}{\\text{C}}_{2}{\\text{H}}_{6}}=\\frac{V\\left({\\text{O}}_{2}\\right)}{\\text{7 mol}{\\text{O}}_{2}}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]V\\left({\\text{O}}_{2}\\right)=\\frac{\\text{12.00 L}\\times 7}{2}=\\text{42.00 L}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n26. (a) The scheme to solve this problem is:\r\n<div data-type=\"newline\">[latex]{\\text{volume C}}_{2}{\\text{H}}_{6}\\left(g\\right)\\stackrel{\\begin{array}{l}\\text{ideal gas}\\\\ \\text{equation}\\end{array}}{\\to }{\\text{mol C}}_{2}{\\text{H}}_{6}\\left(g\\right)\\stackrel{\\begin{array}{l}\\text{reaction}\\\\ \\text{stoichiometry}\\end{array}}{\\to }{\\text{mol CO}}_{2}+{\\text{H}}_{2}\\text{O}\\stackrel{\\begin{array}{l}\\text{ideal gas}\\\\ \\text{equation}\\end{array}}{\\to }{\\text{volume CO}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\begin{array}{l}{\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)+3\\frac{1}{2}{\\text{O}}_{2}\\left(g\\right)\\rightarrow 2{\\text{CO}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\text{O}\\left(g\\right)\\\\ \\text{1}.n\\left({\\text{C}}_{2}{\\text{H}}_{6}\\right)=\\frac{PV}{RT}=\\frac{1.00\\cancel{\\text{atm}}\\times 1.00\\cancel{\\text{L}}}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(273.15\\cancel{\\text{K}}\\right)}=\\text{0.0446 mol}\\\\ \\text{2}.0.0446\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}}\\times \\frac{\\text{5 mol products}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}}}=\\text{0.223 mol products}\\\\ \\text{3}.V=nRT=\\frac{\\left(0.223\\cancel{\\text{mol}}\\right)\\left(\\text{0.08206 L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(873.15\\cancel{\\text{K}}\\right)}{0.888\\cancel{\\text{atm}}}=\\text{18.0 L}\\end{array}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(b) First, calculate the mol H<sub>2<\/sub>O produced:<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\text{0.0446 mol}{\\text{C}}_{2}{\\text{H}}_{6}\\times \\frac{\\text{3 mol products}}{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{6}}=\\text{0.1338 mol}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">Second, calculate the pressure of H<sub>2<\/sub>O:<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]P=\\frac{nRT}{V}=\\frac{\\left(0.1338\\cancel{\\text{mol}}\\right)\\left(0.8206\\cancel{\\text{L}}\\text{atm}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(873.15\\cancel{\\text{K}}\\right)}{18.0\\cancel{\\text{L}}}=\\text{0.533 atm}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n28. First, we must write a balanced equation to establish the stoichiometry of the reaction:\r\n<div data-type=\"newline\">\r\n<div data-type=\"newline\">[latex]2{\\text{BaO}}_{2}\\rightarrow 2\\text{BaO}+{\\text{O}}_{2}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">We are given the mass of BaO<sub>2<\/sub> that decomposes, so the scheme for solving this problem will be:<\/div>\r\n<span id=\"fs-idp92165488\" data-type=\"media\" data-alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled \u201cMass of B a O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cMolar mass B a O subscript 2\u201d pointing right to a second rectangle which is shaded pink and is labeled \u201cMoles of B a O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cReaction stoichiometry\u201d pointing right to a third rectangle which is shaded pink and is labeled \u201cMoles of O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cIdeal gas equation\u201d pointing right to a fourth rectangle which is shaded lavender and is labeled \u201cVolume of O subscript 2.\u201d\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212055\/CNX_Chem_09_03_Exercise29_img1.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled \u201cMass of B a O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cMolar mass B a O subscript 2\u201d pointing right to a second rectangle which is shaded pink and is labeled \u201cMoles of B a O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cReaction stoichiometry\u201d pointing right to a third rectangle which is shaded pink and is labeled \u201cMoles of O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cIdeal gas equation\u201d pointing right to a fourth rectangle which is shaded lavender and is labeled \u201cVolume of O subscript 2.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<div data-type=\"newline\">Mass (BaO<sub>2<\/sub>) = 137.33 + 2(15.9994) = 169.33 g\/mol<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]n\\left({\\text{O}}_{2}\\right)=\\text{129.7 g}{\\text{BaO}}_{2}\\times \\frac{\\text{1 mol}{\\text{BaO}}_{2}}{\\text{169.33 g}{\\text{BaO}}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{1 mol}{\\text{O}}_{2}}{\\text{2 mol}{\\text{BaO}}_{2}}=\\text{0.3830 mol}{\\text{O}}_{2}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]V\\left({\\text{O}}_{2}\\right)=\\frac{nRT}{P}=\\frac{\\text{0.3830 mol}\\left(\\text{8.314 L kPa}{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}\\right)\\left(\\text{423.0 K}\\right)}{\\text{127.4 kPa}}=\\text{10.57 L}{\\text{O}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n30. At 90.1% conversion, a 1.000 \u00d7 10<sup>6<\/sup> g final yield would require a [latex]\\left(\\frac{1.000\\times {10}^{6}}{0.901}\\right)=1.1099\\times {10}^{6}\\text{g}[\/latex] theoretical yield.\r\n<div data-type=\"newline\">3C<sub>2<\/sub>H<sub>4<\/sub> produces 3C<sub>2<\/sub>H<sub>5<\/sub>OH, giving a 1:1 ratio:<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\text{mol}\\left({\\text{C}}_{2}{\\text{H}}_{4}\\right)=1.1099\\times {10}^{6}\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}\\text{OH}}}{46.069\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}\\text{OH}}}=2.409\\times {10}^{4}\\text{mol}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\"><em data-effect=\"italics\">V<\/em> (C<sub>2<\/sub> H<sub>4<\/sub>) = 22.4 L\/mol \u00d7 2.409 \u00d7 10<sup>4<\/sup> mol = 5.40 \u00d7 10<sup>5<\/sup> L<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n32. The reaction is:\r\n<div data-type=\"newline\">[latex]{\\text{XeF}}_{x}+\\frac{x}{2}{\\text{H}}_{2}\\rightarrow\\text{Xe}+x\\text{HF}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">The pressure of H<sub>2<\/sub> that reacts is:<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">48 torr \u2013 24 torr = 24 torr<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">The number of moles of gas is proportional to the partial pressures. The reaction used 24 torr of XeF<sub>x<\/sub> and 24 torr of H<sub>2<\/sub> so:<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\frac{x}{2}=1[\/latex] and <em data-effect=\"italics\">x<\/em> = 2<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">The formula for the xenon compound is XeF<sub>2<\/sub>.<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">Immediately after the H<sub>2<\/sub> is added (before the reaction):<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\begin{array}{lll}\\\\ \\\\ \\hfill {P}_{\\text{Total}}&amp; =&amp; {P}_{{\\text{XeF}}_{2}}+{P}_{{\\text{H}}_{2}}\\hfill \\\\ \\hfill {P}_{{\\text{H}}_{2}}&amp; =&amp; {P}_{\\text{Total}}-{P}_{{\\text{XeF}}_{2}}\\hfill \\\\ &amp; =&amp; \\text{72 torr}-\\text{24 torr}\\hfill \\\\ &amp; =&amp; 48\\text{torr}\\hfill \\end{array}[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">After the reaction:<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]{P}_{\\text{Xe}}=\\text{24 torr}\\text{ }\\left(\\text{1 mol}{\\text{XeF}}_{\\text{x}}\\rightarrow\\text{1 mol Xe}\\right)[\/latex]<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">And the partial pressure of unreacted H<sub>2<\/sub> is:<\/div>\r\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]\\begin{array}{ll}\\\\ \\hfill {P}_{{\\text{H}}_{2}}&amp; ={P}_{\\text{Total}}-{P}_{\\text{Xe}}\\hfill \\\\ &amp; =\\text{48 torr}-\\text{24 torr}\\hfill \\\\ &amp; =\\text{24 torr}\\hfill \\end{array}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"bcc-box bcc-success\"><section id=\"glossary\">\r\n<h3>Glossary<\/h3>\r\n<div data-type=\"definition\">\r\n<div id=\"fs-idm8143856\" data-type=\"definition\">\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">Dalton\u2019s law of partial pressures\r\n<\/span><\/strong>total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">mole fraction\r\n<\/span><\/strong>concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">partial pressure\r\n<\/span><\/strong>pressure exerted by an individual gas in a mixture<\/p>\r\n\r\n<div id=\"fs-idp57901520\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">vapor pressure of water\r\n<\/span><\/strong>pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Use the ideal gas law to compute gas densities and molar masses<\/li>\n<li>Perform stoichiometric calculations involving gaseous substances<\/li>\n<li>State Dalton\u2019s law of partial pressures and use it in calculations involving gaseous mixtures<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp201619456\">The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine <span class=\"no-emphasis\" data-type=\"term\">Lavoisier<\/span>, widely regarded as the \u201cfather of modern chemistry,\u201d changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, \u201cIt took the mob only a moment to remove his head; a century will not suffice to reproduce it.\u201d<\/p>\n<p id=\"fs-idm31863392\">As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask \u201cHow much?\u201d We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.<\/p>\n<section id=\"fs-idp78729184\" data-depth=\"1\">\n<h2 data-type=\"title\">Density of a Gas<\/h2>\n<p id=\"fs-idm26134320\">Recall that the density of a gas is its mass to volume ratio, [latex]\\rho =\\frac{m}{V}.[\/latex] Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can used to determine its molar mass and thereby assist in its identification. The ideal gas law, <em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em>, provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in Example 1.