{"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":"2016-10-27T15:56:09","modified_gmt":"2016-10-27T15:56:09","slug":"stoichiometry-of-gaseous-substances-mixtures-and-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/chapter\/stoichiometry-of-gaseous-substances-mixtures-and-reactions\/","title":{"raw":"Stoichiometry of Gaseous Substances, Mixtures, and Reactions","rendered":"Stoichiometry of Gaseous Substances, Mixtures, and Reactions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\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\nThe study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, 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\r\n\r\nAs 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.\r\n<h2>Density of a Gas<\/h2>\r\nRecall 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>PV<\/em> = <em>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.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Derivation of a Density Formula from the Ideal Gas Law<\/h3>\r\nUse <em>PV<\/em> = <em>nRT<\/em> to derive a formula for the density of gas in g\/L.\r\n\r\n[reveal-answer q=\"829686\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"829686\"]\r\n<ul>\r\n \t<li><em>PV = nRT<\/em><\/li>\r\n \t<li><em>Rearrange to get (mol\/L)<\/em>: [latex]\\frac{n}{v}=\\frac{P}{RT}[\/latex]<\/li>\r\n \t<li><em>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[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<\/ul>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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\r\n[reveal-answer q=\"141474\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"141474\"]\r\n<p style=\"text-align: center;\">[latex]\\rho =\\frac{P\\mathcal{M}}{RT}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]0.0847\\text{g\/L}=760\\cancel{\\text{torr}}\\times \\frac{1\\cancel{\\text{atm}}}{760\\cancel{\\text{torr}}}\\times \\frac{\\mathcal{M}}{\\text{0.0821 L}\\cancel{\\text{atm}}\\text{\/mol K}}\\times \\text{290 K}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">\u2133 = 2.02 g\/mol; therefore, the gas must be hydrogen (H<sub>2<\/sub>, 2.02 g\/mol)<\/p>\r\nWe 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.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Empirical\/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas<\/h3>\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\r\n[reveal-answer q=\"812886\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"812886\"]\r\n\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<p style=\"text-align: center;\">[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]\r\n[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]<\/p>\r\nEmpirical formula is CH<sub>2<\/sub> [empirical mass (EM) of 14.03 g\/empirical unit].\r\n\r\nNext, use the density equation related to the ideal gas law to determine the molar mass:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n\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).\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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\r\n[reveal-answer q=\"513758\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"513758\"]Empirical formula, CH; Molecular formula, C<sub>2<\/sub>H<sub>2<\/sub>[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Molar Mass of a Gas<\/h2>\r\nAnother 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>m<\/em>, to its amount in moles, <em>n<\/em>\r\n<p style=\"text-align: center;\">[latex]\\mathcal{M}=\\frac{\\text{grams of substance}}{\\text{moles of substance}}=\\frac{m}{n}[\/latex]<\/p>\r\nThe ideal gas equation can be rearranged to isolate <em>n:<\/em>\r\n<p style=\"text-align: center;\">[latex]n=\\frac{PV}{RT}[\/latex]<\/p>\r\nand then combined with the molar mass equation to yield:\r\n<p style=\"text-align: center;\">[latex]\\mathcal{M}=\\frac{mRT}{PV}[\/latex]<\/p>\r\nThis equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0Determining the Molar Mass of a Volatile Liquid<\/h3>\r\nThe approximate molar mass of a volatile liquid can be determined by:\r\n<ol>\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><img 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\" \/><\/figure>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?\r\n\r\n[reveal-answer q=\"878003\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"878003\"]\r\n\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]\u00a0then\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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\r\n[reveal-answer q=\"459498\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"459498\"]124 g\/mol P<sub>4<\/sub>[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Pressure of a Mixture of Gases: Dalton\u2019s Law<\/h2>\r\nUnless 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>partial pressure<\/strong>. This observation is summarized by <strong>Dalton\u2019s law of partial pressures<\/strong>: <em>The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases<\/em>:\r\n<p style=\"text-align: center;\">[latex]{P}_{Total}={P}_{A}+{P}_{B}+{P}_{C}+\\ldots ={\\Sigma}_{\\text{i}}{P}_{\\text{i}}[\/latex]<\/p>\r\nIn the equation <em>P<sub>Total<\/sub><\/em> is the total pressure of a mixture of gases, <em>P<sub>A<\/sub><\/em> is the partial pressure of gas A; <em>P<sub>B<\/sub><\/em> is the partial pressure of gas B; <em>P<sub>C<\/sub><\/em> is the partial pressure of gas C; and so on.\r\n\r\n<figure>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"575\"]<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\" width=\"575\" height=\"258\" \/> 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>The partial pressure of gas A is related to the total pressure of the gas mixture via its <strong>mole fraction<\/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):\r\n<p style=\"text-align: center;\">[latex]{P}_{A}={X}_{A}\\times {P}_{Total}\\text{ where }{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex]<\/p>\r\nwhere <em>P<sub>A<\/sub><\/em>, <em>X<sub>A<\/sub><\/em>, and <em>n<sub>A<\/sub><\/em> are the partial pressure, mole fraction, and number of moles of gas A, respectively, and <em>n<sub>Total<\/sub><\/em> is the number of moles of all components in the mixture.\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0The Pressure of a Mixture of Gases<\/h3>\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<ol>\r\n \t<li>What are the partial pressures of each of the gases?