{"id":1154,"date":"2018-07-06T03:02:10","date_gmt":"2018-07-06T03:02:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/?post_type=chapter&#038;p=1154"},"modified":"2023-11-20T17:03:30","modified_gmt":"2023-11-20T17:03:30","slug":"reaction-stoichiometry-che100","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/reaction-stoichiometry-che100\/","title":{"raw":"7.4 Reaction Stoichiometry","rendered":"7.4 Reaction Stoichiometry"},"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>Explain the concept of stoichiometry as it pertains to chemical reactions<\/li>\r\n \t<li>Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products<\/li>\r\n \t<li>Perform stoichiometric calculations involving mass and moles.<\/li>\r\n<\/ul>\r\n<\/div>\r\nA balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction\u2019s <strong>stoichiometry<\/strong>, a term derived from the Greek words <em>stoicheion<\/em> (meaning \u201celement\u201d) and <em>metron<\/em> (meaning \u201cmeasure\u201d). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.\r\n\r\nThe general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for one cup pancake mix, 3\/4 cup milk, and one egg. The \u201cequation\u201d representing the preparation of pancakes per this recipe is\r\n<p style=\"text-align: center;\">[latex]1\\text{ cup mix}+\\frac{3}{4}\\text{ cup milk}+1\\text{ egg}\\rightarrow 8\\text{ pancakes}[\/latex]<\/p>\r\nIf two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is\r\n<p style=\"text-align: center;\">[latex]24\\cancel{\\text{ pancakes}}\\times \\frac{1\\text{ egg}}{8\\cancel{\\text{ pancakes}}}=3\\text{ eggs}[\/latex]<\/p>\r\nBalanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive <strong>stoichiometric factors<\/strong> that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:\r\n<p style=\"text-align: center;\">[latex]{\\text{N}}_{2}\\text{(}g\\text{)}+3{\\text{ H}}_{2}\\text{(}g\\text{)}\\rightarrow 2{\\text{ NH}}_{3}\\text{(}g\\text{)}[\/latex]<\/p>\r\nThis equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{2{\\text{ NH}}_{3}\\text{ molecules}}{3{\\text{ H}}_{2}\\text{ molecules}}\\text{ or }\\frac{\\text{2 doz }{\\text{NH}}_{3}\\text{ molecules}}{\\text{3 doz }{\\text{H}}_{2}\\text{ molecules}}\\text{ or }\\frac{\\text{2 mol}{\\text{ NH}}_{3}\\text{ molecules}}{\\text{3 mol}{\\text{ H}}_{2}\\text{ molecules}}[\/latex]<\/p>\r\nThese stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1: <strong>Moles of Reactant Required in a Reaction<\/strong><\/h3>\r\nHow many moles of I<sub>2<\/sub> are required to react with 0.429 mol of Al according to the following equation (see Figure 1)?\r\n<p style=\"text-align: center;\">[latex]2\\text{ Al}+3{\\text{ I}}_{2}\\rightarrow 2{\\text{ AlI}}_{3}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"879\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211233\/CNX_Chem_04_03_iodine1.jpg\" alt=\"This figure shows three photos with an arrow leading from one to the next. The first photo shows a small pile of iodine and aluminum on a white surface. The second photo shows a small amount of purple smoke coming from the pile. The third photo shows a large amount of purple and gray smoke coming from the pile.\" width=\"879\" height=\"164\" \/> Figure 1. Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)[\/caption]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"307193\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"307193\"]\r\n\r\nReferring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is [latex]\\displaystyle\\frac{3\\text{ mol I}_{2}}{2\\text{ mol Al}}[\/latex]. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:\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\/23211234\/CNX_Chem_04_03_moleratio1_img1.jpg\" alt=\"This figure shows two pink rectangles. The first is labeled, \u201cMoles of A l.\u201d This rectangle is followed by an arrow pointing right to a second rectangle labeled, \u201cMoles of I subscript 2.\u201d\" \/>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\hfill {\\text{ mol I}}_{2}&amp; =0.429\\cancel{\\text{ mol Al}}\\times \\frac{\\text{3 mol}{\\text{ I}}_{2}}{2\\cancel{\\text{mol Al}}}\\\\ &amp; =\\text{0.644 mol}{\\text{ I}}_{2}\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nHow many moles of Ca(OH)<sub>2<\/sub> are required to react with 1.36 mol of H<sub>3<\/sub>PO<sub>4<\/sub> to produce Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> according to the equation [latex]3\\text{ Ca}{\\text{(}\\text{OH}\\text{)}}_{2}+2{\\text{ H}}_{3}{\\text{PO}}_{4}\\rightarrow{\\text{Ca}}_{3}{\\text{(}{\\text{PO}}_{4}\\text{)}}_{2}+6{\\text{ H}}_{2}\\text{O}[\/latex]?\r\n\r\n[reveal-answer q=\"105281\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"105281\"]2.04 mol[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2: <strong>Number of Product Molecules Generated by a Reaction<\/strong><\/h3>\r\nHow many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?\r\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{3}{\\text{H}}_{8}+5{\\text{ O}}_{2}\\rightarrow 3{\\text{ CO}}_{2}+4{\\text{ H}}_{2}\\text{O}[\/latex]<\/p>\r\n[reveal-answer q=\"159596\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"159596\"]\r\n\r\nThe approach here is the same as for Example 1, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro\u2019s number.\r\n\r\nThe balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{\\text{3 mol}{\\text{ CO}}_{2}}{\\text{1 mol}{\\text{ C}}_{3}{\\text{H}}_{8}}[\/latex]<\/p>\r\nUsing this stoichiometric factor, the provided molar amount of propane, and Avogadro\u2019s number,\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\/23211236\/CNX_Chem_04_03_moleratio2_img1.jpg\" alt=\"This figure shows two pink rectangles. The first is labeled, \u201cMoles of C subscript 3 H subscript 8.\u201d This rectangle is followed by an arrow pointing right to a second rectangle labeled, \u201cMoles of C O subscript 2.