<\/p>\n<div id=\"fs-idp128586304\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 1<\/h3>\n<h4 id=\"fs-idm63453648\"><strong><span data-type=\"title\">Derivation of a Density Formula from the Ideal Gas Law<\/span><\/strong><\/h4>\n<p>Use <em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em> to derive a formula for the density of gas in g\/L<\/p>\n<h4 id=\"fs-idp146685296\"><span data-type=\"title\">Solution<\/span><\/h4>\n<ol id=\"fs-idp85675504\" class=\"stepwise\" data-number-style=\"arabic\">\n<li><em data-effect=\"italics\">PV = nRT<\/em><\/li>\n<li><em data-effect=\"italics\">Rearrange to get (mol\/L)<\/em>: [latex]\\frac{n}{v}=\\frac{P}{RT}[\/latex]<\/li>\n<li><em data-effect=\"italics\">Multiply each side of the equation by the molar mass, \u2133.<\/em> When moles are multiplied by \u2133 in g\/mol, g are obtained:\n<div data-type=\"newline\"><\/div>\n<p>[latex]\\left(\\mathcal{M}\\right)\\left(\\frac{n}{V}\\right)=\\left(\\frac{P}{RT}\\right)\\left(\\mathcal{M}\\right)[\/latex]<\/li>\n<li>[latex]g\\text{\/L}=\\rho =\\frac{P\\mathcal{M}}{RT}[\/latex]<\/li>\n<\/ol>\n<h4 id=\"fs-idm8187184\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A gas was found to have a density of 0.0847 g\/L at 17.0 \u00b0C and a pressure of 760 torr. What is its molar mass? What is the gas?<\/p>\n<div id=\"fs-idp1319536\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0[latex]\\rho =\\frac{P\\mathcal{M}}{RT}[\/latex]<\/div>\n<div style=\"text-align: right;\" data-type=\"newline\">[latex]0.0847\\text{g\/L}=760\\cancel{\\text{torr}}\\times \\frac{1\\cancel{\\text{atm}}}{760\\cancel{\\text{torr}}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\mathcal{M}}{\\text{0.0821 L}\\cancel{\\text{atm}}\\text{\/mol K}}\\times \\text{290 K}[\/latex]<\/div>\n<div style=\"text-align: right;\" data-type=\"newline\">\u2133 = 2.02 g\/mol; therefore, the gas must be hydrogen (H<sub>2<\/sub>, 2.02 g\/mol)<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idp140587168\">We must specify both the temperature and the pressure of a gas when calculating its density, because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or pressure. Gas densities are often reported at STP.<\/p>\n<div id=\"fs-idp74274064\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 2<\/h3>\n<h4 id=\"fs-idm20853312\"><strong><span data-type=\"title\">Empirical\/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas<\/span><\/strong><\/h4>\n<p>Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 \u00b0C, what is the molecular formula for cyclopropane?<\/p>\n<h4 id=\"fs-idp234218112\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>Strategy: First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:<\/p>\n<div id=\"fs-idm18588864\" data-type=\"equation\">[latex]\\text{85.7 g C}\\times \\frac{\\text{1 mol C}}{\\text{12.01 g C}}=\\text{7.136 mol C}\\frac{7.136}{7.136}=\\text{1.00 mol C}[\/latex]<\/div>\n<div id=\"fs-idm150258896\" data-type=\"equation\">[latex]\\text{14.3 g H}\\times \\frac{\\text{1 mol H}}{\\text{1.01 g H}}=\\text{14.158 mol H}\\frac{14.158}{7.136}=\\text{1.98 mol H}[\/latex]<\/div>\n<p id=\"fs-idp144237600\">Empirical formula is CH<sub>2<\/sub> [empirical mass (EM) of 14.03 g\/empirical unit].<\/p>\n<p id=\"fs-idp40237504\">Next, use the density equation related to the ideal gas law to determine the molar mass:<\/p>\n<div id=\"fs-idp110330000\" data-type=\"equation\">[latex]\\text{d}=\\frac{\\text{P}\\mathcal{M}}{\\text{RT}}\\frac{\\text{1.56 g}}{\\text{1.00 L}}=\\text{0.984 atm}\\times \\frac{\\mathcal{M}}{\\text{0.0821 L atm\/mol K}}\\times \\text{323 K}[\/latex]<\/div>\n<p id=\"fs-idp221616976\">\u2133 = 42.0 g\/mol, [latex]\\frac{\\mathcal{M}}{\\text{E}\\mathcal{M}}=\\frac{42.0}{14.03}=2.99,[\/latex] so (3)(CH<sub>2<\/sub>) = C<sub>3<\/sub>H<sub>6<\/sub> (molecular formula)<\/p>\n<h4 id=\"fs-idp259910672\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>Acetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 \u00b0C, what is the molecular formula for acetylene?<\/p>\n<div id=\"fs-idp130463408\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0Empirical formula, CH; Molecular formula, C<sub>2<\/sub>H<sub>2<\/sub><\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-idp6528928\" data-depth=\"2\">\n<h2 data-type=\"title\">Molar Mass of a Gas<\/h2>\n<p id=\"fs-idp56968688\">Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, <em data-effect=\"italics\">m<\/em>, to its amount in moles, <em data-effect=\"italics\">n<\/em>:<\/p>\n<div id=\"fs-idp45609344\" data-type=\"equation\">[latex]\\mathcal{M}=\\frac{\\text{grams of substance}}{\\text{moles of substance}}=\\frac{m}{n}[\/latex]<\/div>\n<p id=\"fs-idp44359360\">The ideal gas equation can be rearranged to isolate <em data-effect=\"italics\">n<\/em>:<\/p>\n<div id=\"fs-idp78574320\" data-type=\"equation\">[latex]n=\\frac{PV}{RT}[\/latex]<\/div>\n<p id=\"fs-idp18382336\">and then combined with the molar mass equation to yield:<\/p>\n<div id=\"fs-idm6613536\" data-type=\"equation\">[latex]\\mathcal{M}=\\frac{mRT}{PV}[\/latex]<\/div>\n<p id=\"fs-idm17646432\">This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.<\/p>\n<div id=\"fs-idp207136496\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 3<\/h3>\n<h4 id=\"fs-idp30589168\"><strong><span data-type=\"title\">Determining the Molar Mass of a Volatile Liquid<\/span><\/strong><\/h4>\n<p>The approximate molar mass of a volatile liquid can be determined by:<\/p>\n<ol id=\"fs-idp104043696\" data-number-style=\"arabic\">\n<li>Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole<\/li>\n<li>Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure<\/li>\n<li>Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample\u2019s mass (see\u00a0Figure 1)<\/li>\n<\/ol>\n<figure id=\"CNX_Chem_09_03_liquidgas\">\n<div style=\"width: 849px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212041\/CNX_Chem_09_03_liquidgas1.jpg\" alt=\"This figure shows four photos each connected by a right-facing arrow. The first photo shows a glass flask with aluminum foil covering the top sitting on a scale. The scale reads 89.516. The second photo shows a syringe being inserted into the flask through the aluminum foil covering. The third photo shows the glass flask being inserted into a beaker of water. The water appears to be heated at 100. The fourth photo shows the glass flask being weighed again. This time the scale reads 89.512.\" width=\"839\" height=\"203\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At [latex]{t}_{l\\rightarrow g},[\/latex] the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-idm10831984\">Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm<sup>3<\/sup> at 99.6 \u00b0C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?<\/p>\n<h4 id=\"fs-idp60723232\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>Since [latex]\\mathcal{M}=\\frac{m}{n}[\/latex] and [latex]n=\\frac{PV}{RT},[\/latex] substituting and rearranging gives [latex]\\mathcal{M}=\\frac{mRT}{PV},[\/latex]<\/p>\n<p id=\"fs-idp43964944\">then<\/p>\n<div id=\"fs-idm11022240\" data-type=\"equation\">[latex]\\mathcal{M}=\\frac{mRT}{PV}=\\frac{\\left(\\text{0.494 g}\\right)\\times \\text{0.08206 L\\cdot atm\/mol K}\\times \\text{372.8 K}}{\\text{0.976 atm}\\times \\text{0.129 L}}=120\\text{g\/mol}.[\/latex]<\/div>\n<h4 id=\"fs-idp146723968\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A sample of phosphorus that weighs 3.243 \u00d7 10<sup>-2<\/sup> g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 \u00b0C. What are the molar mass and molecular formula of phosphorus vapor?<\/p>\n<div id=\"fs-idp108389616\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0124 g\/mol P<sub>4<\/sub><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idp148167024\" data-depth=\"2\">\n<h2 data-type=\"title\">The Pressure of a Mixture of Gases: Dalton\u2019s Law<\/h2>\n<p id=\"fs-idp70155072\">Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other\u2019s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it present alone in the container (Figure 2). The pressure exerted by each individual gas in a mixture is called its <strong><span data-type=\"term\">partial pressure<\/span><\/strong>. This observation is summarized by <strong><span data-type=\"term\">Dalton\u2019s law of partial pressures<\/span><\/strong>: <em data-effect=\"italics\">The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases<\/em>:<\/p>\n<div id=\"fs-idp65742144\" data-type=\"equation\">[latex]{P}_{Total}={P}_{A}+{P}_{B}+{P}_{C}+\\ldots ={\\text{\\Sigma}}_{\\text{i}}{P}_{\\text{i}}[\/latex]<\/div>\n<p id=\"fs-idp97232880\">In the equation <em data-effect=\"italics\">P<sub>Total<\/sub><\/em> is the total pressure of a mixture of gases, <em data-effect=\"italics\">P<sub>A<\/sub><\/em> is the partial pressure of gas A; <em data-effect=\"italics\">P<sub>B<\/sub><\/em> is the partial pressure of gas B; <em data-effect=\"italics\">P<sub>C<\/sub><\/em> is the partial pressure of gas C; and so on.<\/p>\n<figure id=\"CNX_Chem_09_03_DaltonLaw1\">\n<div style=\"width: 891px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212042\/CNX_Chem_09_03_DaltonLaw11.jpg\" alt=\"This figure includes images of four gas-filled cylinders or tanks. Each has a valve at the top. The interior of the first cylinder is shaded blue. This region contains 5 small blue circles that are evenly distributed. The label \u201c300 k P a\u201d is on the cylinder. The second cylinder is shaded lavender. This region contains 8 small purple circles that are evenly distributed. The label \u201c600 k P a\u201d is on the cylinder. To the right of these cylinders is a third cylinder. Its interior is shaded pale yellow. This region contains 12 small yellow circles that are evenly distributed. The label \u201c450 k P a\u201d is on this region of the cylinder. An arrow labeled \u201cTotal pressure combined\u201d appears to the right of these three cylinders. This arrow points to a fourth cylinder. The interior of this cylinder is shaded a pale green. It contains evenly distributed small circles in the following quantities and colors; 5 blue, 8 purple, and 12 yellow. This cylinder is labeled \u201c1350 k P a.