<\/li>\r\n \t<li>What is the total pressure in atmospheres?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"786760\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"786760\"]\r\n\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<p style=\"text-align: center;\">[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]<\/p>\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nThe total pressure is given by the sum of the partial pressures:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/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\r\n[reveal-answer q=\"442708\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"442708\"]1.137 atm[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is another example of this concept, but dealing with mole fraction calculations.\r\n<div class=\"textbox examples\">\r\n<h3>Example 5:\u00a0The Pressure of a Mixture of Gases<\/h3>\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<ol>\r\n \t<li>What are the mole fractions of O<sub>2<\/sub> and N<sub>2<\/sub>O?<\/li>\r\n \t<li>What are the partial pressures of O<sub>2<\/sub> and N<sub>2<\/sub>O?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"715641\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"715641\"]\r\n\r\nThe mole fraction is given by [latex]{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex] and the partial pressure is <em>P<sub>A<\/sub><\/em> = <em>X<sub>A<\/sub><\/em> \u00d7 <em>P<sub>Total<\/sub><\/em>.\r\n\r\nFor O<sub>2<\/sub>,\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=0.252\\times \\text{192 kPa}=\\text{48.4 kPa}[\/latex]<\/p>\r\nFor N<sub>2<\/sub>O,\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=\\left(0.252\\right)\\times \\text{192 kPa}[\/latex]<\/p>\r\n[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=0.252\\times \\text{192 kPa}=\\text{48.4 kPa}[\/latex]\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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\r\n[reveal-answer q=\"952300\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"952300\"]1.87 atm[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Collection of Gases over Water<\/h2>\r\n<figure>\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"400\"]<img 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=\"400\" height=\"342\" \/> 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>vapor pressure of water<\/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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img class=\"\" 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\" width=\"500\" height=\"389\" \/> Figure 4. This graph shows the vapor pressure of water at sea level as a function of temperature.[\/caption]\r\n<table class=\" undefined\"><caption>Table 1. Vapor Pressure of Ice and Water in Various Temperatures at Sea Level<\/caption>\r\n<tbody>\r\n<tr>\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>\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>\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>\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>\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>\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>\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>\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>\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>\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>\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>\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>\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 class=\"textbox examples\">\r\n<h3>Example\u00a06:\u00a0Pressure of a Gas Collected Over Water<\/h3>\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\r\n[reveal-answer q=\"331413\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"331413\"]\r\n\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<p style=\"text-align: center;\">[latex]{P}_{\\text{T}}={P}_{\\text{Ar}}+{P}_{{\\text{H}}_{2}\\text{O}}[\/latex]<\/p>\r\nRearranging this equation to solve for the pressure of argon gives:\r\n<p style=\"text-align: center;\">[latex]{P}_{\\text{Ar}}={P}_{\\text{T}}-{P}_{{\\text{H}}_{2}\\text{O}}[\/latex]<\/p>\r\nThe pressure of water vapor above a sample of liquid water at 26 \u00b0C is 25.2 torr (<a href=\".\/chapter\/water-properties-2\/\">Water Properties<\/a>), so:\r\n<p style=\"text-align: center;\">[latex]{P}_{\\text{Ar}}=750\\text{ torr}-25.2\\text{ torr}=725\\text{ torr}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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\r\n[reveal-answer q=\"328726\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"328726\"]734 torr[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Chemical Stoichiometry and Gases<\/h2>\r\nChemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.\r\n\r\nWe 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.\r\n<h2>Avogadro\u2019s Law Revisited<\/h2>\r\nSometimes 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.\r\n\r\nWe 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)\\rightarrow{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.\r\n\r\nThe 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>.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"630\"]<img class=\"\" 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\" width=\"630\" height=\"314\" \/> 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 class=\"textbox examples\">\r\n<h3>Example 7:\u00a0Reaction of Gases<\/h3>\r\nPropane, C<sub>3<\/sub>H<sub>8<\/sub>(<em>g<\/em>), is used in gas grills to provide the heat for cooking. What volume of O<sub>2<\/sub>(<em>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\r\n[reveal-answer q=\"966323\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"966323\"]\r\n\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<p style=\"text-align: center;\">[latex]\\begin{array}{ccccccc}{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)&amp;+&amp;5{\\text{O}}_{2}\\left(g\\right)&amp;\\longrightarrow&amp;3{\\text{CO}}_{2}\\left(g\\right)&amp;+&amp;4{\\text{H}}_{2}\\text{O}\\left(l\\right)\\\\ \\text{1 volume}&amp;+&amp;\\text{5 volumes}&amp;{}&amp;\\text{3 volumes}&amp;+&amp;\\text{4 volumes}\\end{array}[\/latex]<\/p>\r\nFrom 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>:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nA 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>.