\u201d\" \/>\r\n<p style=\"text-align: center;\">[latex]0.75\\cancel{\\text{ mol}{\\text{ C}}_{3}{\\text{H}}_{8}}\\times \\frac{3\\cancel{\\text{ mol}{\\text{ CO}}_{2}}}{1\\cancel{\\text{ mol}{\\text{ C}}_{3}{\\text{H}}_{8}}}\\times \\frac{6.022\\times {10}^{23}{\\text{ CO}}_{2}\\text{ molecules}}{\\cancel{\\text{1 mol}{\\text{ CO}}_{2}}}=1.4\\times {10}^{24}{\\text{ CO}}_{2}\\text{ molecules}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nHow many NH<sub>3<\/sub> molecules are produced by the reaction of 4.0 mol of Ca(OH)<sub>2<\/sub> according to the following equation:\r\n<p style=\"text-align: center;\">[latex]{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}{\\text{SO}}_{4}+\\text{Ca}{\\text{(}\\text{OH}\\text{)}}_{2}\\rightarrow 2{\\text{ NH}}_{3}+{\\text{CaSO}}_{4}+2{\\text{ H}}_{2}\\text{O}[\/latex]<\/p>\r\n[reveal-answer q=\"488302\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"488302\"]4.8 \u00d7 10<sup>24<\/sup> NH<sub>3<\/sub> molecules[\/hidden-answer]\r\n\r\n<\/div>\r\nThese examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3: <strong>Relating Masses of Reactants and Products<\/strong><\/h3>\r\nWhat mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)<sub>2<\/sub>] by the following reaction?\r\n<p style=\"text-align: center;\">[latex]{\\text{MgCl}}_{2}\\text{(}aq\\text{)}+2\\text{ NaOH}\\text{(}aq\\text{)}\\rightarrow\\text{Mg}{\\text{(}\\text{OH}\\text{)}}_{2}\\text{(}s\\text{)}+2\\text{ NaCl}\\text{(}aq\\text{)}[\/latex]<\/p>\r\n[reveal-answer q=\"973760\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"973760\"]\r\n\r\nThe approach used previously in Example 1 and Example 2 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:\r\n\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211237\/CNX_Chem_04_03_map2_img1.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled, \u201cMass of M g ( O H ) subscript 2.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, \u201cMoles of M g ( O H ) subscript 2.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, \u201cMoles of N a O H.\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, \u201cMass of N a O H.\u201d\" width=\"881\" height=\"391\" \/>\r\n<p style=\"text-align: center;\">[latex]16\\cancel{\\text{ g Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}\\times \\frac{1\\cancel{\\text{ mol Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}{58.3\\cancel{\\text{ g Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}\\times \\frac{2\\cancel{\\text{ mol NaOH}}}{1\\cancel{\\text{ mol Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}\\times \\frac{\\text{40.0 g NaOH}}{\\cancel{\\text{1 mol NaOH}}}=\\text{22 g NaOH}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nWhat mass of gallium oxide, Ga<sub>2<\/sub>O<sub>3<\/sub>, can be prepared from 29.0 g of gallium metal? The equation for the reaction is [latex]4\\text{ Ga}+3{\\text{ O}}_{2}\\rightarrow 2{\\text{ Ga}}_{2}{\\text{O}}_{3}\\text{.}[\/latex]\r\n\r\n[reveal-answer q=\"24015\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"24015\"]39.0 g[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 4: <strong>Relating Masses of Reactants<\/strong><\/h3>\r\nWhat mass of oxygen gas, O<sub>2<\/sub>, from the air is consumed in the combustion of 702 g of octane, C<sub>8<\/sub>H<sub>18<\/sub>, one of the principal components of gasoline?\r\n<p style=\"text-align: center;\">[latex]2{\\text{ C}}_{8}{\\text{H}}_{18}+25{\\text{ O}}_{2}\\rightarrow 16{\\text{ CO}}_{2}+18{\\text{ H}}_{2}\\text{O}[\/latex]<\/p>\r\n[reveal-answer q=\"976311\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"976311\"]\r\n\r\nThe approach required here is the same as for the Example 3, differing only in that the provided and requested masses are both for reactant species.\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\/23211238\/CNX_Chem_04_03_map3_img1.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled, \u201cMass of C subscript 8 H subscript 18.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, \u201cMoles of C subscript 8 H subscript 18.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, \u201cMoles of O subscript 2.\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, \u201cMass of O subscript 2.\u201d\" width=\"878\" height=\"390\" \/>\r\n<p style=\"text-align: center;\">[latex]702\\cancel{\\text{ g}{\\text{ C}}_{8}{\\text{H}}_{18}}\\times \\frac{1\\cancel{\\text{ mol}{\\text{ C}}_{8}{\\text{H}}_{18}}}{114.23\\cancel{\\text{ g}{\\text{ C}}_{8}{\\text{H}}_{18}}}\\times \\frac{25\\cancel{\\text{ mol}{\\text{ O}}_{2}}}{2\\cancel{\\text{ mol}{\\text{ C}}_{8}{\\text{H}}_{18}}}\\times \\frac{\\text{32.00 g}{\\text{ O}}_{2}}{\\cancel{\\text{1 mol}{\\text{ O}}_{2}}}=2.46\\times {10}^{3}\\text{ g}{\\text{ O}}_{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nWhat mass of CO is required to react with 25.13 g of Fe<sub>2<\/sub>O<sub>3<\/sub> according to the equation [latex]{\\text{Fe}}_{2}{\\text{O}}_{3}+3\\text{ CO}\\rightarrow 2\\text{ Fe}+3{\\text{CO}}_{2}?[\/latex]\r\n\r\n[reveal-answer q=\"459310\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"459310\"]13.22 g[\/hidden-answer]\r\n\r\n<\/div>\r\nThese examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 2 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.\r\n\r\n[caption id=\"attachment_1209\" align=\"aligncenter\" width=\"648\"]<img class=\"wp-image-1209 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2018\/07\/18022601\/mole-concept-map.jpg\" alt=\"\" width=\"648\" height=\"241\" \/> Figure 2. The flow chart depicts the various computational steps involved in most reaction stoichiometry calculations seen in Preparatory Chemistry.[\/caption]\r\n<h3>Airbags<\/h3>\r\n[caption id=\"\" align=\"alignright\" width=\"300\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211241\/CNX_Chem_04_03_airbag1.jpg\" alt=\"This photograph shows the inside of an automobile from the driver\u2019s side area. The image shows inflated airbags positioned just in front of the driver\u2019s and passenger\u2019s seats and along the length of the passenger side over the windows. A large, round airbag covers the steering wheel.\" width=\"300\" height=\"225\" \/> Figure 3. Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)[\/caption]\r\n\r\nAirbags (Figure 3) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN<sub>3<\/sub>. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN<sub>3<\/sub> to initiate its decomposition:\r\n<p style=\"text-align: center;\">[latex]2{\\text{ NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 3{\\text{ N}}_{2}\\text{(}g\\text{)}+2\\text{ Na}\\text{(}s\\text{)}[\/latex]<\/p>\r\nThis reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03\u20130.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN<sub>3<\/sub> will generate approximately 50 L of N<sub>2<\/sub>.\r\n\r\nFor more information about the chemistry and physics behind airbags and for helpful diagrams on how airbags work, <a href=\"http:\/\/auto.howstuffworks.com\/car-driving-safety\/safety-regulatory-devices\/airbag.htm\" target=\"_blank\" rel=\"noopener noreferrer\">go to How Stuff Works' \"How Airbags Work\" article<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nA balanced chemical equation may be used to describe a reaction\u2019s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>End of Module Problems<\/h3>\r\n<ol>\r\n \t<li>Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The number of moles and the mass of chlorine, Cl<sub>2<\/sub>, required to react with 10.0 g of sodium metal, Na, to produce sodium chloride, NaCl.<\/li>\r\n \t<li>The number of moles and the mass of oxygen, O<sub>2<\/sub>, formed by the decomposition of 1.252 g of mercury(II) oxide, HgO. (Hg is the other product.)<\/li>\r\n \t<li>The number of moles and the mass of sodium nitrate, NaNO<sub>3<\/sub>, required to produce 128 g of oxygen, O<sub>2.<\/sub> (NaNO<sub>2<\/sub> is the other product.)<\/li>\r\n \t<li>The number of moles and the mass of carbon dioxide, CO<sub>2<\/sub>, formed by the combustion of 2.0 \u00d7 10<sup>4<\/sup>\u00a0of carbon, C, in an excess of oxygen, O<sub>2<\/sub>.<\/li>\r\n \t<li>The number of moles and the mass of copper(II) carbonate, Cu(CO<sub>3<\/sub>)<sub>2<\/sub> needed to produce 1.500 \u00d7 10<sup>3\u00a0<\/sup>g\u00a0of copper(II) oxide, CuO<sub>2<\/sub>. (CO<sub>2<\/sub> is the other product.)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine the number of moles and the mass requested for each reaction in Exercise 1.<\/li>\r\n \t<li>Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl<sub>2<\/sub> and H<sub>2<\/sub>.<\/li>\r\n \t<li>The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.<\/li>\r\n \t<li>The number of moles and the mass of magnesium carbonate, MgCO<sub>3<\/sub>, required to produce 283 g of carbon dioxide. (MgO is the other product.)<\/li>\r\n \t<li>The number of moles and the mass of water formed by the combustion of\u00a02.0 \u00d7 10<sup>4\u00a0<\/sup>g of acetylene, C<sub>2<\/sub>H<sub>2<\/sub>, in an excess of oxygen.<\/li>\r\n \t<li>The number of moles and the mass of barium peroxide, BaO<sub>2<\/sub>, needed to produce\u00a02.5 \u00d7 10<sup>4\u00a0<\/sup>g of barium oxide, BaO (O<sub>2<\/sub> is the other product.)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine the number of moles and the mass requested for each reaction in Exercise 3.<\/li>\r\n \t<li>H<sub>2<\/sub> is produced by the reaction of 10.19 g of H<sub>3<\/sub>PO<sub>4<\/sub> according to the following equation: [latex]2\\text{ Cr}+2{\\text{ H}}_{3}{\\text{PO}}_{4}\\rightarrow 3{\\text{ H}}_{2}+2{\\text{ CrPO}}_{4}\\text{.}[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Outline the steps necessary to determine the number of moles and mass of H<sub>2<\/sub>.<\/li>\r\n \t<li>Perform the calculations outlined.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Gallium chloride is formed by the reaction of 0.102 g of HCl according to the following equation: [latex]2\\text{ Ga}+6\\text{ HCl}\\rightarrow 2{\\text{ GaCl}}_{3}+3{\\text{ H}}_{2}\\text{.}[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Outline the steps necessary to determine the number of moles and mass of gallium chloride.<\/li>\r\n \t<li>Perform the calculations outlined.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>I<sub>2<\/sub> is produced by the reaction of 0.4235 mol of CuCl<sub>2<\/sub> according to the following equation: [latex]2{\\text{ CuCl}}_{2}+4\\text{ KI}\\rightarrow 2\\text{ CuI}+4\\text{ KCl}+{\\text{ I}}_{2}\\text{.}[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>How many molecules of I<sub>2<\/sub> are produced?<\/li>\r\n \t<li>What mass of I<sub>2<\/sub> is produced?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Aluminum reacts with iodine to produce aluminum iodide according to the reaction equation below:\r\n<p style=\"text-align: center;\">[latex]2\\text{ Al}+3{\\text{ I}}_{2}\\rightarrow 2{\\text{ AlI}}_{3}[\/latex]<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>How many formula units of AlI<sub>3<\/sub> are produced by the reaction of 35.27 g of Al with excess iodine?<\/li>\r\n \t<li>What mass (in grams) of AlI<sub>3<\/sub>\u00a0is produced?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What mass of silver oxide, Ag<sub>2<\/sub>O, is required to produce 25.0 g of silver sulfadiazine, AgC<sub>10<\/sub>H<sub>9<\/sub>N<sub>4<\/sub>SO<sub>2<\/sub>, from the reaction of silver oxide and sulfadiazine? [latex]2{\\text{C}}_{10}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{Ag}}_{2}\\text{O}\\rightarrow 2{\\text{AgC}}_{10}{\\text{H}}_{9}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\r\n \t<li>Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO<sub>2<\/sub>, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO<sub>2<\/sub> is required to produce [latex]3.00\\times {10}^{3}{\\text{ g SiC}}[\/latex].<\/li>\r\n \t<li>Automotive air bags inflate when a sample of sodium azide, NaN<sub>3<\/sub>, is very rapidly decomposed via the following reaction [latex]2{\\text{NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 2\\text{Na}\\text{(}s\\text{)}+3{\\text{N}}_{2}\\text{(}g\\text{)}[\/latex]. What mass of sodium azide is required to produce 92.0 g of nitrogen gas?