\u201d\" width=\"881\" height=\"395\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. If equal-volume cylinders containing gas A at a pressure of 300 kPa, gas B at a pressure of 600 kPa, and gas C at a pressure of 450 kPa are all combined in the same-size cylinder, the total pressure of the mixture is 1350 kPa.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-idp110759216\">The partial pressure of gas A is related to the total pressure of the gas mixture via its <strong><span data-type=\"term\">mole fraction<\/span><\/strong>, a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components):<\/p>\n<div id=\"fs-idp18188304\" data-type=\"equation\">[latex]{P}_{A}={X}_{A}\\times {P}_{Total}\\text{where}{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex]<\/div>\n<p id=\"fs-idp75739968\">where <em data-effect=\"italics\">P<sub>A<\/sub><\/em>, <em data-effect=\"italics\">X<sub>A<\/sub><\/em>, and <em data-effect=\"italics\">n<sub>A<\/sub><\/em> are the partial pressure, mole fraction, and number of moles of gas A, respectively, and <em data-effect=\"italics\">n<sub>Total<\/sub><\/em> is the number of moles of all components in the mixture.<\/p>\n<div id=\"fs-idp143452448\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 4<\/h3>\n<h4 id=\"fs-idp147731856\"><strong><span data-type=\"title\">The Pressure of a Mixture of Gases<\/span><\/strong><\/h4>\n<p>A 10.0-L vessel contains 2.50 \u00d7 10<sup>-3<\/sup> mol of H<sub>2<\/sub>, 1.00 \u00d7 10<sup>-3<\/sup> mol of He, and 3.00 \u00d7 10<sup>-4<\/sup> mol of Ne at 35 \u00b0C.<\/p>\n<p id=\"fs-idp100506960\">(a) What are the partial pressures of each of the gases?<\/p>\n<p id=\"fs-idp147676624\">(b) What is the total pressure in atmospheres?<\/p>\n<h4 id=\"fs-idp17315648\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using [latex]P=\\frac{nRT}{V}:[\/latex]<\/p>\n<div id=\"fs-idp201684064\" data-type=\"equation\">[latex]{P}_{{\\text{H}}_{2}}=\\frac{\\left(2.50\\times {10}^{\\text{-3}}\\cancel{\\text{mol}}\\right)\\left(0.08206\\cancel{\\text{L}}\\text{atm}\\cancel{{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}}\\right)\\left(\\text{308 K}\\right)}{10.0\\cancel{\\text{L}}}=6.32\\times {10}^{\\text{-3}}\\text{atm}[\/latex]<\/div>\n<div id=\"fs-idp100280736\" data-type=\"equation\">[latex]{P}_{\\text{He}}=\\frac{\\left(1.00\\times {10}^{\\text{-3}}\\cancel{\\text{mol}}\\right)\\left(\\text{0.08206 L atm}\\cancel{{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}}\\right)\\left(\\text{308 K}\\right)}{10.0\\cancel{\\text{L}}}=2.53\\times {10}^{\\text{-3}}\\text{atm}[\/latex]<\/div>\n<div id=\"fs-idp36264816\" data-type=\"equation\">[latex]{P}_{\\text{Ne}}=\\frac{\\left(3.00\\times {10}^{\\text{-4}}\\cancel{\\text{mol}}\\right)\\left(0.08206\\cancel{\\text{L}}\\text{atm}\\cancel{{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}}\\right)\\left(\\text{308 K}\\right)}{10.0\\cancel{\\text{L}}}=7.58\\times {10}^{\\text{-4}}\\text{atm}[\/latex]<\/div>\n<p id=\"fs-idp49915536\">The total pressure is given by the sum of the partial pressures:<\/p>\n<div id=\"fs-idp207663808\" data-type=\"equation\">[latex]{P}_{\\text{T}}={P}_{{\\text{H}}_{2}}+{P}_{\\text{He}}+{P}_{\\text{Ne}}=\\left(0.00632+0.00253+0.00076\\right)\\text{atm}=9.61\\times {10}^{\\text{-3}}\\text{atm}[\/latex]<\/div>\n<h4 id=\"fs-idp1986144\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A 5.73-L flask at 25 \u00b0C contains 0.0388 mol of N<sub>2<\/sub>, 0.147 mol of CO, and 0.0803 mol of H<sub>2<\/sub>. What is the total pressure in the flask in atmospheres?<\/p>\n<div id=\"fs-idm10367824\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a01.137 atm<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idp107652512\">Here is another example of this concept, but dealing with mole fraction calculations.<\/p>\n<div id=\"fs-idp107854880\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 5<\/h3>\n<h4 id=\"fs-idp279668416\"><strong><span data-type=\"title\">The Pressure of a Mixture of Gases<\/span><\/strong><\/h4>\n<p>A gas mixture used for anesthesia contains 2.83 mol oxygen, O<sub>2<\/sub>, and 8.41 mol nitrous oxide, N<sub>2<\/sub>O. The total pressure of the mixture is 192 kPa.<\/p>\n<p id=\"fs-idp41662496\">(a) What are the mole fractions of O<sub>2<\/sub> and N<sub>2<\/sub>O?<\/p>\n<p id=\"fs-idp132439904\">(b) What are the partial pressures of O<sub>2<\/sub> and N<sub>2<\/sub>O?<\/p>\n<h4 id=\"fs-idp86077792\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>The mole fraction is given by [latex]{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex] and the partial pressure is <em data-effect=\"italics\">P<sub>A<\/sub><\/em> = <em data-effect=\"italics\">X<sub>A<\/sub><\/em> \u00d7 <em data-effect=\"italics\">P<sub>Total<\/sub><\/em>.<\/p>\n<p id=\"fs-idm32258688\">For O<sub>2<\/sub>,<\/p>\n<div id=\"fs-idm52855552\" data-type=\"equation\">[latex]{X}_{{O}_{2}}=\\frac{{n}_{{O}_{2}}}{{n}_{Total}}=\\frac{\\text{2.83 mol}}{\\left(2.83+8.41\\right)\\text{mol}}=0.252[\/latex]<\/div>\n<p id=\"fs-idp192169392\">and [latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=0.252\\times \\text{192 kPa}=\\text{48.4 kPa}[\/latex]<\/p>\n<p id=\"fs-idp263870464\">For N<sub>2<\/sub>O,<\/p>\n<div id=\"fs-idp202334992\" data-type=\"equation\">[latex]{X}_{{O}_{2}}=\\frac{{n}_{{O}_{2}}}{{n}_{Total}}=\\frac{\\text{2.83 mol}}{\\left(2.83+8.41\\right)\\text{mol}}=0.252[\/latex]<\/div>\n<p id=\"fs-idp259944368\">and<\/p>\n<p>[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=\\left(0.252\\right)\\times \\text{192 kPa}[\/latex]<\/p>\n<p id=\"fs-idp66867744\">[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=0.252\\times \\text{192 kPa}=\\text{48.4 kPa}[\/latex]<\/p>\n<h4 id=\"fs-idm31767024\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What is the pressure of a mixture of 0.200 g of H<sub>2<\/sub>, 1.00 g of N<sub>2<\/sub>, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 \u00b0C?<\/p>\n<div id=\"fs-idp69846704\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a01.87 atm<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idp221977104\" data-depth=\"2\">\n<h2 data-type=\"title\">Collection of Gases over Water<\/h2>\n<figure id=\"CNX_Chem_09_03_WaterVapor\">\n<div style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212044\/CNX_Chem_09_03_WaterVapor1.jpg\" alt=\"This figure shows a diagram of equipment used for collecting a gas over water. To the left is an Erlenmeyer flask. It is approximately two thirds full of a lavender colored liquid. Bubbles are evident in the liquid. The label \u201cReaction Producing Gas\u201d appears below the flask. A line segment connects this label to the liquid in the flask. The flask has a stopper in it through which a single glass tube extends from the open region above the liquid in the flask up, through the stopper, to the right, then angles down into a pan that is nearly full of light blue water. This tube again extends right once it is well beneath the water\u2019s surface. It then bends up into an inverted flask which is labeled \u201cCollection Flask.\u201d This collection flask is positioned with its mouth beneath the surface of the light blue water and appears approximately half full. Bubbles are evident in the water in the inverted flask. The open space above the water in the inverted flask is labeled \u201ccollected gas.\u201d\" width=\"300\" height=\"257\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).<\/p>\n<\/div>\n<\/figure>\n<p>A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 3), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.<\/p>\n<p>However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor\u2014this is referred to as the \u201cdry\u201d gas pressure, that is, the pressure of the gas only, without water vapor. The <strong><span data-type=\"term\">vapor pressure of water<\/span><\/strong>, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 4); more detailed information on the temperature dependence of water vapor can be found in Table 1, and vapor pressure will be discussed in more detail in the next chapter on liquids.<span data-type=\"media\" data-alt=\"A graph is shown. The horizontal axis is labeled \u201cTemperature ( degrees C )\u201d with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled \u201cVapor pressure ( torr )\u201d with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, \u201cVapor pressure at ( 100 degrees C ).\u201d\"><span data-type=\"media\" data-alt=\"A graph is shown. The horizontal axis is labeled \u201cTemperature ( degrees C )\u201d with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled \u201cVapor pressure ( torr )\u201d with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, \u201cVapor pressure at ( 100 degrees C ).\u201d\"><br \/>\n<\/span><\/span><\/p>\n<div style=\"width: 660px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212045\/CNX_Chem_09_03_WaterVapor21.jpg\" alt=\"A graph is shown. The horizontal axis is labeled \u201cTemperature ( degrees C )\u201d with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled \u201cVapor pressure ( torr )\u201d with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, \u201cVapor pressure at ( 100 degrees C ).\u201d\" width=\"650\" height=\"506\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. This graph shows the vapor pressure of water at sea level as a function of temperature.