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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<p style=\"text-align: center;\">[latex]2{\\text{C}}_{2}{\\text{H}}_{2}+5{\\text{O}}_{2}\\rightarrow 4{\\text{CO}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\r\n[reveal-answer q=\"444294\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"444294\"]3.34 tanks (2.34 \u00d7 10<sup>4<\/sup> L)[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 8:\u00a0Volumes of Reacting Gases<\/h3>\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>g<\/em>), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N<sub>2<\/sub>?\r\n<p style=\"text-align: center;\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightarrow 2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"209868\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"209868\"]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>:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nThe 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.)\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nWhat volume of O<sub>2<\/sub>(<em>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>g<\/em>), measured under the same conditions of temperature and pressure? The products are CO<sub>2<\/sub> and water vapor.\r\n\r\n[reveal-answer q=\"97143\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"97143\"]51.0 L[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 9:\u00a0Volume of Gaseous Product<\/h3>\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<p style=\"text-align: center;\">[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]<\/p>\r\n[reveal-answer q=\"327604\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"327604\"]To convert from the mass of gallium to the volume of H<sub>2<\/sub>(<em>g<\/em>), we need to do something like this:\r\n\r\n<img class=\"aligncenter\" 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\" \/>\r\n\r\nThe first two conversions are:\r\n<p style=\"text-align: center;\">[latex]8.88\\cancel{\\text{g Ga}}\\times \\frac{1\\cancel{\\text{mol Ga}}}{69.723\\cancel{\\text{g Ga}}}\\times \\frac{\\text{3 mol}{\\text{H}}_{2}}{2\\cancel{\\text{mol Ga}}}=0.191{\\text{mol H}}_{2}[\/latex]<\/p>\r\nFinally, we can use the ideal gas law:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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\r\n[reveal-answer q=\"810770\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"810770\"]1.30 \u00d7 10<sup>3<\/sup> L[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Greenhouse Gases and Climate Change<\/h3>\r\nThe 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).\r\n\r\n<figure>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"538\"]<img class=\"\" 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\" width=\"538\" height=\"361\" \/> 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>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).\r\n\r\n<img 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\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox\"><a href=\"https:\/\/www3.epa.gov\/climatechange\/kids\/basics\/today\/greenhouse-effect.html\">Click here to see a 2-minute video from the Environmental Protection Agency <\/a>explaining greenhouse gases and global warming.<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>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\" \/> 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\r\nAtmospheric and climate scientist Susan Solomon (Figure 8) is the author of one of <em>The New York Times<\/em> books of the year (<em>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.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key\u00a0Concepts and Summary<\/h3>\r\nThe 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.\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li><em>P<sub>Total<\/sub><\/em> = <em>P<sub>A<\/sub><\/em> + <em>P<sub>B<\/sub><\/em> + <em>P<sub>C<\/sub><\/em> + \u2026 = \u01a9<sub>i<\/sub><em>P<\/em><sub>i<\/sub><\/li>\r\n \t<li><em>P<sub>A<\/sub><\/em> = <em>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<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>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>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 style=\"list-style-type: lower-alpha;\">\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: \u00a0[latex]2\\text{HgO}\\left(s\\right)\\rightarrow 2\\text{Hg}\\left(l\\right)+{\\text{O}}_{2}\\left(g\\right)[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Outline the steps necessary to answer the following question: What volume of O<sub>2<\/sub> at 23\u00b0 C 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 style=\"list-style-type: lower-alpha;\">\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>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 style=\"list-style-type: lower-alpha;\">\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 style=\"list-style-type: lower-alpha;\">\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 style=\"list-style-type: lower-alpha;\">\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 style=\"list-style-type: lower-alpha;\">\r\n \t<li>What is the total volume of the CO<sub>2<\/sub>(<em>g<\/em>) and H<sub>2<\/sub>O(<em>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>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:\r\n[latex]\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)\\,\\,\\,{\\xrightarrow{\\text{copper catalyst }300^{\\circ}\\text{ C}, 300\\text{ atm}}}\\,\\,\\,{\\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:\r\n[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]\r\nWhat 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?\r\n[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[reveal-answer q=\"588380\"]Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"588380\"]\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\n6. From the ideal gas law, <em>PV = nRT<\/em>, set [latex]n=\\frac{\\text{mass}}{\\text{molar mass}}[\/latex] and solve the molar mass. [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] [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] \u2133<sub>phosphorous<\/sub> = 30.97376 g\/mol\r\n\u2133<sub>fluorine<\/sub> = 18.998403 g\/mol\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrl}\\text{molecular formula:}&amp;\\text{ phosphorous: }&amp;30.97376\\\\&amp;{}\\text{flourine:}&amp;\\underline{3\\left(18.998403\\right)}\\\\{}&amp;{}&amp;87.968969\\end{array}[\/latex]<\/p>\r\nThe molecular formula is PF<sub>3<\/sub>.