<\/li>\r\n \t<li>Urea, CO(NH<sub>2<\/sub>)<sub>2<\/sub>, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO<sub>2<\/sub> produced by combustion of [latex]1.00\\times {10}^{3}\\text{kg}[\/latex] of carbon followed by the reaction? [latex]{\\text{CO}}_{2}\\text{(}g\\text{)}+2{\\text{ NH}}_{3}\\text{(}g\\text{)}\\rightarrow\\text{CO}{\\text{(}{\\text{NH}}_{2}\\text{)}}_{2}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"132180\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"132180\"]\r\n\r\n2. (a) 0.435 mol Na, 0.271 mol Cl<sub>2<\/sub>, 15.4 g Cl<sub>2<\/sub>\r\n\r\n(b) 0.005780 mol HgO, 2.890 \u00d7 10<sup>\u22123<\/sup> mol O<sub>2<\/sub>, 9.248 \u00d7 10<sup>\u22122<\/sup> g O<sub>2<\/sub>\r\n\r\n(c) 8.00 mol NaNO<sub>3<\/sub>, 6.8 \u00d7 10<sup>2<\/sup> g NaNO<sub>3<\/sub>\r\n\r\n(d) 1665 mol CO<sub>2<\/sub>, 73.3 kg CO<sub>2<\/sub>\r\n\r\n(e) 18.86 mol CuO, 2.330 kg CuCO<sub>3<\/sub>\r\n\r\n(f) 0.4580 mol C<sub>2<\/sub>H<sub>4<\/sub>Br<sub>2<\/sub>, 86.05 g C<sub>2<\/sub>H<sub>4<\/sub>Br<sub>2<\/sub>\r\n\r\n4. (a) [latex]\\text{mol Mg}=5.00\\cancel{\\text{g HCl}}\\times \\frac{1\\cancel{\\text{mol HCl}}}{36.4606\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol Mg}}{2\\cancel{\\text{mol HCl}}}=0.0686\\text{mol,}[\/latex] [latex]\\text{g Mg}=0.0686\\cancel{\\text{mol Mg}}\\times \\frac{\\text{24.305 g}}{1\\cancel{\\text{mol Mg}}}=\\text{1.67 g};[\/latex]\r\n(b) [latex]{\\text{mol O}}_{2}=1.252\\cancel{\\text{g}{\\text{Ag}}_{2}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}{231.7358\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{2\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}=2.701\\times {10}^{-3}[\/latex], [latex]{\\text{g O}}_{2}=2.701\\times {10}^{-3}\\cancel{{\\text{mol O}}_{2}}\\times \\frac{\\text{31.9988 g}}{1\\cancel{\\text{mol}{\\text{O}}_{2}}}=0.08644\\text{g}[\/latex]\r\n\r\n(c) [latex]{\\text{mol MgCO}}_{3}=283\\cancel{\\text{g}{\\text{CO}}_{2}}\\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{44.010\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{MgCO}}_{3}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=6.43\\text{mol,}[\/latex] [latex]{\\text{g MgCO}}_{3}=6.43\\cancel{{\\text{mol MgCO}}_{3}}\\times \\frac{\\text{84.314 g}}{1\\cancel{{\\text{mol MgCO}}_{3}}}=542\\text{g}[\/latex]\r\n\r\n(d) [latex]{\\text{mol H}}_{2}\\text{O}=2.00\\times {10}^{4}\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}{28.054\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{H}}_{2}\\text{O}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}=713\\text{mol,}[\/latex] [latex]{\\text{g H}}_{2}\\text{O}=713\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}\\times \\frac{18.01528\\cancel{\\text{g}}}{1\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}}\\times \\frac{\\text{1 kg}}{1000\\cancel{\\text{g}}}=\\text{12.8 kg}[\/latex]\r\n\r\n(e) [latex]2.500\\cancel{\\text{kg BaO}}\\times \\frac{1000\\cancel{\\text{g BaO}}}{1\\cancel{\\text{kg BaO}}}\\times \\frac{1\\cancel{\\text{mol BaO}}}{153.326\\cancel{\\text{g BaO}}}\\times \\frac{\\text{2 mol}{\\text{BaO}}_{2}}{2\\cancel{\\text{mol BaO}}}=\\text{16.31 mol}{\\text{BaO}}_{2}[\/latex] [latex]16.31\\cancel{\\text{mol}{\\text{BaO}}_{2}}\\times \\frac{\\text{169.326 g}{\\text{BaO}}_{2}}{1\\cancel{\\text{mol}{\\text{BaO}}_{2}}}=\\text{2762 g}{\\text{BaO}}_{2}[\/latex]\r\n\r\n(f) [latex]9.55\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}{46.068\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}\\times \\frac{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}=\\text{0.207 mol}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex] [latex]0.207\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{\\text{28.053 g}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}}=\\text{5.81 g}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex]\r\n\r\n6. (a) [latex]\\text{g HCl solution}\\rightarrow\\text{mol HCl}\\rightarrow{\\text{mol GaCl}}_{3}[\/latex]\r\n\r\n(b) [latex]\\text{0.102 }\\cancel{\\text{g HCl}}\\times \\frac{\\text{1 }\\cancel{\\text{mol HCl}}}{\\text{36.46 }\\cancel{\\text{g HCl}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}{6\\cancel{\\text{mol HCl}}}\\times \\frac{\\text{180.079 g}{\\text{GaCl}}_{3}}{1\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}=2.3\\times {10}^{2}\\text{g}{\\text{GaCl}}_{3}[\/latex]\r\n\r\n8. (a)\u00a0[latex]35.27\\cancel{\\text{g Al}}\\times \\frac{1\\cancel{\\text{mol Al}}}{26.98\\cancel{\\text{g Al}}}\\times \\frac{2\\cancel{\\text{mol Al}{\\text{I}}_{3}}}{2\\cancel{\\text{mol Al}}}\\times \\frac{6.022\\times {10}^{23}\\text{forumula units}\\text{ Al}{\\text{I}}_{3}}{1\\cancel{\\text{mol Al}{\\text{I}}_{3}}}=7.872\\times {10}^{23}\\text{formula units}\\text{ Al}{\\text{I}}_{3}[\/latex]\r\n\r\n(b)\u00a0[latex]35.27\\cancel{\\text{g Al}}\\times \\frac{1\\cancel{\\text{mol Al}}}{26.98\\cancel{\\text{g Al}}}\\times \\frac{2\\cancel{\\text{mol Al}{\\text{I}}_{3}}}{2\\cancel{\\text{mol Al}}}\\times \\frac{\\text{407.68 g}\\text{ Al}{\\text{I}}_{3}}{1\\cancel{\\text{mol Al}{\\text{I}}_{3}}}=532.9 \\text{ g Al}{\\text{I}}_{3}[\/latex]\r\n\r\n10. [latex]{\\text{SiO}}_{2}+3\\text{C}\\rightarrow\\text{SiC}+2\\text{CO.