<\/p>\n<\/div>\n<table id=\"fs-idm68841392\" summary=\"This table has six columns and 13 rows. The first row is a header and it labels each column, \u201cTemperature (degree sign C),\u201d \u201cPressure (torr),\u201d \u201cTemperature (degree sign C),\u201d \u201cPressure (torr),\u201d \u201cTemperature (degree sign C),\u201d and \u201cPressure (torr).\u201d Under the first column are the following: negative 10, negative 5, negative 2, 0, 2, 4, 6, 8, 10, 12, 14, and 16. Under the second column are the following: 1.95, 3.0, 3.9, 4.6, 5.3, 6.1, 7.0, 8.0, 9.2, 10.5, 12.0, and 13.6. Under the third column are the following: 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, and 29. Under the fourth column are the following: 15.5, 16.5, 17.5, 18.7, 19.8, 21.1, 22.4, 23.8, 25.2, 26.7, 28.3, and 30.0. Under the fifth column are the following: 30, 35, 40, 50, 60, 70, 80, 90, 95, 99, 100.0, and 101.0. Under the sixth column are the following: 31.8, 42.2, 55.3, 92.5, 149.4, 233.7, 355.1, 525.8, 633.9, 733.2, 760.0, and 787.6.\">\n<thead>\n<tr>\n<th colspan=\"6\">Table 1. Vapor Pressure of Ice and Water in Various Temperatures at Sea Level<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<th>Temperature (\u00b0C)<\/th>\n<th>Pressure (torr)<\/th>\n<th>Temperature (\u00b0C)<\/th>\n<th>Pressure (torr)<\/th>\n<th>Temperature (\u00b0C)<\/th>\n<th>Pressure (torr)<\/th>\n<\/tr>\n<tr valign=\"top\">\n<td>\u201310<\/td>\n<td>1.95<\/td>\n<td>18<\/td>\n<td>15.5<\/td>\n<td>30<\/td>\n<td>31.8<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>\u20135<\/td>\n<td>3.0<\/td>\n<td>19<\/td>\n<td>16.5<\/td>\n<td>35<\/td>\n<td>42.2<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>\u20132<\/td>\n<td>3.9<\/td>\n<td>20<\/td>\n<td>17.5<\/td>\n<td>40<\/td>\n<td>55.3<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>0<\/td>\n<td>4.6<\/td>\n<td>21<\/td>\n<td>18.7<\/td>\n<td>50<\/td>\n<td>92.5<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>5.3<\/td>\n<td>22<\/td>\n<td>19.8<\/td>\n<td>60<\/td>\n<td>149.4<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4<\/td>\n<td>6.1<\/td>\n<td>23<\/td>\n<td>21.1<\/td>\n<td>70<\/td>\n<td>233.7<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>6<\/td>\n<td>7.0<\/td>\n<td>24<\/td>\n<td>22.4<\/td>\n<td>80<\/td>\n<td>355.1<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>8<\/td>\n<td>8.0<\/td>\n<td>25<\/td>\n<td>23.8<\/td>\n<td>90<\/td>\n<td>525.8<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>10<\/td>\n<td>9.2<\/td>\n<td>26<\/td>\n<td>25.2<\/td>\n<td>95<\/td>\n<td>633.9<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>12<\/td>\n<td>10.5<\/td>\n<td>27<\/td>\n<td>26.7<\/td>\n<td>99<\/td>\n<td>733.2<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>14<\/td>\n<td>12.0<\/td>\n<td>28<\/td>\n<td>28.3<\/td>\n<td>100.0<\/td>\n<td>760.0<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>16<\/td>\n<td>13.6<\/td>\n<td>29<\/td>\n<td>30.0<\/td>\n<td>101.0<\/td>\n<td>787.6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-idp46681200\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example\u00a06<\/h3>\n<h4 id=\"fs-idp35760544\"><strong><span data-type=\"title\">Pressure of a Gas Collected Over Water<\/span><\/strong><\/h4>\n<p>If 0.200 L of argon is collected over water at a temperature of 26 \u00b0C and a pressure of 750 torr in a system like that shown in Figure 3, what is the partial pressure of argon?<\/p>\n<h4 id=\"fs-idp18515888\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>According to Dalton\u2019s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:<\/p>\n<div id=\"fs-idp297283264\" data-type=\"equation\">[latex]{P}_{\\text{T}}={P}_{\\text{Ar}}+{P}_{{\\text{H}}_{2}\\text{O}}[\/latex]<\/div>\n<p id=\"fs-idp135677552\">Rearranging this equation to solve for the pressure of argon gives:<\/p>\n<div id=\"fs-idp33156768\" data-type=\"equation\">[latex]{P}_{\\text{Ar}}={P}_{\\text{T}}-{P}_{{\\text{H}}_{2}\\text{O}}[\/latex]<\/div>\n<p id=\"fs-idp144314112\">The pressure of water vapor above a sample of liquid water at 26 \u00b0C is 25.2 torr (<a class=\"target-chapter\" title=\"Water Properties\" href=\".\/chapter\/water-properties-2\/\" target=\"_blank\">Water Properties<\/a>), so:<\/p>\n<div id=\"fs-idm40258608\" data-type=\"equation\">[latex]{P}_{\\text{Ar}}=750\\text{torr}-25.2\\text{torr}=725\\text{torr}[\/latex]<\/div>\n<h4 id=\"fs-idp98186880\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A sample of oxygen collected over water at a temperature of 29.0 \u00b0C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?<\/p>\n<div id=\"fs-idp98213776\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0734 torr<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n<section id=\"fs-idp132626656\" data-depth=\"1\">\n<h2 data-type=\"title\">Chemical Stoichiometry and Gases<\/h2>\n<p id=\"fs-idp137887648\">Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.<\/p>\n<p id=\"fs-idp221629776\">We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.<\/p>\n<\/section>\n<section id=\"fs-idp45488016\" data-depth=\"1\">\n<h2 data-type=\"title\">Avogadro\u2019s Law Revisited<\/h2>\n<p id=\"fs-idm7424624\">Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.<\/p>\n<p id=\"fs-idp21344160\">We can extend Avogadro\u2019s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to [latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\phantom{\\rule{0.4em}{0ex}}\\rightarrow\\phantom{\\rule{0.4em}{0ex}}2{\\text{NH}}_{3}\\left(g\\right),[\/latex] a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.<\/p>\n<p id=\"fs-idp16937680\">The explanation for this is illustrated in Figure 5. According to Avogadro\u2019s law, equal volumes of gaseous N<sub>2<\/sub>, H<sub>2<\/sub>, and NH<sub>3<\/sub>, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N<sub>2<\/sub> reacts with three molecules of H<sub>2<\/sub> to produce two molecules of NH<sub>3<\/sub>, the volume of H<sub>2<\/sub> required is three times the volume of N<sub>2<\/sub>, and the volume of NH<sub>3<\/sub> produced is two times the volume of N<sub>2<\/sub>.<span data-type=\"media\" data-alt=\"This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled \u201cN subscript 2\u201d. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled \u201cH subscript 2.\u201d Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled \u201cN H subscript 3.\u201d Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.\"><span data-type=\"media\" data-alt=\"This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled \u201cN subscript 2\u201d. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled \u201cH subscript 2.\u201d Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled \u201cN H subscript 3.\u201d Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.\"><br \/>\n<\/span><\/span><\/p>\n<div style=\"width: 890px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212046\/CNX_Chem_09_03_Ammonia1.jpg\" alt=\"This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled \u201cN subscript 2\u201d. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled \u201cH subscript 2.\u201d Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled \u201cN H subscript 3.\u201d Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.\" width=\"880\" height=\"438\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. One volume of N<sub>2<\/sub> combines with three volumes of H<sub>2<\/sub> to form two volumes of NH<sub>3<\/sub>.<\/p>\n<\/div>\n<div id=\"fs-idp89809920\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 7<\/h3>\n<h4 id=\"fs-idp31720640\"><strong><span data-type=\"title\">Reaction of Gases<\/span><\/strong><\/h4>\n<p>Propane, C<sub>3<\/sub>H<sub>8<\/sub>(<em data-effect=\"italics\">g<\/em>), is used in gas grills to provide the heat for cooking. What volume of O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) measured at 25 \u00b0C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.<\/p>\n<h4 id=\"fs-idp137922560\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>The ratio of the volumes of C<sub>3<\/sub>H<sub>8<\/sub> and O<sub>2<\/sub> will be equal to the ratio of their coefficients in the balanced equation for the reaction:<\/p>\n<div id=\"fs-idp57291280\" data-type=\"equation\">[latex]\\begin{array}{l}{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\text{ }\\text{ }\\rightarrow\\text{ }\\text{ }3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(l\\right)\\\\ \\text{1 volume}+\\text{5 volumes}\\phantom{\\rule{3.5em}{0ex}}\\text{3 volumes}+\\text{4 volumes}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idp90728656\">From the equation, we see that one volume of C<sub>3<\/sub>H<sub>8<\/sub> will react with five volumes of O<sub>2<\/sub>:<\/p>\n<div id=\"fs-idp8505312\" data-type=\"equation\">[latex]2.7\\cancel{\\text{L}{\\text{C}}_{3}{\\text{H}}_{8}}\\times \\frac{\\text{5 L}{\\text{O}}_{2}}{1\\cancel{\\text{L}{\\text{C}}_{3}{\\text{H}}_{8}}}=\\text{13.5 L}{\\text{O}}_{2}[\/latex]<\/div>\n<p id=\"fs-idp78474976\">A volume of 13.5 L of O<sub>2<\/sub> will be required to react with 2.7 L of C<sub>3<\/sub>H<sub>8<\/sub>.<\/p>\n<h4 id=\"fs-idp109294464\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C<sub>2<\/sub>H<sub>2<\/sub>, at 0 \u00b0C and 1 atm. How many tanks of oxygen, each providing 7.00 \u00d7 10<sup>3<\/sup> L of O<sub>2<\/sub> at 0 \u00b0C and 1 atm, will be required to burn the acetylene?<\/p>\n<div id=\"fs-idp172649872\" data-type=\"equation\">[latex]2{\\text{C}}_{2}{\\text{H}}_{2}+5{\\text{O}}_{2}\\rightarrow 4{\\text{CO}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]<\/div>\n<div id=\"fs-idm19557184\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a03.34 tanks (2.34 \u00d7 10<sup>4<\/sup> L)<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8<\/h3>\n<h4 id=\"fs-idp19707328\"><strong><span data-type=\"title\">Volumes of Reacting Gases<\/span><\/strong><\/h4>\n<p>Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 \u00b0C and 1 atm, was manufactured. What volume of H<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N<sub>2<\/sub>?