\r\n\r\nTo 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.\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<p style=\"text-align: center;\">[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]\r\n[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]\r\n[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]<\/p>\r\nTotal moles = 7.953 + 25.157 + 174.202 = 207.312 mol\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n12. Since these are percentages of the total pressure, the partial pressure can be calculated as follows:\r\n<ul>\r\n \t<li>CH<sup>4<\/sup>: 90% of 307.2 kPa = 0.900 \u00d7 307.2 = 276 kPa<\/li>\r\n \t<li>C<sup>2<\/sup> H<sup>6<\/sup>: 8.9% of 307.2 kPa = 0.089 \u00d7 307.2 = 27 kPa<\/li>\r\n \t<li>C<sup>3<\/sup> H<sup>8<\/sup>: 1.1% of 307.2 kPa = 0.011 \u00d7 307.2 = 3.4 kPa<\/li>\r\n<\/ul>\r\n14.\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.\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<p style=\"text-align: center;\"><em>P<\/em><sub>T<\/sub> = <em>P<\/em><sub>gas<\/sub> + <em>P<\/em><sub>water<\/sub><\/p>\r\nRearrangement gives:\u00a0<em>P<\/em><sub>T<\/sub> \u2013<em> P<\/em><sub>water =<\/sub><em>P<\/em><sub>gas<\/sub>\r\n<p style=\"text-align: center;\">756 torr \u2013 15.5 torr = 740 torr<\/p>\r\n18. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>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.<\/li>\r\n \t<li>[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]<\/li>\r\n<\/ol>\r\n<em>PV<\/em> = <em>nRT<\/em>\r\n<em>P<\/em> = 0.975 atm\r\n<em>T<\/em> = (23.0 + 273.15) K\r\n<p style=\"text-align: center;\">[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}}}=0.308\\text{ L}[\/latex]<\/p>\r\n20. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>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.<\/li>\r\n \t<li>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\r\n[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}}\\times \\frac{\\text{4 mol}{\\text{H}}_{2}}{\\text{1 mol}{\\text{CCl}}_{2}{\\text{F}}_{2}}=3.308\\times {10}^{4}\\text{mol}[\/latex]\r\n[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]<\/li>\r\n<\/ol>\r\n22.\u00a0The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>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.<\/li>\r\n \t<li>[latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightarrow\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]\r\n[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}}\\times \\frac{\\text{44.01 g}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{CO}}_{2}}\\times \\frac{\\text{1 mol}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{CaCO}}_{2}}=4.397\\times {10}^{5}\\text{g}[\/latex]\r\n[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]\r\n[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]<\/li>\r\n<\/ol>\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\r\nFrom 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,\r\n<p style=\"text-align: center;\">[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]\r\n[latex]V\\left({\\text{O}}_{2}\\right)=\\frac{\\text{12.00 L}\\times 7}{2}=\\text{42.00 L}[\/latex]<\/p>\r\n26.\u00a0The answers are as follows:\r\n\r\n(a) The scheme to solve this problem is:\r\n<p style=\"text-align: center;\">[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]\r\n[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]<\/p>\r\n(b) First, calculate the mol H<sub>2<\/sub>O produced:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nSecond, calculate the pressure of H<sub>2<\/sub>O:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n28. First, we must write a balanced equation to establish the stoichiometry of the reaction:\r\n\r\n[latex]2{\\text{BaO}}_{2}\\rightarrow 2\\text{BaO}+{\\text{O}}_{2}[\/latex]\r\nWe are given the mass of BaO<sub>2<\/sub> that decomposes, so the scheme for solving this problem will be:\r\n<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\" \/>\r\nMass (BaO<sub>2<\/sub>) = 137.33 + 2(15.9994) = 169.33 g\/mol\r\n<p style=\"text-align: center;\">[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}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{\\text{2 mol}{\\text{BaO}}_{2}}=\\text{0.3830 mol}{\\text{O}}_{2}[\/latex]\r\n[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]<\/p>\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\n3C<sub>2<\/sub>H<sub>4<\/sub> produces 3C<sub>2<\/sub>H<sub>5<\/sub>OH, giving a 1:1 ratio:\r\n<p style=\"text-align: center;\">[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}}}\\times \\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]<\/p>\r\n<em>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\r\n\r\n32. The reaction is:\r\n<p style=\"text-align: center;\">[latex]{\\text{XeF}}_{x}+\\frac{x}{2}{\\text{H}}_{2}\\rightarrow\\text{Xe}+x\\text{HF}[\/latex]<\/p>\r\nThe pressure of H<sub>2<\/sub> that reacts is\u00a048 torr \u2013 24 torr = 24 torr\r\n\r\nThe 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:\u00a0[latex]\\frac{x}{2}=1[\/latex] and <em>x<\/em> = 2\r\n\r\nThe formula for the xenon compound is XeF<sub>2<\/sub>.\r\n\r\nImmediately after the H<sub>2<\/sub> is added (before the reaction):\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nAfter the reaction:\r\n<p style=\"text-align: center;\">[latex]{P}_{\\text{Xe}}=\\text{24 torr}\\text{ }\\left(\\text{1 mol}{\\text{XeF}}_{\\text{x}}\\rightarrow\\text{1 mol Xe}\\right)[\/latex]<\/p>\r\nAnd the partial pressure of unreacted H<sub>2<\/sub> is:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>Dalton\u2019s law of partial pressures<\/strong>:\u00a0total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.\r\n\r\n<strong>mole fraction<\/strong>:\u00a0concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components\r\n\r\n<strong>partial pressure<\/strong>:\u00a0pressure exerted by an individual gas in a mixture\r\n\r\n<strong>vapor pressure of water<\/strong>:\u00a0pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature","rendered":"<div class=\"textbox learning-objectives\">\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>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 Lavoisier, 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>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<h2>Density of a Gas<\/h2>\n<p>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>PV<\/em> = <em>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 class=\"textbox examples\">\n<h3>Example 1:\u00a0Derivation of a Density Formula from the Ideal Gas Law<\/h3>\n<p>Use <em>PV<\/em> = <em>nRT<\/em> to derive a formula for the density of gas in g\/L.