}[\/latex] From the balanced equation, 1 mol of SiO<sub>2<\/sub> produces 1 mol of SiC. The unknown is the mass of SiO<sub>2<\/sub> required to produce 3.00 kg (3000 g) of SiC. To calculate the mass of SiO<sub>2<\/sub> required, determine the molar masses of SiO<sub>2<\/sub> and SiC. Then calculate the number of moles of SiC required, and through the mole relation of SiO<sub>2<\/sub> to SiC, find the mass of SiO<sub>2<\/sub> required. The conversions required are: [latex]\\text{g SiC}\\rightarrow\\text{mol SiC}\\rightarrow\\text{mol}{\\text{SiO}}_{2}\\rightarrow\\text{g}{\\text{SiO}}_{2}[\/latex]\r\nMolar masses: SiO<sub>2<\/sub> = 60.0843 g mol<sup>\u20131<\/sup>; SiC = 40.0955 g mol<sup>\u20131<\/sup>\r\n[latex]\\text{mass SiO}2=3000\\cancel{\\text{g SiC}}\\times \\frac{1\\cancel{\\text{mol SiC}}}{40.955\\cancel{\\text{g SiC}}}\\times \\frac{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}{1\\cancel{\\text{mol SiC}}}\\times \\frac{\\text{60.843 g}{\\text{SiO}}_{2}}{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}=\\text{4496 g}{\\text{SiO}}_{2}=\\text{4.50 kg}{\\text{SiO}}_{2}[\/latex]\r\n\r\n12. Molar mass urea = 12.011 + 15.9994 + 2(14.0067) + 4(1.0079) = 60.054 g mol<sup>\u20131<\/sup>\r\n[latex]\\text{1 mol C}\\rightarrow 1{\\text{mol CO}}_{2}\\rightarrow 1\\text{mol urea}[\/latex]\r\n[latex]\\begin{array}{ll}\\hfill \\text{mass urea}&amp; =1.00\\times {10}^{3}\\cancel{\\text{kg}}\\times \\frac{1000\\cancel{\\text{g}}}{\\cancel{\\text{kg}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.0\\cancel{\\text{g C}}}\\times \\frac{1\\cancel{\\text{mol urea}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{60.054 g urea}}{1\\cancel{\\text{mol urea}}}\\\\ &amp; =5.00\\times {10}^{6}\\text{g or}5.00\\times {10}^{3}\\text{kg}\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>stoichiometric factor: <\/strong>ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products\r\n\r\n<strong>stoichiometry: <\/strong>relationships between the amounts of reactants and products of a chemical reaction","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>Explain the concept of stoichiometry as it pertains to chemical reactions<\/li>\n<li>Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products<\/li>\n<li>Perform stoichiometric calculations involving mass and moles.<\/li>\n<\/ul>\n<\/div>\n<p>A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction\u2019s <strong>stoichiometry<\/strong>, a term derived from the Greek words <em>stoicheion<\/em> (meaning \u201celement\u201d) and <em>metron<\/em> (meaning \u201cmeasure\u201d). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.<\/p>\n<p>The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for one cup pancake mix, 3\/4 cup milk, and one egg. The \u201cequation\u201d representing the preparation of pancakes per this recipe is<\/p>\n<p style=\"text-align: center;\">[latex]1\\text{ cup mix}+\\frac{3}{4}\\text{ cup milk}+1\\text{ egg}\\rightarrow 8\\text{ pancakes}[\/latex]<\/p>\n<p>If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is<\/p>\n<p style=\"text-align: center;\">[latex]24\\cancel{\\text{ pancakes}}\\times \\frac{1\\text{ egg}}{8\\cancel{\\text{ pancakes}}}=3\\text{ eggs}[\/latex]<\/p>\n<p>Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive <strong>stoichiometric factors<\/strong> that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{N}}_{2}\\text{(}g\\text{)}+3{\\text{ H}}_{2}\\text{(}g\\text{)}\\rightarrow 2{\\text{ NH}}_{3}\\text{(}g\\text{)}[\/latex]<\/p>\n<p>This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{2{\\text{ NH}}_{3}\\text{ molecules}}{3{\\text{ H}}_{2}\\text{ molecules}}\\text{ or }\\frac{\\text{2 doz }{\\text{NH}}_{3}\\text{ molecules}}{\\text{3 doz }{\\text{H}}_{2}\\text{ molecules}}\\text{ or }\\frac{\\text{2 mol}{\\text{ NH}}_{3}\\text{ molecules}}{\\text{3 mol}{\\text{ H}}_{2}\\text{ molecules}}[\/latex]<\/p>\n<p>These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1: <strong>Moles of Reactant Required in a Reaction<\/strong><\/h3>\n<p>How many moles of I<sub>2<\/sub> are required to react with 0.429 mol of Al according to the following equation (see Figure 1)?<\/p>\n<p style=\"text-align: center;\">[latex]2\\text{ Al}+3{\\text{ I}}_{2}\\rightarrow 2{\\text{ AlI}}_{3}[\/latex]<\/p>\n<div style=\"width: 889px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211233\/CNX_Chem_04_03_iodine1.jpg\" alt=\"This figure shows three photos with an arrow leading from one to the next. The first photo shows a small pile of iodine and aluminum on a white surface. The second photo shows a small amount of purple smoke coming from the pile. The third photo shows a large amount of purple and gray smoke coming from the pile.\" width=\"879\" height=\"164\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q307193\">Show Answer<\/span><\/p>\n<div id=\"q307193\" class=\"hidden-answer\" style=\"display: none\">\n<p>Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is [latex]\\displaystyle\\frac{3\\text{ mol I}_{2}}{2\\text{ mol Al}}[\/latex]. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:<\/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\/23211234\/CNX_Chem_04_03_moleratio1_img1.jpg\" alt=\"This figure shows two pink rectangles. The first is labeled, \u201cMoles of A l.\u201d This rectangle is followed by an arrow pointing right to a second rectangle labeled, \u201cMoles of I subscript 2.\u201d\" \/><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\hfill {\\text{ mol I}}_{2}& =0.429\\cancel{\\text{ mol Al}}\\times \\frac{\\text{3 mol}{\\text{ I}}_{2}}{2\\cancel{\\text{mol Al}}}\\\\ & =\\text{0.644 mol}{\\text{ I}}_{2}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>How many moles of Ca(OH)<sub>2<\/sub> are required to react with 1.36 mol of H<sub>3<\/sub>PO<sub>4<\/sub> to produce Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> according to the equation [latex]3\\text{ Ca}{\\text{(}\\text{OH}\\text{)}}_{2}+2{\\text{ H}}_{3}{\\text{PO}}_{4}\\rightarrow{\\text{Ca}}_{3}{\\text{(}{\\text{PO}}_{4}\\text{)}}_{2}+6{\\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=\"q105281\">Show Answer<\/span><\/p>\n<div id=\"q105281\" class=\"hidden-answer\" style=\"display: none\">2.04 mol<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2: <strong>Number of Product Molecules Generated by a Reaction<\/strong><\/h3>\n<p>How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{3}{\\text{H}}_{8}+5{\\text{ O}}_{2}\\rightarrow 3{\\text{ CO}}_{2}+4{\\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=\"q159596\">Show Answer<\/span><\/p>\n<div id=\"q159596\" class=\"hidden-answer\" style=\"display: none\">\n<p>The approach here is the same as for Example 1, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro\u2019s number.