{"id":1457,"date":"2018-08-16T04:11:46","date_gmt":"2018-08-16T04:11:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/?post_type=chapter&#038;p=1457"},"modified":"2023-11-21T19:18:44","modified_gmt":"2023-11-21T19:18:44","slug":"8-4-applications-of-the-ideal-gas-law","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/8-4-applications-of-the-ideal-gas-law\/","title":{"raw":"8.4 Applications of the Ideal Gas Law","rendered":"8.4 Applications of the Ideal Gas Law"},"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 molar masses<\/li>\r\n \t<li>Perform stoichiometric calculations involving gaseous substances<\/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>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]\\large\\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]\\large 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]\\large\\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 1: Determining 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 Figure 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>\r\nUsing 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]\\large\\mathcal{M}=\\frac{m}{n}[\/latex] and [latex]n=\\frac{PV}{RT},[\/latex] substituting and rearranging gives [latex]\\mathcal{M}=\\frac{mRT}{PV},[\/latex] then\r\n<p style=\"text-align: center;\">[latex]\\large\\mathcal{M}=\\frac{mRT}{PV}=\\frac{\\left(0.494\\cancel{\\text{ g}}\\right){(0.0821\\frac{\\cancel{\\text{L}}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\text{mol}})}\\left(372.8\\cancel{\\text{ K}}\\right)}{(0.976\\cancel{\\text{ atm}}){(0.129\\cancel{\\text{ 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 is the molar mass 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>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; 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<div class=\"textbox examples\">\r\n<h3>Example 2: Gas Stoichiometry<\/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\r\n[latex]\\large2\\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]\r\n\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\nThe first two conversions are:\r\n\r\n[latex]\\large8.88\\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]\r\n\r\nFinally, we can use the ideal gas law:\r\n\r\n[latex]\\large {V}=\\frac{nRT}{P}=\\frac{\\left(0.191\\cancel{\\text{ mol}}\\right){(0.0821\\frac{\\text{L}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\cancel{\\text{mol}}})}\\left(300\\cancel{\\text{ K}}\\right)}{(0.951\\cancel{\\text{ atm}})}=4.94\\text{ L}[\/latex]\r\n\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 1.00 kg of sulfur in oxygen?\r\n\r\n[latex]\\large\\text{S}_{8}\\left(s\\right)+\\text{8 O}_{2}\\left(g\\right)\\rightarrow {\\text{8 SO}}_{2}\\left(g\\right)[\/latex]\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 examples\">\r\n<h3>Example 3: Gas Stoichiometry<\/h3>\r\n<p id=\"ball-ch06_s05_p22\" class=\"para\">How many grams of Zn is required in the following reaction to produce 19.6 L of hydrogen gas at 299 K and 1.07 atm?<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large \\text{Zn}\\left(s\\right)+2\\text{ HCl}\\left(aq\\right)\\rightarrow 2{\\text{ ZnCl}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/span><\/span>\r\n\r\n[reveal-answer q=\"966323\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"966323\"]\r\n<p id=\"ball-ch06_s05_p23\" class=\"para\">From the question, we know we are looking for grams of Zn, therefore at some point we will need the molar mass of Zn (65.41 g\/mol). We are given 19.6 L of H<sub>2<\/sub>, the pressure (1.07 atm), and temperature (299 K).\u00a0 We will also need a mole to mole ration between Zn and H<sub>2<\/sub>, since we are essentially converting units of H<sub>2<\/sub> (19.6 L) to units of Zn (? g). Due to the pressure and temperature being given we will need the ideal gas law (PV = nRT).\u00a0 Once you realize you will be using the ideal gas law, identify how you will be using it based off the information provided.\u00a0 For PV = nRT, we have pressure, temperature, and volume given (we also know that R is the gas constant), therefore the only term we do not know is n (moles of gas).\u00a0 We now have all the pieces we need to solve the problem, so where do we start?<\/p>\r\nWe start by using the ideal gas law to solve for moles of H<sub>2<\/sub>. Once we have moles of H<sub>2<\/sub>, we can then convert it to moles of Zn by using the mole to mole ratio from the chemical equation, and then to grams of Zn using the molar mass of Zn.\r\n\r\nUse ideal gas law to determine moles of hydrogen gas:\r\n<p style=\"text-align: center;\">[latex]\\Large {n}=\\frac{PV}{RT}=\\frac{\\left(1.07\\cancel{\\text{ atm}}\\right)\\left(19.6\\cancel{\\text{ L}}\\right)}{(0.0821\\frac{\\cancel{\\text{ L}}\\cdot\\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\text{mol}})(299\\cancel{\\text{ K}})}=0.853\\text{ mol H}_{2}[\/latex]<\/p>\r\n<p id=\"ball-ch06_s05_p25\" class=\"para\">All the units cancel except for moles, which means <span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">n<\/em> = 0.853 moles H<sub>2<\/sub>.