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q829686\">Show Answer<\/span><\/p>\n<div id=\"q829686\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li><em>PV = nRT<\/em><\/li>\n<li><em>Rearrange to get (mol\/L)<\/em>: [latex]\\frac{n}{v}=\\frac{P}{RT}[\/latex]<\/li>\n<li><em>Multiply each side of the equation by the molar mass, \u2133.<\/em> When moles are multiplied by \u2133 in g\/mol, g are obtained:<br \/>\n[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<\/ul>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q141474\">Show Answer<\/span><\/p>\n<div id=\"q141474\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\rho =\\frac{P\\mathcal{M}}{RT}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]0.0847\\text{g\/L}=760\\cancel{\\text{torr}}\\times \\frac{1\\cancel{\\text{atm}}}{760\\cancel{\\text{torr}}}\\times \\frac{\\mathcal{M}}{\\text{0.0821 L}\\cancel{\\text{atm}}\\text{\/mol K}}\\times \\text{290 K}[\/latex]<\/p>\n<p style=\"text-align: center;\">\u2133 = 2.02 g\/mol; therefore, the gas must be hydrogen (H<sub>2<\/sub>, 2.02 g\/mol)<\/p>\n<p>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>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Empirical\/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas<\/h3>\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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q812886\">Show Answer<\/span><\/p>\n<div id=\"q812886\" class=\"hidden-answer\" style=\"display: none\">\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<p style=\"text-align: center;\">[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]<br \/>\n[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]<\/p>\n<p>Empirical formula is CH<sub>2<\/sub> [empirical mass (EM) of 14.03 g\/empirical unit].<\/p>\n<p>Next, use the density equation related to the ideal gas law to determine the molar mass:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>\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<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q513758\">Show Answer<\/span><\/p>\n<div id=\"q513758\" class=\"hidden-answer\" style=\"display: none\">Empirical formula, CH; Molecular formula, C<sub>2<\/sub>H<sub>2<\/sub><\/div>\n<\/div>\n<\/div>\n<h2>Molar Mass of a Gas<\/h2>\n<p>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>m<\/em>, to its amount in moles, <em>n<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\mathcal{M}=\\frac{\\text{grams of substance}}{\\text{moles of substance}}=\\frac{m}{n}[\/latex]<\/p>\n<p>The ideal gas equation can be rearranged to isolate <em>n:<\/em><\/p>\n<p style=\"text-align: center;\">[latex]n=\\frac{PV}{RT}[\/latex]<\/p>\n<p>and then combined with the molar mass equation to yield:<\/p>\n<p style=\"text-align: center;\">[latex]\\mathcal{M}=\\frac{mRT}{PV}[\/latex]<\/p>\n<p>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 class=\"textbox examples\">\n<h3>Example 3:\u00a0Determining the Molar Mass of a Volatile Liquid<\/h3>\n<p>The approximate molar mass of a volatile liquid can be determined by:<\/p>\n<ol>\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><img decoding=\"async\" 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=\"image\" \/><\/figure>\n<p>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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q878003\">Show Answer<\/span><\/p>\n<div id=\"q878003\" class=\"hidden-answer\" style=\"display: none\">\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]\u00a0then<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q459498\">Show Answer<\/span><\/p>\n<div id=\"q459498\" class=\"hidden-answer\" style=\"display: none\">124 g\/mol P<sub>4<\/sub><\/div>\n<\/div>\n<\/div>\n<h2>The Pressure of a Mixture of Gases: Dalton\u2019s Law<\/h2>\n<p>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>partial pressure<\/strong>. This observation is summarized by <strong>Dalton\u2019s law of partial pressures<\/strong>: <em>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<p style=\"text-align: center;\">[latex]{P}_{Total}={P}_{A}+{P}_{B}+{P}_{C}+\\ldots ={\\Sigma}_{\\text{i}}{P}_{\\text{i}}[\/latex]<\/p>\n<p>In the equation <em>P<sub>Total<\/sub><\/em> is the total pressure of a mixture of gases, <em>P<sub>A<\/sub><\/em> is the partial pressure of gas A; <em>P<sub>B<\/sub><\/em> is the partial pressure of gas B; <em>P<sub>C<\/sub><\/em> is the partial pressure of gas C; and so on.<\/p>\n<figure>\n<div style=\"width: 585px\" class=\"wp-caption aligncenter\"><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\" width=\"575\" height=\"258\" alt=\"image\" \/><\/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>The partial pressure of gas A is related to the total pressure of the gas mixture via its <strong>mole fraction<\/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<p style=\"text-align: center;\">[latex]{P}_{A}={X}_{A}\\times {P}_{Total}\\text{ where }{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex]<\/p>\n<p>where <em>P<sub>A<\/sub><\/em>, <em>X<sub>A<\/sub><\/em>, and <em>n<sub>A<\/sub><\/em> are the partial pressure, mole fraction, and number of moles of gas A, respectively, and <em>n<sub>Total<\/sub><\/em> is the number of moles of all components in the mixture.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0The Pressure of a Mixture of Gases<\/h3>\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<ol>\n<li>What are the partial pressures of each of the gases?<\/li>\n<li>What is the total pressure in atmospheres?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q786760\">Show Answer<\/span><\/p>\n<div id=\"q786760\" class=\"hidden-answer\" style=\"display: none\">\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<p style=\"text-align: center;\">[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]<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>The total pressure is given by the sum of the partial pressures:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q442708\">Show Answer<\/span><\/p>\n<div id=\"q442708\" class=\"hidden-answer\" style=\"display: none\">1.137 atm<\/div>\n<\/div>\n<\/div>\n<p>Here is another example of this concept, but dealing with mole fraction calculations.