<\/p>\n<p>The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{\\text{3 mol}{\\text{ CO}}_{2}}{\\text{1 mol}{\\text{ C}}_{3}{\\text{H}}_{8}}[\/latex]<\/p>\n<p>Using this stoichiometric factor, the provided molar amount of propane, and Avogadro\u2019s number,<\/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\/23211236\/CNX_Chem_04_03_moleratio2_img1.jpg\" alt=\"This figure shows two pink rectangles. The first is labeled, \u201cMoles of C subscript 3 H subscript 8.\u201d This rectangle is followed by an arrow pointing right to a second rectangle labeled, \u201cMoles of C O subscript 2.\u201d\" \/><\/p>\n<p style=\"text-align: center;\">[latex]0.75\\cancel{\\text{ mol}{\\text{ C}}_{3}{\\text{H}}_{8}}\\times \\frac{3\\cancel{\\text{ mol}{\\text{ CO}}_{2}}}{1\\cancel{\\text{ mol}{\\text{ C}}_{3}{\\text{H}}_{8}}}\\times \\frac{6.022\\times {10}^{23}{\\text{ CO}}_{2}\\text{ molecules}}{\\cancel{\\text{1 mol}{\\text{ CO}}_{2}}}=1.4\\times {10}^{24}{\\text{ CO}}_{2}\\text{ molecules}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>How many NH<sub>3<\/sub> molecules are produced by the reaction of 4.0 mol of Ca(OH)<sub>2<\/sub> according to the following equation:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}{\\text{SO}}_{4}+\\text{Ca}{\\text{(}\\text{OH}\\text{)}}_{2}\\rightarrow 2{\\text{ NH}}_{3}+{\\text{CaSO}}_{4}+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=\"q488302\">Show Answer<\/span><\/p>\n<div id=\"q488302\" class=\"hidden-answer\" style=\"display: none\">4.8 \u00d7 10<sup>24<\/sup> NH<sub>3<\/sub> molecules<\/div>\n<\/div>\n<\/div>\n<p>These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3: <strong>Relating Masses of Reactants and Products<\/strong><\/h3>\n<p>What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)<sub>2<\/sub>] by the following reaction?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{MgCl}}_{2}\\text{(}aq\\text{)}+2\\text{ NaOH}\\text{(}aq\\text{)}\\rightarrow\\text{Mg}{\\text{(}\\text{OH}\\text{)}}_{2}\\text{(}s\\text{)}+2\\text{ NaCl}\\text{(}aq\\text{)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q973760\">Show Answer<\/span><\/p>\n<div id=\"q973760\" class=\"hidden-answer\" style=\"display: none\">\n<p>The approach used previously in Example 1 and Example 2 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:<\/p>\n<p><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\/23211237\/CNX_Chem_04_03_map2_img1.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled, \u201cMass of M g ( O H ) subscript 2.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, \u201cMoles of M g ( O H ) subscript 2.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, \u201cMoles of N a O H.\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, \u201cMass of N a O H.\u201d\" width=\"881\" height=\"391\" \/><\/p>\n<p style=\"text-align: center;\">[latex]16\\cancel{\\text{ g Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}\\times \\frac{1\\cancel{\\text{ mol Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}{58.3\\cancel{\\text{ g Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}\\times \\frac{2\\cancel{\\text{ mol NaOH}}}{1\\cancel{\\text{ mol Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}\\times \\frac{\\text{40.0 g NaOH}}{\\cancel{\\text{1 mol NaOH}}}=\\text{22 g NaOH}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>What mass of gallium oxide, Ga<sub>2<\/sub>O<sub>3<\/sub>, can be prepared from 29.0 g of gallium metal? The equation for the reaction is [latex]4\\text{ Ga}+3{\\text{ O}}_{2}\\rightarrow 2{\\text{ Ga}}_{2}{\\text{O}}_{3}\\text{.}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q24015\">Show Answer<\/span><\/p>\n<div id=\"q24015\" class=\"hidden-answer\" style=\"display: none\">39.0 g<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 4: <strong>Relating Masses of Reactants<\/strong><\/h3>\n<p>What mass of oxygen gas, O<sub>2<\/sub>, from the air is consumed in the combustion of 702 g of octane, C<sub>8<\/sub>H<sub>18<\/sub>, one of the principal components of gasoline?<\/p>\n<p style=\"text-align: center;\">[latex]2{\\text{ C}}_{8}{\\text{H}}_{18}+25{\\text{ O}}_{2}\\rightarrow 16{\\text{ CO}}_{2}+18{\\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=\"q976311\">Show Answer<\/span><\/p>\n<div id=\"q976311\" class=\"hidden-answer\" style=\"display: none\">\n<p>The approach required here is the same as for the Example 3, differing only in that the provided and requested masses are both for reactant species.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211238\/CNX_Chem_04_03_map3_img1.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled, \u201cMass of C subscript 8 H subscript 18.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, \u201cMoles of C subscript 8 H subscript 18.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, \u201cMoles of O subscript 2.\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, \u201cMass of O subscript 2.\u201d\" width=\"878\" height=\"390\" \/><\/p>\n<p style=\"text-align: center;\">[latex]702\\cancel{\\text{ g}{\\text{ C}}_{8}{\\text{H}}_{18}}\\times \\frac{1\\cancel{\\text{ mol}{\\text{ C}}_{8}{\\text{H}}_{18}}}{114.23\\cancel{\\text{ g}{\\text{ C}}_{8}{\\text{H}}_{18}}}\\times \\frac{25\\cancel{\\text{ mol}{\\text{ O}}_{2}}}{2\\cancel{\\text{ mol}{\\text{ C}}_{8}{\\text{H}}_{18}}}\\times \\frac{\\text{32.00 g}{\\text{ O}}_{2}}{\\cancel{\\text{1 mol}{\\text{ O}}_{2}}}=2.46\\times {10}^{3}\\text{ g}{\\text{ O}}_{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>What mass of CO is required to react with 25.13 g of Fe<sub>2<\/sub>O<sub>3<\/sub> according to the equation [latex]{\\text{Fe}}_{2}{\\text{O}}_{3}+3\\text{ CO}\\rightarrow 2\\text{ Fe}+3{\\text{CO}}_{2}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q459310\">Show Answer<\/span><\/p>\n<div id=\"q459310\" class=\"hidden-answer\" style=\"display: none\">13.22 g<\/div>\n<\/div>\n<\/div>\n<p>These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 2 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.