<\/p>\n<div id=\"fs-idp109603280\" data-type=\"equation\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightarrow 2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/div>\n<h4 id=\"fs-idp212669488\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>Because equal volumes of H<sub>2<\/sub> and NH<sub>3<\/sub> contain equal numbers of molecules and each three molecules of H<sub>2<\/sub> that react produce two molecules of NH<sub>3<\/sub>, the ratio of the volumes of H<sub>2<\/sub> and NH<sub>3<\/sub> will be equal to 3:2. Two volumes of NH<sub>3<\/sub>, in this case in units of billion ft<sup>3<\/sup>, will be formed from three volumes of H<sub>2<\/sub>:<\/p>\n<div id=\"fs-idp171418912\" data-type=\"equation\">[latex]683\\cancel{\\text{billion}{\\text{ft}}^{3}{\\text{NH}}_{3}}\\times \\frac{\\text{3 billion}{\\text{ft}}^{3}{\\text{H}}_{2}}{2\\cancel{\\text{billion}{\\text{ft}}^{3}{\\text{NH}}_{3}}}=1.02\\times {10}^{3}\\text{billion}{\\text{ft}}^{3}{\\text{H}}_{2}[\/latex]<\/div>\n<p id=\"fs-idp138098944\">The manufacture of 683 billion ft<sup>3<\/sup> of NH<sub>3<\/sub> required 1020 billion ft<sup>3<\/sup> of H<sub>2<\/sub>. (At 25 \u00b0C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)<\/p>\n<h4 id=\"fs-idp37301360\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What volume of O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) measured at 25 \u00b0C and 760 torr is required to react with 17.0 L of ethylene, C<sub>2<\/sub>H<sub>4<\/sub>(<em data-effect=\"italics\">g<\/em>), measured under the same conditions of temperature and pressure? The products are CO<sub>2<\/sub> and water vapor.<\/p>\n<div id=\"fs-idp105257584\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a051.0 L<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9<\/h3>\n<div id=\"fs-idp13049024\" data-type=\"example\">\n<h4 id=\"fs-idm2007840\"><strong><span data-type=\"title\">Volume of Gaseous Product<\/span><\/strong><\/h4>\n<p>What volume of hydrogen at 27 \u00b0C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?<\/p>\n<div id=\"fs-idm20890912\" data-type=\"equation\">[latex]2\\text{Ga}\\left(s\\right)+6\\text{HCl}\\left(aq\\right)\\rightarrow 2{\\text{GaCl}}_{3}\\left(aq\\right)+3{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/div>\n<h4 id=\"fs-idp98177696\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>To convert from the mass of gallium to the volume of H<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), we need to do something like this:<span id=\"fs-idp205351920\" data-type=\"media\" data-alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled \u201cMass of G a.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled \u201cMoles of G a.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled \u201cMoles of H subscript 2 ( g ).\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded lavender and is labeled \u201cVolume of H subscript 2 ( g ).\u201d\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212048\/CNX_Chem_09_03_Example3_img1.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled \u201cMass of G a.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled \u201cMoles of G a.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled \u201cMoles of H subscript 2 ( g ).\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded lavender and is labeled \u201cVolume of H subscript 2 ( g ).\u201d\" width=\"881\" height=\"104\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idp47678000\">The first two conversions are:<\/p>\n<div id=\"fs-idm46693760\" data-type=\"equation\">[latex]8.88\\cancel{\\text{g Ga}}\\times \\frac{1\\cancel{\\text{mol Ga}}}{69.723\\cancel{\\text{g Ga}}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{3 mol}{\\text{H}}_{2}}{2\\cancel{\\text{mol Ga}}}=0.191{\\text{mol H}}_{2}[\/latex]<\/div>\n<p id=\"fs-idp35037744\">Finally, we can use the ideal gas law:<\/p>\n<div id=\"fs-idp48121168\" data-type=\"equation\">[latex]{V}_{{\\text{H}}_{2}}={\\left(\\frac{nRT}{P}\\right)}_{{\\text{H}}_{2}}=\\frac{0.191\\cancel{\\text{mol}}\\times \\text{0.08206 L}\\cancel{\\text{atm}{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}}\\times \\text{300 K}}{0.951\\cancel{\\text{atm}}}=\\text{4.94 L}[\/latex]<\/div>\n<h4 id=\"fs-idm17626704\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO<sub>2<\/sub> at 343 \u00b0C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen?<\/p>\n<div id=\"fs-idp30333040\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a01.30 \u00d7 10<sup>3<\/sup> L<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm13395728\" class=\"chemistry sciences-interconnect\" data-type=\"note\"><\/div>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"title\">\n<div id=\"fs-idm152886256\" class=\"chemistry sciences-interconnect\" data-type=\"note\">\n<div id=\"fs-idp13049024\" data-type=\"example\">\n<h3 id=\"fs-idm2007840\">Greenhouse Gases and Climate Change<\/h3>\n<\/div>\n<div id=\"fs-idm13395728\" class=\"chemistry sciences-interconnect\" data-type=\"note\">\n<p id=\"fs-idp143406944\">The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost [latex]\\frac{1}{3}[\/latex] is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions\u2014without atmosphere, the average global average temperature of 14 \u00b0C (57 \u00b0F) would be about \u201319 \u00b0C (\u20132 \u00b0F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth\u2019s climate (Figure 6).<\/p>\n<figure id=\"CNX_Chem_09_03_GlobalWarming\">\n<div style=\"width: 660px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212049\/CNX_Chem_09_03_GlobalWarming1.jpg\" alt=\"This diagram shows half of a two dimensional view of the earth in blue and green at the left of the image. A slight distance outside the hemisphere is a grey arc. A line segment connects the label \u201cAtmosphere\u201d to the region between the hemisphere and the grey arc. In this region, near the surface of the earth the chemical formulas C O subscript 2, C H subscript 3, and N subscript 2 O appear. Five red arrows formed from wavy lines extend from green regions on the earth out into and just beyond the region labeled \u201cAtmosphere.\u201d The label \u201cInfrared radiation\u201d points to one of these red arrows. At a fair distance outside of the grey arc appears a yellow circle with a jagged boundary. This circle is labeled \u201cSun.\u201d From it extend yellow arrows with wavy lines which extend toward the earth. Three of the arrows extend to the green region on the earth. One of the arrows appears to be reflected off the grey arc, causing its path to turn away from the earth.\" width=\"650\" height=\"436\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Greenhouse gases trap enough of the sun\u2019s energy to make the planet habitable\u2014this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-idm12984240\">There is strong evidence from multiple sources that higher atmospheric levels of CO<sub>2<\/sub> are caused by human activity, with fossil fuel burning accounting for about [latex]\\frac{3}{4}[\/latex] of the recent increase in CO<sub>2<\/sub>. Reliable data from ice cores reveals that CO<sub>2<\/sub> concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO<sub>2<\/sub> concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure 7).<span data-type=\"media\" data-alt=\"This figure has the heading \u201cCarbon Dioxide in the Atmosphere.\u201d The first graph has a horizontal axis label \u201cYear ( B C )\u201d and a vertical axis label \u201cCarbon dioxide concentration ( p p m ).\u201d The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.\"><span data-type=\"media\" data-alt=\"This figure has the heading \u201cCarbon Dioxide in the Atmosphere.\u201d The first graph has a horizontal axis label \u201cYear ( B C )\u201d and a vertical axis label \u201cCarbon dioxide concentration ( p p m ).\u201d The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.\"><br \/>\n<\/span><\/span><\/p>\n<div style=\"width: 889px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212050\/CNX_Chem_09_03_GlobalWarming21.jpg\" alt=\"This figure has the heading \u201cCarbon Dioxide in the Atmosphere.\u201d The first graph has a horizontal axis label \u201cYear ( B C )\u201d and a vertical axis label \u201cCarbon dioxide concentration ( p p m ).\u201d The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.\" width=\"879\" height=\"329\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. CO<sub>2<\/sub> levels over the past 700,000 years were typically from 200\u2013300 ppm, with a steep, unprecedented increase over the past 50 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<p><a href=\"http:\/\/www.epa.gov\/climatestudents\/basics\/today\/greenhouse-effect.html\" target=\"_blank\">Click here to see a 2-minute video from the Environmental Protection Agency <\/a>explaining greenhouse gases and global warming.<\/p>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"title\">\n<div id=\"fs-idm152886256\" class=\"chemistry sciences-interconnect\" data-type=\"note\">\n<h3 data-type=\"title\">Portrait of a Chemist: Susan Solomon<\/h3>\n<div style=\"width: 335px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212054\/CNX_Chem_09_03_SusanSolom1.jpg\" alt=\"A photograph is shown of Susan Solomon sitting next to a globe.\" width=\"325\" height=\"274\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. Susan Solomon\u2019s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)<\/p>\n<\/div>\n<p id=\"CNX_Chem_09_03_SusanSolom\">Atmospheric and climate scientist Susan <span class=\"no-emphasis\" data-type=\"term\">Solomon<\/span> (Figure 8) is the author of one of <em data-effect=\"italics\">The New York Times<\/em> books of the year (<em data-effect=\"italics\">The Coldest March<\/em>, 2001), one of Time magazine\u2019s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.