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 5:\u00a0The Pressure of a Mixture of Gases<\/h3>\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<ol>\n<li>What are the mole fractions of O<sub>2<\/sub> and N<sub>2<\/sub>O?<\/li>\n<li>What are the partial pressures of O<sub>2<\/sub> and N<sub>2<\/sub>O?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q715641\">Show Answer<\/span><\/p>\n<div id=\"q715641\" class=\"hidden-answer\" style=\"display: none\">\n<p>The mole fraction is given by [latex]{X}_{A}=\\frac{{n}_{A}}{{n}_{Total}}[\/latex] and the partial pressure is <em>P<sub>A<\/sub><\/em> = <em>X<sub>A<\/sub><\/em> \u00d7 <em>P<sub>Total<\/sub><\/em>.<\/p>\n<p>For O<sub>2<\/sub>,<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=0.252\\times \\text{192 kPa}=\\text{48.4 kPa}[\/latex]<\/p>\n<p>For N<sub>2<\/sub>O,<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=\\left(0.252\\right)\\times \\text{192 kPa}[\/latex]<\/p>\n<p>[latex]{P}_{{O}_{2}}={X}_{{O}_{2}}\\times {P}_{Total}=0.252\\times \\text{192 kPa}=\\text{48.4 kPa}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q952300\">Show Answer<\/span><\/p>\n<div id=\"q952300\" class=\"hidden-answer\" style=\"display: none\">1.87 atm<\/div>\n<\/div>\n<\/div>\n<h2>Collection of Gases over Water<\/h2>\n<figure>\n<div style=\"width: 410px\" 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\/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=\"400\" height=\"342\" \/><\/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>vapor pressure of water<\/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.<\/p>\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><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\/23212045\/CNX_Chem_09_03_WaterVapor21.jpg\" width=\"500\" height=\"389\" alt=\"image\" \/><\/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 class=\"undefined\">\n<caption>Table 1. Vapor Pressure of Ice and Water in Various Temperatures at Sea Level<\/caption>\n<tbody>\n<tr>\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>\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>\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>\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>\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>\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>\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>\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>\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>\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>\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>\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>\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 class=\"textbox examples\">\n<h3>Example\u00a06:\u00a0Pressure of a Gas Collected Over Water<\/h3>\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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q331413\">Show Answer<\/span><\/p>\n<div id=\"q331413\" class=\"hidden-answer\" style=\"display: none\">\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<p style=\"text-align: center;\">[latex]{P}_{\\text{T}}={P}_{\\text{Ar}}+{P}_{{\\text{H}}_{2}\\text{O}}[\/latex]<\/p>\n<p>Rearranging this equation to solve for the pressure of argon gives:<\/p>\n<p style=\"text-align: center;\">[latex]{P}_{\\text{Ar}}={P}_{\\text{T}}-{P}_{{\\text{H}}_{2}\\text{O}}[\/latex]<\/p>\n<p>The pressure of water vapor above a sample of liquid water at 26 \u00b0C is 25.2 torr (<a href=\".\/chapter\/water-properties-2\/\">Water Properties<\/a>), so:<\/p>\n<p style=\"text-align: center;\">[latex]{P}_{\\text{Ar}}=750\\text{ torr}-25.2\\text{ torr}=725\\text{ torr}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q328726\">Show Answer<\/span><\/p>\n<div id=\"q328726\" class=\"hidden-answer\" style=\"display: none\">734 torr<\/div>\n<\/div>\n<\/div>\n<h2>Chemical Stoichiometry and Gases<\/h2>\n<p>Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.<\/p>\n<p>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<h2>Avogadro\u2019s Law Revisited<\/h2>\n<p>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>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)\\rightarrow{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>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>.<\/p>\n<div style=\"width: 640px\" class=\"wp-caption aligncenter\"><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\/23212046\/CNX_Chem_09_03_Ammonia1.jpg\" width=\"630\" height=\"314\" alt=\"image\" \/><\/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 class=\"textbox examples\">\n<h3>Example 7:\u00a0Reaction of Gases<\/h3>\n<p>Propane, C<sub>3<\/sub>H<sub>8<\/sub>(<em>g<\/em>), is used in gas grills to provide the heat for cooking. What volume of O<sub>2<\/sub>(<em>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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q966323\">Show Answer<\/span><\/p>\n<div id=\"q966323\" class=\"hidden-answer\" style=\"display: none\">\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<p style=\"text-align: center;\">[latex]\\begin{array}{ccccccc}{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)&+&5{\\text{O}}_{2}\\left(g\\right)&\\longrightarrow&3{\\text{CO}}_{2}\\left(g\\right)&+&4{\\text{H}}_{2}\\text{O}\\left(l\\right)\\\\ \\text{1 volume}&+&\\text{5 volumes}&{}&\\text{3 volumes}&+&\\text{4 volumes}\\end{array}[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[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]<\/p>\n<p>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<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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<p style=\"text-align: center;\">[latex]2{\\text{C}}_{2}{\\text{H}}_{2}+5{\\text{O}}_{2}\\rightarrow 4{\\text{CO}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q444294\">Show Answer<\/span><\/p>\n<div id=\"q444294\" class=\"hidden-answer\" style=\"display: none\">3.34 tanks (2.34 \u00d7 10<sup>4<\/sup> L)<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 8:\u00a0Volumes of Reacting Gases<\/h3>\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>g<\/em>), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N<sub>2<\/sub>?