<\/p>\n<div id=\"attachment_1209\" style=\"width: 658px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1209\" class=\"wp-image-1209 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2018\/07\/18022601\/mole-concept-map.jpg\" alt=\"\" width=\"648\" height=\"241\" \/><\/p>\n<p id=\"caption-attachment-1209\" class=\"wp-caption-text\">Figure 2. The flow chart depicts the various computational steps involved in most reaction stoichiometry calculations seen in Preparatory Chemistry.<\/p>\n<\/div>\n<h3>Airbags<\/h3>\n<div style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211241\/CNX_Chem_04_03_airbag1.jpg\" alt=\"This photograph shows the inside of an automobile from the driver\u2019s side area. The image shows inflated airbags positioned just in front of the driver\u2019s and passenger\u2019s seats and along the length of the passenger side over the windows. A large, round airbag covers the steering wheel.\" width=\"300\" height=\"225\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)<\/p>\n<\/div>\n<p>Airbags (Figure 3) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN<sub>3<\/sub>. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN<sub>3<\/sub> to initiate its decomposition:<\/p>\n<p style=\"text-align: center;\">[latex]2{\\text{ NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 3{\\text{ N}}_{2}\\text{(}g\\text{)}+2\\text{ Na}\\text{(}s\\text{)}[\/latex]<\/p>\n<p>This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03\u20130.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN<sub>3<\/sub> will generate approximately 50 L of N<sub>2<\/sub>.<\/p>\n<p>For more information about the chemistry and physics behind airbags and for helpful diagrams on how airbags work, <a href=\"http:\/\/auto.howstuffworks.com\/car-driving-safety\/safety-regulatory-devices\/airbag.htm\" target=\"_blank\" rel=\"noopener noreferrer\">go to How Stuff Works&#8217; &#8220;How Airbags Work&#8221; article<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>A balanced chemical equation may be used to describe a reaction\u2019s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>End of Module Problems<\/h3>\n<ol>\n<li>Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The number of moles and the mass of chlorine, Cl<sub>2<\/sub>, required to react with 10.0 g of sodium metal, Na, to produce sodium chloride, NaCl.<\/li>\n<li>The number of moles and the mass of oxygen, O<sub>2<\/sub>, formed by the decomposition of 1.252 g of mercury(II) oxide, HgO. (Hg is the other product.)<\/li>\n<li>The number of moles and the mass of sodium nitrate, NaNO<sub>3<\/sub>, required to produce 128 g of oxygen, O<sub>2.<\/sub> (NaNO<sub>2<\/sub> is the other product.)<\/li>\n<li>The number of moles and the mass of carbon dioxide, CO<sub>2<\/sub>, formed by the combustion of 2.0 \u00d7 10<sup>4<\/sup>\u00a0of carbon, C, in an excess of oxygen, O<sub>2<\/sub>.<\/li>\n<li>The number of moles and the mass of copper(II) carbonate, Cu(CO<sub>3<\/sub>)<sub>2<\/sub> needed to produce 1.500 \u00d7 10<sup>3\u00a0<\/sup>g\u00a0of copper(II) oxide, CuO<sub>2<\/sub>. (CO<sub>2<\/sub> is the other product.)<\/li>\n<\/ol>\n<\/li>\n<li>Determine the number of moles and the mass requested for each reaction in Exercise 1.<\/li>\n<li>Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl<sub>2<\/sub> and H<sub>2<\/sub>.<\/li>\n<li>The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.<\/li>\n<li>The number of moles and the mass of magnesium carbonate, MgCO<sub>3<\/sub>, required to produce 283 g of carbon dioxide. (MgO is the other product.)<\/li>\n<li>The number of moles and the mass of water formed by the combustion of\u00a02.0 \u00d7 10<sup>4\u00a0<\/sup>g of acetylene, C<sub>2<\/sub>H<sub>2<\/sub>, in an excess of oxygen.<\/li>\n<li>The number of moles and the mass of barium peroxide, BaO<sub>2<\/sub>, needed to produce\u00a02.5 \u00d7 10<sup>4\u00a0<\/sup>g of barium oxide, BaO (O<sub>2<\/sub> is the other product.)<\/li>\n<\/ol>\n<\/li>\n<li>Determine the number of moles and the mass requested for each reaction in Exercise 3.<\/li>\n<li>H<sub>2<\/sub> is produced by the reaction of 10.19 g of H<sub>3<\/sub>PO<sub>4<\/sub> according to the following equation: [latex]2\\text{ Cr}+2{\\text{ H}}_{3}{\\text{PO}}_{4}\\rightarrow 3{\\text{ H}}_{2}+2{\\text{ CrPO}}_{4}\\text{.}[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Outline the steps necessary to determine the number of moles and mass of H<sub>2<\/sub>.<\/li>\n<li>Perform the calculations outlined.<\/li>\n<\/ol>\n<\/li>\n<li>Gallium chloride is formed by the reaction of 0.102 g of HCl according to the following equation: [latex]2\\text{ Ga}+6\\text{ HCl}\\rightarrow 2{\\text{ GaCl}}_{3}+3{\\text{ H}}_{2}\\text{.}[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Outline the steps necessary to determine the number of moles and mass of gallium chloride.<\/li>\n<li>Perform the calculations outlined.<\/li>\n<\/ol>\n<\/li>\n<li>I<sub>2<\/sub> is produced by the reaction of 0.4235 mol of CuCl<sub>2<\/sub> according to the following equation: [latex]2{\\text{ CuCl}}_{2}+4\\text{ KI}\\rightarrow 2\\text{ CuI}+4\\text{ KCl}+{\\text{ I}}_{2}\\text{.}[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>How many molecules of I<sub>2<\/sub> are produced?<\/li>\n<li>What mass of I<sub>2<\/sub> is produced?<\/li>\n<\/ol>\n<\/li>\n<li>Aluminum reacts with iodine to produce aluminum iodide according to the reaction equation below:\n<p style=\"text-align: center;\">[latex]2\\text{ Al}+3{\\text{ I}}_{2}\\rightarrow 2{\\text{ AlI}}_{3}[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>How many formula units of AlI<sub>3<\/sub> are produced by the reaction of 35.27 g of Al with excess iodine?<\/li>\n<li>What mass (in grams) of AlI<sub>3<\/sub>\u00a0is produced?<\/li>\n<\/ol>\n<\/li>\n<li>What mass of silver oxide, Ag<sub>2<\/sub>O, is required to produce 25.0 g of silver sulfadiazine, AgC<sub>10<\/sub>H<sub>9<\/sub>N<sub>4<\/sub>SO<sub>2<\/sub>, from the reaction of silver oxide and sulfadiazine? [latex]2{\\text{C}}_{10}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{Ag}}_{2}\\text{O}\\rightarrow 2{\\text{AgC}}_{10}{\\text{H}}_{9}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\n<li>Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO<sub>2<\/sub>, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO<sub>2<\/sub> is required to produce [latex]3.00\\times {10}^{3}{\\text{ g SiC}}[\/latex].