<\/span><\/span><\/p>\r\n<p id=\"ball-ch06_s05_p24\" class=\"para\">Now that we know the number of moles of gas, we can use stoichiometry to find grams of Zn:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large 0.853\\cancel{\\text{ mol H}_{2}}\\times \\frac{1\\cancel{\\text{ mol Zn}}}{1\\cancel{\\text{ mol H}_{2}}}\\times \\frac{\\text{65.41 g}{\\text{ Zn}}}{1\\cancel{\\text{ mol Zn}}}=55.8{\\text{ g Zn}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\n<p id=\"ball-ch06_s05_p26\" class=\"para\">What pressure of HCl is generated if 3.44 g of Cl<sub class=\"subscript\">2<\/sub> are reacted in 4.55 L at 455 K?<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large {\\text{H}}_{2}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightarrow 2{\\text{ HCl}}\\left(g\\right)[\/latex]<\/span><\/span>\r\n\r\n[reveal-answer q=\"444294\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"444294\"]0.796 atm[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts 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 molar masses. The ideal gas law can also be used in relation with stoichiometry.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\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>Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO: [latex]2\\text{ HgO}\\left(s\\right)\\rightarrow 2\\text{ Hg}\\left(l\\right)+{\\text{ O}}_{2}\\left(g\\right)[\/latex]\r\n<ol 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:\u00a0 [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 \u00b0C 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<\/ol>\r\n[reveal-answer q=\"588380\"]Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"588380\"]\r\n\r\n1. From the ideal gas law, <em>PV = nRT<\/em>, set [latex]\\large 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\n3. 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]\\large\\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]\\large 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\n5. 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]\\large\\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]\\large 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\n7. The 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]\\large{\\text{CaCO}}_{3}\\left(s\\right)\\rightarrow\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]\r\n[latex]\\large\\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]\\large\\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\n9. [latex]\\large 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]\\large\\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\n[\/hidden-answer]\r\n\r\n<\/div>","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 molar masses<\/li>\n<li>Perform stoichiometric calculations involving gaseous substances<\/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>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]\\large\\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]\\large 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]\\large\\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 1: Determining 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 Figure 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]\\large\\mathcal{M}=\\frac{m}{n}[\/latex] and [latex]n=\\frac{PV}{RT},[\/latex] substituting and rearranging gives [latex]\\mathcal{M}=\\frac{mRT}{PV},[\/latex] then<\/p>\n<p style=\"text-align: center;\">[latex]\\large\\mathcal{M}=\\frac{mRT}{PV}=\\frac{\\left(0.494\\cancel{\\text{ g}}\\right){(0.0821\\frac{\\cancel{\\text{L}}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\text{mol}})}\\left(372.8\\cancel{\\text{ K}}\\right)}{(0.976\\cancel{\\text{ atm}}){(0.129\\cancel{\\text{ 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 is the molar mass 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>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; 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<div class=\"textbox examples\">\n<h3>Example 2: Gas Stoichiometry<\/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>[latex]\\large2\\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>The first two conversions are:<\/p>\n<p>[latex]\\large8.88\\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>[latex]\\large {V}=\\frac{nRT}{P}=\\frac{\\left(0.191\\cancel{\\text{ mol}}\\right){(0.0821\\frac{\\text{L}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\cancel{\\text{mol}}})}\\left(300\\cancel{\\text{ K}}\\right)}{(0.951\\cancel{\\text{ atm}})}=4.94\\text{ 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 1.00 kg of sulfur in oxygen?<\/p>\n<p>[latex]\\large\\text{S}_{8}\\left(s\\right)+\\text{8 O}_{2}\\left(g\\right)\\rightarrow {\\text{8 SO}}_{2}\\left(g\\right)[\/latex]<\/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 examples\">\n<h3>Example 3: Gas Stoichiometry<\/h3>\n<p id=\"ball-ch06_s05_p22\" class=\"para\">How many grams of Zn is required in the following reaction to produce 19.6 L of hydrogen gas at 299 K and 1.07 atm?<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large \\text{Zn}\\left(s\\right)+2\\text{ HCl}\\left(aq\\right)\\rightarrow 2{\\text{ ZnCl}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/span><\/span><\/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 id=\"ball-ch06_s05_p23\" class=\"para\">From the question, we know we are looking for grams of Zn, therefore at some point we will need the molar mass of Zn (65.41 g\/mol). We are given 19.6 L of H<sub>2<\/sub>, the pressure (1.07 atm), and temperature (299 K).\u00a0 We will also need a mole to mole ration between Zn and H<sub>2<\/sub>, since we are essentially converting units of H<sub>2<\/sub> (19.6 L) to units of Zn (? g). Due to the pressure and temperature being given we will need the ideal gas law (PV = nRT).