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightarrow 2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q209868\">Show Answer<\/span><\/p>\n<div id=\"q209868\" class=\"hidden-answer\" style=\"display: none\">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<p style=\"text-align: center;\">[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]<\/p>\n<p>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<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>What volume of O<sub>2<\/sub>(<em>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>g<\/em>), measured under the same conditions of temperature and pressure? The products are CO<sub>2<\/sub> and water vapor.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q97143\">Show Answer<\/span><\/p>\n<div id=\"q97143\" class=\"hidden-answer\" style=\"display: none\">51.0 L<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 9:\u00a0Volume of Gaseous Product<\/h3>\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<p style=\"text-align: center;\">[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]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q327604\">Show Answer<\/span><\/p>\n<div id=\"q327604\" class=\"hidden-answer\" style=\"display: none\">To convert from the mass of gallium to the volume of H<sub>2<\/sub>(<em>g<\/em>), we need to do something like this:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" 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=\"image\" \/><\/p>\n<p>The first two conversions are:<\/p>\n<p style=\"text-align: center;\">[latex]8.88\\cancel{\\text{g Ga}}\\times \\frac{1\\cancel{\\text{mol Ga}}}{69.723\\cancel{\\text{g Ga}}}\\times \\frac{\\text{3 mol}{\\text{H}}_{2}}{2\\cancel{\\text{mol Ga}}}=0.191{\\text{mol H}}_{2}[\/latex]<\/p>\n<p>Finally, we can use the ideal gas law:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q810770\">Show Answer<\/span><\/p>\n<div id=\"q810770\" class=\"hidden-answer\" style=\"display: none\">1.30 \u00d7 10<sup>3<\/sup> L<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Greenhouse Gases and Climate Change<\/h3>\n<p>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>\n<div style=\"width: 548px\" class=\"wp-caption aligncenter\"><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\/23212049\/CNX_Chem_09_03_GlobalWarming1.jpg\" width=\"538\" height=\"361\" alt=\"image\" \/><\/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>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).<\/p>\n<p><img decoding=\"async\" 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=\"image\" \/><\/p>\n<\/div>\n<div class=\"textbox\"><a href=\"https:\/\/www3.epa.gov\/climatechange\/kids\/basics\/today\/greenhouse-effect.html\">Click here to see a 2-minute video from the Environmental Protection Agency <\/a>explaining greenhouse gases and global warming.<\/div>\n<div class=\"textbox shaded\">\n<h3>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\" \/><\/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>Atmospheric and climate scientist Susan Solomon (Figure 8) is the author of one of <em>The New York Times<\/em> books of the year (<em>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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key\u00a0Concepts and Summary<\/h3>\n<p>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<h4>Key Equations<\/h4>\n<ul>\n<li><em>P<sub>Total<\/sub><\/em> = <em>P<sub>A<\/sub><\/em> + <em>P<sub>B<\/sub><\/em> + <em>P<sub>C<\/sub><\/em> + \u2026 = \u01a9<sub>i<\/sub><em>P<\/em><sub>i<\/sub><\/li>\n<li><em>P<sub>A<\/sub><\/em> = <em>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<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>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>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 style=\"list-style-type: lower-alpha;\">\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: \u00a0[latex]2\\text{HgO}\\left(s\\right)\\rightarrow 2\\text{Hg}\\left(l\\right)+{\\text{O}}_{2}\\left(g\\right)[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Outline the steps necessary to answer the following question: What volume of O<sub>2<\/sub> at 23\u00b0 C 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 style=\"list-style-type: lower-alpha;\">\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>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 style=\"list-style-type: lower-alpha;\">\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 style=\"list-style-type: lower-alpha;\">\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 style=\"list-style-type: lower-alpha;\">\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 style=\"list-style-type: lower-alpha;\">\n<li>What is the total volume of the CO<sub>2<\/sub>(<em>g<\/em>) and H<sub>2<\/sub>O(<em>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>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:<br \/>\n[latex]\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)\\,\\,\\,{\\xrightarrow{\\text{copper catalyst }300^{\\circ}\\text{ C}, 300\\text{ atm}}}\\,\\,\\,{\\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:<br \/>\n[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]<br \/>\nWhat 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?<br \/>\n[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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q588380\">Selected Answers<\/span><\/p>\n<div id=\"q588380\" class=\"hidden-answer\" style=\"display: none\">\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]<\/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>6. From the ideal gas law, <em>PV = nRT<\/em>, set [latex]n=\\frac{\\text{mass}}{\\text{molar mass}}[\/latex] and solve the molar mass. [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] [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] \u2133<sub>phosphorous<\/sub> = 30.97376 g\/mol<br \/>\n\u2133<sub>fluorine<\/sub> = 18.998403 g\/mol<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrl}\\text{molecular formula:}&\\text{ phosphorous: }&30.97376\\\\&{}\\text{flourine:}&\\underline{3\\left(18.998403\\right)}\\\\{}&{}&87.968969\\end{array}[\/latex]<\/p>\n<p>The molecular formula is PF<sub>3<\/sub>.<\/p>\n<p>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.<\/p>\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<p style=\"text-align: center;\">[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]<br \/>\n[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]<br \/>\n[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]<\/p>\n<p>Total moles = 7.953 + 25.157 + 174.202 = 207.312 mol<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>12. Since these are percentages of the total pressure, the partial pressure can be calculated as follows:<\/p>\n<ul>\n<li>CH<sup>4<\/sup>: 90% of 307.2 kPa = 0.900 \u00d7 307.2 = 276 kPa<\/li>\n<li>C<sup>2<\/sup> H<sup>6<\/sup>: 8.9% of 307.2 kPa = 0.089 \u00d7 307.2 = 27 kPa<\/li>\n<li>C<sup>3<\/sup> H<sup>8<\/sup>: 1.1% of 307.2 kPa = 0.011 \u00d7 307.2 = 3.4 kPa<\/li>\n<\/ul>\n<p>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.