<\/p>\n<p>For more information, watch this <a href=\"http:\/\/techtv.mit.edu\/videos\/17040-introducing-atmospheric-chemist-prof-susan-solomon\" target=\"_blank\">video\u00a0from MIT about Susan Solomon<\/a>.<\/p>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<section>\n<div data-type=\"note\">\n<p id=\"fs-idp236281408\">The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton\u2019s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro\u2019s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Equations<\/h3>\n<section>\n<div data-type=\"note\">\n<ul>\n<li><em data-effect=\"italics\">P<sub>Total<\/sub><\/em> = <em data-effect=\"italics\">P<sub>A<\/sub><\/em> + <em data-effect=\"italics\">P<sub>B<\/sub><\/em> + <em data-effect=\"italics\">P<sub>C<\/sub><\/em> + \u2026 = \u01a9<sub>i<\/sub><em data-effect=\"italics\">P<\/em><sub>i<\/sub><\/li>\n<li><em data-effect=\"italics\">P<sub>A<\/sub><\/em> = <em data-effect=\"italics\">X<sub>A<\/sub> P<sub>Total<\/sub><\/em><\/li>\n<li>[latex]{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<\/div>\n<section id=\"fs-idm40478192\" class=\"summary\" data-depth=\"1\">\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\n<div class=\"entry-content\">\n<div class=\"im_section\">\n<div class=\"im_section\">\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<section id=\"fs-idp118960256\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idp231956592\" data-type=\"exercise\">\n<div id=\"fs-idp8716432\" data-type=\"problem\">\n<ol>\n<li id=\"fs-idp5305888\">What is the density of laughing gas, dinitrogen monoxide, N<sub>2<\/sub>O, at a temperature of 325 K and a pressure of 113.0 kPa?<\/li>\n<li>Calculate the density of Freon 12, CF<sub>2<\/sub>Cl<sub>2<\/sub>, at 30.0 \u00b0C and 0.954 atm.<\/li>\n<li>Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.<\/li>\n<li>A cylinder of O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) used in breathing by emphysema patients has a volume of 3.00 L at a pressure of 10.0 atm. If the temperature of the cylinder is 28.0 \u00b0C, what mass of oxygen is in the cylinder?<\/li>\n<li>What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 \u00b0C and a pressure of 307 torr?<\/li>\n<li>What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 \u00b0C and a pressure of 777 torr?<\/li>\n<li>How could you show experimentally that the molecular formula of propene is C<sub>3<\/sub>H<sub>6<\/sub>, not CH<sub>2<\/sub>?<\/li>\n<li>The density of a certain gaseous fluoride of phosphorus is 3.93 g\/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.<\/li>\n<li>Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies l25 mL with a pressure of 99.5 kPa at 22 \u00b0C?\n<ol>\n<li>Outline the steps necessary to answer the question.<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>A 36.0\u2013L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO<sub>2<\/sub>, 805 g O<sub>2<\/sub>, and 4,880 g N<sub>2<\/sub>. What is the pressure in the flask in atmospheres, in torr, and in kilopascals?<\/li>\n<li>A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO<sub>2<\/sub>, 12.0% O<sub>2<\/sub>, and the remainder N<sub>2<\/sub> at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)<\/li>\n<li>A sample of gas isolated from unrefined petroleum contains 90.0% CH<sub>4<\/sub>, 8.9% C<sub>2<\/sub>H<sub>6<\/sub>, and 1.1% C<sub>3<\/sub>H<sub>8<\/sub> at a total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)<\/li>\n<li>A mixture of 0.200 g of H<sub>2<\/sub>, 1.00 g of N<sub>2<\/sub>, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.<\/li>\n<li>Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than 3.0 % O<sub>2<\/sub> is not. If enough O<sub>2<\/sub> is added to a cylinder of H<sub>2<\/sub> at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive?<\/li>\n<li>A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 \u00d7 10<sup>-6<\/sup> mg\/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 \u00b0C?<\/li>\n<li>A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of 18 \u00b0C. What is the pressure of the carbon monoxide? (See Table 9.2\u00a0for the vapor pressure of water.)<\/li>\n<li>In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 \u00b0C. The mass of the gas was 0.472 g. What was the molar mass of the gas?<\/li>\n<li>Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO:[latex]2\\text{HgO}\\left(s\\right)\\rightarrow 2\\text{Hg}\\left(l\\right)+{\\text{O}}_{2}\\left(g\\right)[\/latex]\n<ol>\n<li>Outline the steps necessary to answer the following question: What volume of O<sub>2<\/sub> at 23 \u00b0C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:[latex]4{\\text{H}}_{2}\\text{O}\\left(g\\right)+3\\text{Fe}\\left(s\\right)\\rightarrow{\\text{Fe}}_{3}{\\text{O}}_{4}\\left(s\\right)+4{\\text{H}}_{2}\\left(g\\right)[\/latex]\n<ol>\n<li>Outline the steps necessary to answer the following question: What volume of H<sub>2<\/sub> at a pressure of 745 torr and a temperature of 20 \u00b0C can be prepared from the reaction of 15.O g of H<sub>2<\/sub>O?<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>The chlorofluorocarbon CCl<sub>2<\/sub>F<sub>2<\/sub> can be recycled into a different compound by reaction with hydrogen to produce CH<sub>2<\/sub>F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), a compound useful in chemical manufacturing: [latex]{\\text{CCl}}_{2}{\\text{F}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{CH}}_{2}{\\text{F}}_{2}\\left(g\\right)+2\\text{HCl}\\left(g\\right).[\/latex]\n<ol>\n<li>Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and 35.5 \u00b0C would be required to react with 1 ton (1.000 \u00d7 10<sup>3<\/sup> kg) of CCl<sub>2<\/sub>F<sub>2<\/sub>?<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN<sub>3<\/sub>). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 \u00b0C and 756 torr formed by the decomposition of 125 g of sodium azide.<\/li>\n<li>Lime, CaO, is produced by heating calcium carbonate, CaCO<sub>3<\/sub>; carbon dioxide is the other product.\n<ol>\n<li>Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875\u00b0 and 0.966 atm is produced by the decomposition of 1 ton (1.000 \u00d7 10<sup>3<\/sup> kg) of calcium carbonate?<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C<sub>2<\/sub>H<sub>2<\/sub>, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC<sub>2<\/sub>, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.\n<ol>\n<li>Outline the steps necessary to answer the following question: What volume of C<sub>2<\/sub>H<sub>2<\/sub> at 1.005 atm and 12.2 \u00b0C is formed by the reaction of 15.48 g of CaC<sub>2<\/sub> with water?<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C<sub>2<\/sub>H<sub>6<\/sub>, to produce carbon dioxide and water, if the volumes of C<sub>2<\/sub>H<sub>6<\/sub> and O<sub>2<\/sub> are measured under the same conditions of temperature and pressure.<\/li>\n<li>What volume of O<sub>2<\/sub> at STP is required to oxidize 8.0 L of NO at STP to NO<sub>2<\/sub>? What volume of NO<sub>2<\/sub> is produced at STP?<\/li>\n<li>Consider the following questions:\n<ol>\n<li>What is the total volume of the CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) and H<sub>2<\/sub>O(<em data-effect=\"italics\">g<\/em>) at 600 \u00b0C and 0.888 atm produced by the combustion of 1.00 L of C<sub>2<\/sub>H<sub>6<\/sub>(<em data-effect=\"italics\">g<\/em>) measured at STP?<\/li>\n<li>What is the partial pressure of H<sub>2<\/sub>O in the product gases?<\/li>\n<\/ol>\n<\/li>\n<li>Methanol, CH<sub>3<\/sub>OH, is produced industrially by the following reaction:[latex]\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)\\stackrel{\\text{copper catalyst 300 \\textdegree C, 300 atm}}{\\to }{\\text{CH}}_{3}\\text{OH}\\left(g\\right)[\/latex]<br \/>\nAssuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.<\/li>\n<li>What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO<sub>2<\/sub> to BaO and O<sub>2<\/sub>?<\/li>\n<li>A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N<sub>2<\/sub> and 1.25 L of O<sub>2<\/sub> at STP. What is the colorless gas?<\/li>\n<li>Ethanol, C<sub>2<\/sub>H<sub>5<\/sub>OH, is produced industrially from ethylene, C<sub>2<\/sub>H<sub>4<\/sub>, by the following sequence of reactions: [latex]3{\\text{C}}_{2}{\\text{H}}_{4}+2{\\text{H}}_{2}{\\text{SO}}_{4}\\rightarrow{\\text{C}}_{2}{\\text{H}}_{5}{\\text{HSO}}_{4}+{\\left({\\text{C}}_{2}{\\text{H}}_{5}\\right)}_{2}{\\text{SO}}_{4}[\/latex] [latex]{\\text{C}}_{2}{\\text{H}}_{5}{\\text{HSO}}_{4}+{\\left({\\text{C}}_{2}{\\text{H}}_{5}\\right)}_{2}{\\text{SO}}_{4}+3{\\text{H}}_{2}\\text{O}\\rightarrow 3{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}+2{\\text{H}}_{2}{\\text{SO}}_{4}[\/latex] What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?<\/li>\n<li>One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature (37 \u00b0C) and a pressure of 743 torr, what is the molar mass of hemoglobin?<\/li>\n<li>A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)<\/li>\n<li>One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (-NH<sub>2<\/sub>) in protein material are allowed to react with nitrous acid, HNO<sub>2<\/sub>, to form N<sub>2<\/sub> gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH<sub>2<\/sub>(NH<sub>2<\/sub>)COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N<sub>2<\/sub> collected over water at a pressure of 735 torr and 29 \u00b0C. What was the percentage of glycine in the sample? [latex]{\\text{CH}}_{2}\\left({\\text{NH}}_{2}\\right){\\text{CO}}_{2}\\text{H}+{\\text{HNO}}_{2}\\rightarrow{\\text{CH}}_{2}\\left(\\text{OH}\\right){\\text{CO}}_{2}\\text{H}+{\\text{H}}_{2}\\text{O}+{\\text{N}}_{2}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\n<div class=\"entry-content\">\n<div class=\"im_section\">\n<div class=\"im_section\">\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\n<div class=\"bcc-box bcc-info\">\n<h4>Selected Answers<\/h4>\n<p>2.