\u00a0 Once you realize you will be using the ideal gas law, identify how you will be using it based off the information provided.\u00a0 For PV = nRT, we have pressure, temperature, and volume given (we also know that R is the gas constant), therefore the only term we do not know is n (moles of gas).\u00a0 We now have all the pieces we need to solve the problem, so where do we start?<\/p>\n<p>We start by using the ideal gas law to solve for moles of H<sub>2<\/sub>. Once we have moles of H<sub>2<\/sub>, we can then convert it to moles of Zn by using the mole to mole ratio from the chemical equation, and then to grams of Zn using the molar mass of Zn.<\/p>\n<p>Use ideal gas law to determine moles of hydrogen gas:<\/p>\n<p style=\"text-align: center;\">[latex]\\Large {n}=\\frac{PV}{RT}=\\frac{\\left(1.07\\cancel{\\text{ atm}}\\right)\\left(19.6\\cancel{\\text{ L}}\\right)}{(0.0821\\frac{\\cancel{\\text{ L}}\\cdot\\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\text{mol}})(299\\cancel{\\text{ K}})}=0.853\\text{ mol H}_{2}[\/latex]<\/p>\n<p id=\"ball-ch06_s05_p25\" class=\"para\">All the units cancel except for moles, which means <span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">n<\/em> = 0.853 moles H<sub>2<\/sub>.<\/span><\/span><\/p>\n<p id=\"ball-ch06_s05_p24\" class=\"para\">Now that we know the number of moles of gas, we can use stoichiometry to find grams of Zn:<\/p>\n<p style=\"text-align: center;\">[latex]\\Large 0.853\\cancel{\\text{ mol H}_{2}}\\times \\frac{1\\cancel{\\text{ mol Zn}}}{1\\cancel{\\text{ mol H}_{2}}}\\times \\frac{\\text{65.41 g}{\\text{ Zn}}}{1\\cancel{\\text{ mol Zn}}}=55.8{\\text{ g Zn}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p id=\"ball-ch06_s05_p26\" class=\"para\">What pressure of HCl is generated if 3.44 g of Cl<sub class=\"subscript\">2<\/sub> are reacted in 4.55 L at 455 K?<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large {\\text{H}}_{2}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightarrow 2{\\text{ HCl}}\\left(g\\right)[\/latex]<\/span><\/span><\/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\">0.796 atm<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts 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 molar masses. The ideal gas law can also be used in relation with stoichiometry.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\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>Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO: [latex]2\\text{ HgO}\\left(s\\right)\\rightarrow 2\\text{ Hg}\\left(l\\right)+{\\text{ O}}_{2}\\left(g\\right)[\/latex]\n<ol 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:\u00a0 [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 \u00b0C 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<\/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\">\n<p>1. From the ideal gas law, <em>PV = nRT<\/em>, set [latex]\\large 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>3. 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]\\large\\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]\\large 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>5. 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]\\large\\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]\\large 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>7. The 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]\\large{\\text{CaCO}}_{3}\\left(s\\right)\\rightarrow\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<br \/>\n[latex]\\large\\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]\\large\\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>9. [latex]\\large 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]\\large\\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<\/div>\n<\/div>\n<\/div>\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-1457\">\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>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/li><li>Chemistry. <strong>Provided by<\/strong>: OpenStaxCollege. <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":6181,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. Ball\",\"organization\":\"BCCampus\",\"url\":\"https:\/\/opentextbc.ca\/introductorychemistry\/\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Download this book for free at http:\/\/open.bccampus.ca\"},{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStaxCollege\",\"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-1457","chapter","type-chapter","status-publish","hentry"],"part":196,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1457","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/users\/6181"}],"version-history":[{"count":24,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1457\/revisions"}],"predecessor-version":[{"id":1968,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1457\/revisions\/1968"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/196"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1457\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/media?parent=1457"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1457"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/contributor?post=1457"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/license?post=1457"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}