<\/p>\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<p style=\"text-align: center;\"><em>P<\/em><sub>T<\/sub> = <em>P<\/em><sub>gas<\/sub> + <em>P<\/em><sub>water<\/sub><\/p>\n<p>Rearrangement gives:\u00a0<em>P<\/em><sub>T<\/sub> \u2013<em> P<\/em><sub>water =<\/sub><em>P<\/em><sub>gas<\/sub><\/p>\n<p style=\"text-align: center;\">756 torr \u2013 15.5 torr = 740 torr<\/p>\n<p>18. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>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.<\/li>\n<li>[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]<\/li>\n<\/ol>\n<p><em>PV<\/em> = <em>nRT<\/em><br \/>\n<em>P<\/em> = 0.975 atm<br \/>\n<em>T<\/em> = (23.0 + 273.15) K<\/p>\n<p style=\"text-align: center;\">[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}}}=0.308\\text{ L}[\/latex]<\/p>\n<p>20. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>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.<\/li>\n<li>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<br \/>\n[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}}\\times \\frac{\\text{4 mol}{\\text{H}}_{2}}{\\text{1 mol}{\\text{CCl}}_{2}{\\text{F}}_{2}}=3.308\\times {10}^{4}\\text{mol}[\/latex]<br \/>\n[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]<\/li>\n<\/ol>\n<p>22.\u00a0The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>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.<\/li>\n<li>[latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightarrow\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<br \/>\n[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}}\\times \\frac{\\text{44.01 g}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{CO}}_{2}}\\times \\frac{\\text{1 mol}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{CaCO}}_{2}}=4.397\\times {10}^{5}\\text{g}[\/latex]<br \/>\n[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]<br \/>\n[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]<\/li>\n<\/ol>\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<p>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,<\/p>\n<p style=\"text-align: center;\">[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]<br \/>\n[latex]V\\left({\\text{O}}_{2}\\right)=\\frac{\\text{12.00 L}\\times 7}{2}=\\text{42.00 L}[\/latex]<\/p>\n<p>26.\u00a0The answers are as follows:<\/p>\n<p>(a) The scheme to solve this problem is:<\/p>\n<p style=\"text-align: center;\">[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]<br \/>\n[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]<\/p>\n<p>(b) First, calculate the mol H<sub>2<\/sub>O produced:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>Second, calculate the pressure of H<sub>2<\/sub>O:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>28. First, we must write a balanced equation to establish the stoichiometry of the reaction:<\/p>\n<p>[latex]2{\\text{BaO}}_{2}\\rightarrow 2\\text{BaO}+{\\text{O}}_{2}[\/latex]<br \/>\nWe are given the mass of BaO<sub>2<\/sub> that decomposes, so the scheme for solving this problem will be:<br \/>\n<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=\"image\" \/><br \/>\nMass (BaO<sub>2<\/sub>) = 137.33 + 2(15.9994) = 169.33 g\/mol<\/p>\n<p style=\"text-align: center;\">[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}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{\\text{2 mol}{\\text{BaO}}_{2}}=\\text{0.3830 mol}{\\text{O}}_{2}[\/latex]<br \/>\n[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]<\/p>\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.<br \/>\n3C<sub>2<\/sub>H<sub>4<\/sub> produces 3C<sub>2<\/sub>H<sub>5<\/sub>OH, giving a 1:1 ratio:<\/p>\n<p style=\"text-align: center;\">[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}}}\\times \\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]<\/p>\n<p><em>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<\/p>\n<p>32. The reaction is:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{XeF}}_{x}+\\frac{x}{2}{\\text{H}}_{2}\\rightarrow\\text{Xe}+x\\text{HF}[\/latex]<\/p>\n<p>The pressure of H<sub>2<\/sub> that reacts is\u00a048 torr \u2013 24 torr = 24 torr<\/p>\n<p>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:\u00a0[latex]\\frac{x}{2}=1[\/latex] and <em>x<\/em> = 2<\/p>\n<p>The formula for the xenon compound is XeF<sub>2<\/sub>.<\/p>\n<p>Immediately after the H<sub>2<\/sub> is added (before the reaction):<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>After the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{P}_{\\text{Xe}}=\\text{24 torr}\\text{ }\\left(\\text{1 mol}{\\text{XeF}}_{\\text{x}}\\rightarrow\\text{1 mol Xe}\\right)[\/latex]<\/p>\n<p>And the partial pressure of unreacted H<sub>2<\/sub> is:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>Dalton\u2019s law of partial pressures<\/strong>:\u00a0total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.<\/p>\n<p><strong>mole fraction<\/strong>:\u00a0concentration 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><strong>partial pressure<\/strong>:\u00a0pressure exerted by an individual gas in a mixture<\/p>\n<p><strong>vapor pressure of water<\/strong>:\u00a0pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature<\/p>\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-fmcc-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2102","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":31,"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2102\/revisions"}],"predecessor-version":[{"id":6016,"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2102\/revisions\/6016"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/pressbooks\/v2\/parts\/3001"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2102\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/wp\/v2\/media?parent=2102"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapter-type?post=2102"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/wp\/v2\/contributor?post=2102"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-fmcc-chem-atoms-first\/wp-json\/wp\/v2\/license?post=2102"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}