\u00a0[latex]\\rho =\\frac{P\\mathcal{M}}{RT}=\\frac{0.954\\cancel{\\text{atm}}\\left[12.011+2\\left(18.9954\\right)+2\\left(35.453\\right)\\right]\\text{g}\\cancel{{\\text{mol}}^{\\text{-1}}}}{\\text{0.08206 L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 303.15\\cancel{\\text{K}}}=\\text{4.64 g}{\\text{L}}^{\\text{-1}}[\/latex]<\/p>\n<p>4.\u00a0[latex]\\text{mass}{\\text{O}}_{2}=\\frac{\\left(\\text{31.9988 g}\\cancel{{\\text{mol}}^{\\text{-1}}}\\right)\\left(10.0\\cancel{\\text{atm}}\\right)\\left(3.00\\cancel{\\text{L}}\\right)}{\\left(0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(301.15\\cancel{\\text{K}}\\right)}=\\text{38.8 g}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>6. From the ideal gas law, <em data-effect=\"italics\">PV = nRT<\/em>, set [latex]n=\\frac{\\text{mass}}{\\text{molar mass}}[\/latex] and solve the molar mass.<\/p>\n<p>[latex]\\text{molar mass}=\\frac{\\left(\\text{0.281 g}\\right)\\left(0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(399.15\\cancel{\\text{K}}\\right)}{\\left(\\frac{777\\cancel{\\text{torr}}}{760\\cancel{\\text{torr}}\\cancel{{\\text{atm}}^{\\text{-1}}}}\\right)\\left(0.125\\cancel{\\text{L}}\\right)}=\\text{72.0 g}{\\text{mol}}^{\\text{-1}}[\/latex]<\/p>\n<p>8. [latex]\\mathcal{M}=\\frac{mRT}{PV}D=\\frac{m}{V}\\mathcal{M}=\\frac{DRT}{P}[\/latex]<\/p>\n<p style=\"text-align: left;\">[latex]\\mathcal{M}=\\frac{\\text{3.93 g}{\\text{L}}^{\\text{-1}}\\left(\\text{0.08206 L}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(273.15\\cancel{\\text{K}}\\right)}{1.00\\cancel{\\text{atm}}}=\\text{88.1 g}{\\text{mol}}^{\\text{-1}}[\/latex]<\/p>\n<p style=\"text-align: left;\">\u2133<sub>phosphorous<\/sub> = 30.97376 g\/mol<\/p>\n<div style=\"text-align: left;\" data-type=\"newline\">\u2133<sub>fluorine<\/sub> = 18.998403 g\/mol<\/div>\n<div style=\"text-align: left;\" data-type=\"newline\"><\/div>\n<div style=\"text-align: left;\" data-type=\"newline\">[latex]\\begin{array}{l}\\text{molecular formula:}\\text{phosphorous:}30.97376\\\\ \\\\ \\phantom{\\rule{7.9em}{0ex}}\\text{flourine:}\\phantom{\\rule{1.3em}{0ex}}\\underline{3\\left(18.998403\\right)}\\\\ \\phantom{\\rule{13.5em}{0ex}}87.968969\\end{array}[\/latex]<\/div>\n<div style=\"text-align: left;\" data-type=\"newline\"><\/div>\n<div style=\"text-align: left;\" data-type=\"newline\">The molecular formula is PF<sub>3<\/sub>.<\/div>\n<div style=\"text-align: left;\" data-type=\"newline\"><\/div>\n<div style=\"text-align: left;\" data-type=\"newline\">To find this answer you can either use trial and error, or you can realize that since phosphorus is in group 5, it can fill its valence shell by forming three bonds. Fluorine, being in group 7, needs to form only one bond to fill its shell. Thus it makes sense to start with PF<sub>3<\/sub> as a probable formula.<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>10. Calculate the moles of each gas present and from that, calculate the pressure from the ideal gas law. Assume 25 \u00b0C. The calibration gas contains:<\/p>\n<div data-type=\"newline\">[latex]\\frac{350\\cancel{\\text{g}}{\\text{CO}}_{2}}{44.0098\\cancel{\\text{g}}{\\text{mol}}^{\\text{-1}}{\\text{CO}}_{2}}=\\text{7.953 mol}{\\text{CO}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\frac{805\\cancel{\\text{g}}{\\text{O}}_{2}}{31.9988\\cancel{\\text{g}}{\\text{mol}}^{\\text{-1}}{\\text{O}}_{2}}=\\text{25.157 mol}{\\text{O}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\frac{4880\\cancel{\\text{g}}{\\text{N}}_{2}}{28.01348\\cancel{\\text{g}}{\\text{mol}}^{\\text{-1}}{\\text{N}}_{2}}=\\text{174.202 mol}{\\text{N}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">Total moles = 7.953 + 25.157 + 174.202 = 207.312 mol<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]P=\\frac{nRT}{V}=\\frac{207.312\\cancel{\\text{mol}}\\times 0.08206\\cancel{\\text{L}}\\text{atm}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 298.15\\cancel{\\text{K}}}{36.0\\cancel{\\text{L}}}=\\text{141 atm}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>12. Since these are percentages of the total pressure, the partial pressure can be calculated as follows:<\/p>\n<div id=\"fs-idp207682832\" data-type=\"exercise\">\n<div id=\"fs-idm99835968\" data-type=\"solution\">\n<div data-type=\"newline\">CH<sup>4<\/sup>: 90% of 307.2 kPa = 0.900 \u00d7 307.2 = 276 kPa<\/div>\n<div data-type=\"newline\">C<sup>2<\/sup> H<sup>6<\/sup>: 8.9% of 307.2 kPa = 0.089 \u00d7 307.2 = 27 kPa<\/div>\n<div data-type=\"newline\">C<sup>3<\/sup> H<sup>8<\/sup>: 1.1% of 307.2 kPa = 0.011 \u00d7 307.2 = 3.4 kPa<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">14.\u00a0The oxygen increases the pressure within the tank to (34.5 atm \u2013 33.2 atm =) 1.3 atm. The percentage O<sub>2<\/sub> on a mole basis is [latex]\\frac{1.3}{34.5}\\times 100\\%=3.77\\%.[\/latex] The mixture is explosive. However, the percentage is given as a weight percent. Converting to a mass basis increases the percentage of oxygen even more, so the mixture is still explosive.<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>16. The vapor pressure of water at 18 \u00b0C is 15.5 torr. Subtract the vapor pressure of water from the total pressure to find the pressure of the carbon monoxide:<\/p>\n<div data-type=\"newline\"><em data-effect=\"italics\">P<\/em><sub>T<\/sub> = <em data-effect=\"italics\">P<\/em><sub>gas<\/sub> + <em data-effect=\"italics\">P<\/em><sub>water<\/sub><\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">Rearrangement gives:<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><em data-effect=\"italics\">P<\/em><sub>T<\/sub> \u2013<em data-effect=\"italics\"> P<\/em><sub>water =<\/sub><em data-effect=\"italics\">P<\/em><sub>gas<\/sub><\/div>\n<div data-type=\"newline\">756 torr \u2013 15.5 torr = 740 torr<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>18. (a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O<sub>2<\/sub> produced by decomposition of this amount of HgO; and determine the volume of O<sub>2<\/sub> from the moles of O<sub>2<\/sub>, temperature, and pressure.<\/p>\n<div data-type=\"newline\">(b) [latex]\\begin{array}{l}\\\\ \\\\ 5.36\\cancel{\\text{g HgO}}\\times \\frac{\\text{1 mol HgO}}{\\left(200.59+15.9994\\right)\\cancel{\\text{g HgO}}}=\\text{0.0247 mol HgO}\\\\ 0.0247\\cancel{\\text{mol HgO}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{2\\cancel{\\text{mol HgO}}}=\\text{0.01235 mol}{\\text{O}}_{2}\\end{array}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em><\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><em data-effect=\"italics\">P<\/em> = 0.975 atm<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><em data-effect=\"italics\">T<\/em> = (23.0 + 273.15) K<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]V=\\frac{nRT}{P}=\\frac{0.01235\\cancel{\\text{mol}}\\left(\\text{0.08206 L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(296.15\\cancel{\\text{K}}\\right)}{0.975\\cancel{\\text{atm}}}\\phantom{\\rule{0.4em}{0ex}}=\\phantom{\\rule{0.4em}{0ex}}\\text{0.308 L}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div id=\"fs-idp46179152\" data-type=\"exercise\">\n<div id=\"fs-idp138325296\" data-type=\"solution\">\n<div data-type=\"newline\">\n<p>20. (a) Determine the molar mass of CCl<sub>2<\/sub>F<sub>2<\/sub>. From the balanced equation, calculate the moles of H<sub>2<\/sub> needed for the complete reaction. From the ideal gas law, convert moles of H<sub>2<\/sub> into volume.<\/p>\n<div id=\"fs-idp39473616\" data-type=\"exercise\">\n<div id=\"fs-idp75359296\" data-type=\"solution\">\n<div data-type=\"newline\">(b) Molar mass of CCl<sub>2<\/sub> F<sub>2<\/sub> = 12.011 + 2 \u00d7 18.9984 + 2 \u00d7 35.4527 = 120.913 g\/mol<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\text{mol}{\\text{H}}_{2}=1.000\\times {10}^{6}\\text{g}\\times \\frac{\\text{1 mol}{\\text{CCL}}_{2}{\\text{F}}_{2}}{\\text{120.913 g}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{4 mol}{\\text{H}}_{2}}{\\text{1 mol}{\\text{CCl}}_{2}{\\text{F}}_{2}}=3.308\\times {10}^{4}\\text{mol}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]V=\\frac{nRT}{P}=\\frac{\\left(3.308\\times {10}^{4}\\cancel{\\text{mol}}\\right)\\left(\\text{0.08206 L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}{\\cancel{\\text{K}}}^{\\text{-1}}\\right)\\left(308.65\\cancel{\\text{K}}\\right)}{225\\cancel{\\text{atm}}}=3.72\\times {10}^{3}\\text{L}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>22. (a) Balance the equation. Determine the grams of CO<sub>2<\/sub> produced and the number of moles. From the ideal gas law, determine the volume of gas.<\/p>\n<div data-type=\"newline\">(b) [latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightarrow\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\text{mass}{\\text{CO}}_{2}=1.00\\times {10}^{6}\\text{g}\\times \\frac{\\text{1 mol}{\\text{CaCO}}_{2}}{\\text{100.087 g}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{44.01 g}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{CO}}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{1 mol}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{CaCO}}_{2}}=4.397\\times {10}^{5}\\text{g}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\text{mol}{\\text{CO}}_{2}=\\frac{4.397\\times {10}^{5}\\text{g}}{\\text{44.01 g}{\\text{mol}}^{\\text{-1}}}=\\text{9991 mol}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]V=\\frac{nRT}{P}=\\frac{\\left(\\text{9991 mol}\\right)\\left(\\text{0.08206 L atm}{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}\\right)\\left(\\text{875 K}\\right)}{\\text{0.966 atm}}=7.43\\times {10}^{5}\\text{L}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>24. [latex]2{\\text{C}}_{2}{\\text{H}}_{6}\\left(\\text{g}\\right)+7{\\text{O}}_{2}\\left(\\text{g}\\right)\\rightarrow 4{\\text{CO}}_{2}\\left(\\text{g}\\right)+6{\\text{H}}_{2}\\text{O}\\left(\\text{g}\\right)[\/latex]<\/p>\n<div data-type=\"newline\">From the balanced equation, we see that 2 mol of C<sub>2<\/sub>H<sub>6<\/sub> requires 7 mol of O<sub>2<\/sub> to burn completely. Gay-Lussac\u2019s law states that gases react in simple proportions by volume. As the number of liters is proportional to the number of moles,<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\frac{\\text{12.00 L}}{\\text{2 mol}{\\text{C}}_{2}{\\text{H}}_{6}}=\\frac{V\\left({\\text{O}}_{2}\\right)}{\\text{7 mol}{\\text{O}}_{2}}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]V\\left({\\text{O}}_{2}\\right)=\\frac{\\text{12.00 L}\\times 7}{2}=\\text{42.00 L}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>26. (a) The scheme to solve this problem is:<\/p>\n<div data-type=\"newline\">[latex]{\\text{volume C}}_{2}{\\text{H}}_{6}\\left(g\\right)\\stackrel{\\begin{array}{l}\\text{ideal gas}\\\\ \\text{equation}\\end{array}}{\\to }{\\text{mol C}}_{2}{\\text{H}}_{6}\\left(g\\right)\\stackrel{\\begin{array}{l}\\text{reaction}\\\\ \\text{stoichiometry}\\end{array}}{\\to }{\\text{mol CO}}_{2}+{\\text{H}}_{2}\\text{O}\\stackrel{\\begin{array}{l}\\text{ideal gas}\\\\ \\text{equation}\\end{array}}{\\to }{\\text{volume CO}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\begin{array}{l}{\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)+3\\frac{1}{2}{\\text{O}}_{2}\\left(g\\right)\\rightarrow 2{\\text{CO}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\text{O}\\left(g\\right)\\\\ \\text{1}.n\\left({\\text{C}}_{2}{\\text{H}}_{6}\\right)=\\frac{PV}{RT}=\\frac{1.00\\cancel{\\text{atm}}\\times 1.00\\cancel{\\text{L}}}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(273.15\\cancel{\\text{K}}\\right)}=\\text{0.0446 mol}\\\\ \\text{2}.0.0446\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}}\\times \\frac{\\text{5 mol products}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}}}=\\text{0.223 mol products}\\\\ \\text{3}.V=nRT=\\frac{\\left(0.223\\cancel{\\text{mol}}\\right)\\left(\\text{0.08206 L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(873.15\\cancel{\\text{K}}\\right)}{0.888\\cancel{\\text{atm}}}=\\text{18.0 L}\\end{array}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(b) First, calculate the mol H<sub>2<\/sub>O produced:<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\text{0.0446 mol}{\\text{C}}_{2}{\\text{H}}_{6}\\times \\frac{\\text{3 mol products}}{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{6}}=\\text{0.1338 mol}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">Second, calculate the pressure of H<sub>2<\/sub>O:<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]P=\\frac{nRT}{V}=\\frac{\\left(0.1338\\cancel{\\text{mol}}\\right)\\left(0.8206\\cancel{\\text{L}}\\text{atm}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(873.15\\cancel{\\text{K}}\\right)}{18.0\\cancel{\\text{L}}}=\\text{0.533 atm}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>28. First, we must write a balanced equation to establish the stoichiometry of the reaction:<\/p>\n<div data-type=\"newline\">\n<div data-type=\"newline\">[latex]2{\\text{BaO}}_{2}\\rightarrow 2\\text{BaO}+{\\text{O}}_{2}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">We are given the mass of BaO<sub>2<\/sub> that decomposes, so the scheme for solving this problem will be:<\/div>\n<p><span id=\"fs-idp92165488\" data-type=\"media\" data-alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled \u201cMass of B a O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cMolar mass B a O subscript 2\u201d pointing right to a second rectangle which is shaded pink and is labeled \u201cMoles of B a O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cReaction stoichiometry\u201d pointing right to a third rectangle which is shaded pink and is labeled \u201cMoles of O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cIdeal gas equation\u201d pointing right to a fourth rectangle which is shaded lavender and is labeled \u201cVolume of O subscript 2.\u201d\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212055\/CNX_Chem_09_03_Exercise29_img1.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled \u201cMass of B a O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cMolar mass B a O subscript 2\u201d pointing right to a second rectangle which is shaded pink and is labeled \u201cMoles of B a O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cReaction stoichiometry\u201d pointing right to a third rectangle which is shaded pink and is labeled \u201cMoles of O subscript 2.\u201d This rectangle is followed by an arrow labeled \u201cIdeal gas equation\u201d pointing right to a fourth rectangle which is shaded lavender and is labeled \u201cVolume of O subscript 2.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<div data-type=\"newline\">Mass (BaO<sub>2<\/sub>) = 137.33 + 2(15.9994) = 169.33 g\/mol<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]n\\left({\\text{O}}_{2}\\right)=\\text{129.7 g}{\\text{BaO}}_{2}\\times \\frac{\\text{1 mol}{\\text{BaO}}_{2}}{\\text{169.33 g}{\\text{BaO}}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{1 mol}{\\text{O}}_{2}}{\\text{2 mol}{\\text{BaO}}_{2}}=\\text{0.3830 mol}{\\text{O}}_{2}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]V\\left({\\text{O}}_{2}\\right)=\\frac{nRT}{P}=\\frac{\\text{0.3830 mol}\\left(\\text{8.314 L kPa}{\\text{mol}}^{\\text{-1}}{\\text{K}}^{\\text{-1}}\\right)\\left(\\text{423.0 K}\\right)}{\\text{127.4 kPa}}=\\text{10.57 L}{\\text{O}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>30. At 90.1% conversion, a 1.000 \u00d7 10<sup>6<\/sup> g final yield would require a [latex]\\left(\\frac{1.000\\times {10}^{6}}{0.901}\\right)=1.1099\\times {10}^{6}\\text{g}[\/latex] theoretical yield.<\/p>\n<div data-type=\"newline\">3C<sub>2<\/sub>H<sub>4<\/sub> produces 3C<sub>2<\/sub>H<sub>5<\/sub>OH, giving a 1:1 ratio:<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\text{mol}\\left({\\text{C}}_{2}{\\text{H}}_{4}\\right)=1.1099\\times {10}^{6}\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}\\text{OH}}}{46.069\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}\\text{OH}}}=2.409\\times {10}^{4}\\text{mol}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><em data-effect=\"italics\">V<\/em> (C<sub>2<\/sub> H<sub>4<\/sub>) = 22.4 L\/mol \u00d7 2.409 \u00d7 10<sup>4<\/sup> mol = 5.40 \u00d7 10<sup>5<\/sup> L<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>32. The reaction is:<\/p>\n<div data-type=\"newline\">[latex]{\\text{XeF}}_{x}+\\frac{x}{2}{\\text{H}}_{2}\\rightarrow\\text{Xe}+x\\text{HF}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">The pressure of H<sub>2<\/sub> that reacts is:<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">48 torr \u2013 24 torr = 24 torr<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">The number of moles of gas is proportional to the partial pressures. The reaction used 24 torr of XeF<sub>x<\/sub> and 24 torr of H<sub>2<\/sub> so:<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\frac{x}{2}=1[\/latex] and <em data-effect=\"italics\">x<\/em> = 2<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">The formula for the xenon compound is XeF<sub>2<\/sub>.<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">Immediately after the H<sub>2<\/sub> is added (before the reaction):<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\begin{array}{lll}\\\\ \\\\ \\hfill {P}_{\\text{Total}}& =& {P}_{{\\text{XeF}}_{2}}+{P}_{{\\text{H}}_{2}}\\hfill \\\\ \\hfill {P}_{{\\text{H}}_{2}}& =& {P}_{\\text{Total}}-{P}_{{\\text{XeF}}_{2}}\\hfill \\\\ & =& \\text{72 torr}-\\text{24 torr}\\hfill \\\\ & =& 48\\text{torr}\\hfill \\end{array}[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">After the reaction:<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]{P}_{\\text{Xe}}=\\text{24 torr}\\text{ }\\left(\\text{1 mol}{\\text{XeF}}_{\\text{x}}\\rightarrow\\text{1 mol Xe}\\right)[\/latex]<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">And the partial pressure of unreacted H<sub>2<\/sub> is:<\/div>\n<div style=\"padding-left: 30px;\" data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]\\begin{array}{ll}\\\\ \\hfill {P}_{{\\text{H}}_{2}}& ={P}_{\\text{Total}}-{P}_{\\text{Xe}}\\hfill \\\\ & =\\text{48 torr}-\\text{24 torr}\\hfill \\\\ & =\\text{24 torr}\\hfill \\end{array}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-success\">\n<section id=\"glossary\">\n<h3>Glossary<\/h3>\n<div data-type=\"definition\">\n<div id=\"fs-idm8143856\" data-type=\"definition\">\n<p data-type=\"definition\"><strong><span data-type=\"term\">Dalton\u2019s law of partial pressures<br \/>\n<\/span><\/strong>total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">mole fraction<br \/>\n<\/span><\/strong>concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">partial pressure<br \/>\n<\/span><\/strong>pressure exerted by an individual gas in a mixture<\/p>\n<div id=\"fs-idp57901520\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">vapor pressure of water<br \/>\n<\/span><\/strong>pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2102\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2102","chapter","type-chapter","status-publish","hentry"],"part":3001,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/2102","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":16,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/2102\/revisions"}],"predecessor-version":[{"id":5509,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/2102\/revisions\/5509"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/parts\/3001"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/2102\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/media?parent=2102"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapter-type?post=2102"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/contributor?post=2102"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/license?post=2102"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}