<\/li>\n<li>Automotive air bags inflate when a sample of sodium azide, NaN<sub>3<\/sub>, is very rapidly decomposed via the following reaction [latex]2{\\text{NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 2\\text{Na}\\text{(}s\\text{)}+3{\\text{N}}_{2}\\text{(}g\\text{)}[\/latex]. What mass of sodium azide is required to produce 92.0 g of nitrogen gas?<\/li>\n<li>Urea, CO(NH<sub>2<\/sub>)<sub>2<\/sub>, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO<sub>2<\/sub> produced by combustion of [latex]1.00\\times {10}^{3}\\text{kg}[\/latex] of carbon followed by the reaction? [latex]{\\text{CO}}_{2}\\text{(}g\\text{)}+2{\\text{ NH}}_{3}\\text{(}g\\text{)}\\rightarrow\\text{CO}{\\text{(}{\\text{NH}}_{2}\\text{)}}_{2}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q132180\">Show Selected Answers<\/span><\/p>\n<div id=\"q132180\" class=\"hidden-answer\" style=\"display: none\">\n<p>2. (a) 0.435 mol Na, 0.271 mol Cl<sub>2<\/sub>, 15.4 g Cl<sub>2<\/sub><\/p>\n<p>(b) 0.005780 mol HgO, 2.890 \u00d7 10<sup>\u22123<\/sup> mol O<sub>2<\/sub>, 9.248 \u00d7 10<sup>\u22122<\/sup> g O<sub>2<\/sub><\/p>\n<p>(c) 8.00 mol NaNO<sub>3<\/sub>, 6.8 \u00d7 10<sup>2<\/sup> g NaNO<sub>3<\/sub><\/p>\n<p>(d) 1665 mol CO<sub>2<\/sub>, 73.3 kg CO<sub>2<\/sub><\/p>\n<p>(e) 18.86 mol CuO, 2.330 kg CuCO<sub>3<\/sub><\/p>\n<p>(f) 0.4580 mol C<sub>2<\/sub>H<sub>4<\/sub>Br<sub>2<\/sub>, 86.05 g C<sub>2<\/sub>H<sub>4<\/sub>Br<sub>2<\/sub><\/p>\n<p>4. (a) [latex]\\text{mol Mg}=5.00\\cancel{\\text{g HCl}}\\times \\frac{1\\cancel{\\text{mol HCl}}}{36.4606\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol Mg}}{2\\cancel{\\text{mol HCl}}}=0.0686\\text{mol,}[\/latex] [latex]\\text{g Mg}=0.0686\\cancel{\\text{mol Mg}}\\times \\frac{\\text{24.305 g}}{1\\cancel{\\text{mol Mg}}}=\\text{1.67 g};[\/latex]<br \/>\n(b) [latex]{\\text{mol O}}_{2}=1.252\\cancel{\\text{g}{\\text{Ag}}_{2}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}{231.7358\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{2\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}=2.701\\times {10}^{-3}[\/latex], [latex]{\\text{g O}}_{2}=2.701\\times {10}^{-3}\\cancel{{\\text{mol O}}_{2}}\\times \\frac{\\text{31.9988 g}}{1\\cancel{\\text{mol}{\\text{O}}_{2}}}=0.08644\\text{g}[\/latex]<\/p>\n<p>(c) [latex]{\\text{mol MgCO}}_{3}=283\\cancel{\\text{g}{\\text{CO}}_{2}}\\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{44.010\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{MgCO}}_{3}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=6.43\\text{mol,}[\/latex] [latex]{\\text{g MgCO}}_{3}=6.43\\cancel{{\\text{mol MgCO}}_{3}}\\times \\frac{\\text{84.314 g}}{1\\cancel{{\\text{mol MgCO}}_{3}}}=542\\text{g}[\/latex]<\/p>\n<p>(d) [latex]{\\text{mol H}}_{2}\\text{O}=2.00\\times {10}^{4}\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}{28.054\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{H}}_{2}\\text{O}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}=713\\text{mol,}[\/latex] [latex]{\\text{g H}}_{2}\\text{O}=713\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}\\times \\frac{18.01528\\cancel{\\text{g}}}{1\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}}\\times \\frac{\\text{1 kg}}{1000\\cancel{\\text{g}}}=\\text{12.8 kg}[\/latex]<\/p>\n<p>(e) [latex]2.500\\cancel{\\text{kg BaO}}\\times \\frac{1000\\cancel{\\text{g BaO}}}{1\\cancel{\\text{kg BaO}}}\\times \\frac{1\\cancel{\\text{mol BaO}}}{153.326\\cancel{\\text{g BaO}}}\\times \\frac{\\text{2 mol}{\\text{BaO}}_{2}}{2\\cancel{\\text{mol BaO}}}=\\text{16.31 mol}{\\text{BaO}}_{2}[\/latex] [latex]16.31\\cancel{\\text{mol}{\\text{BaO}}_{2}}\\times \\frac{\\text{169.326 g}{\\text{BaO}}_{2}}{1\\cancel{\\text{mol}{\\text{BaO}}_{2}}}=\\text{2762 g}{\\text{BaO}}_{2}[\/latex]<\/p>\n<p>(f) [latex]9.55\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}{46.068\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}\\times \\frac{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}=\\text{0.207 mol}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex] [latex]0.207\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{\\text{28.053 g}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}}=\\text{5.81 g}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex]<\/p>\n<p>6. (a) [latex]\\text{g HCl solution}\\rightarrow\\text{mol HCl}\\rightarrow{\\text{mol GaCl}}_{3}[\/latex]<\/p>\n<p>(b) [latex]\\text{0.102 }\\cancel{\\text{g HCl}}\\times \\frac{\\text{1 }\\cancel{\\text{mol HCl}}}{\\text{36.46 }\\cancel{\\text{g HCl}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}{6\\cancel{\\text{mol HCl}}}\\times \\frac{\\text{180.079 g}{\\text{GaCl}}_{3}}{1\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}=2.3\\times {10}^{2}\\text{g}{\\text{GaCl}}_{3}[\/latex]<\/p>\n<p>8. (a)\u00a0[latex]35.27\\cancel{\\text{g Al}}\\times \\frac{1\\cancel{\\text{mol Al}}}{26.98\\cancel{\\text{g Al}}}\\times \\frac{2\\cancel{\\text{mol Al}{\\text{I}}_{3}}}{2\\cancel{\\text{mol Al}}}\\times \\frac{6.022\\times {10}^{23}\\text{forumula units}\\text{ Al}{\\text{I}}_{3}}{1\\cancel{\\text{mol Al}{\\text{I}}_{3}}}=7.872\\times {10}^{23}\\text{formula units}\\text{ Al}{\\text{I}}_{3}[\/latex]<\/p>\n<p>(b)\u00a0[latex]35.27\\cancel{\\text{g Al}}\\times \\frac{1\\cancel{\\text{mol Al}}}{26.98\\cancel{\\text{g Al}}}\\times \\frac{2\\cancel{\\text{mol Al}{\\text{I}}_{3}}}{2\\cancel{\\text{mol Al}}}\\times \\frac{\\text{407.68 g}\\text{ Al}{\\text{I}}_{3}}{1\\cancel{\\text{mol Al}{\\text{I}}_{3}}}=532.9 \\text{ g Al}{\\text{I}}_{3}[\/latex]<\/p>\n<p>10. [latex]{\\text{SiO}}_{2}+3\\text{C}\\rightarrow\\text{SiC}+2\\text{CO.}[\/latex] From the balanced equation, 1 mol of SiO<sub>2<\/sub> produces 1 mol of SiC. The unknown is the mass of SiO<sub>2<\/sub> required to produce 3.00 kg (3000 g) of SiC. To calculate the mass of SiO<sub>2<\/sub> required, determine the molar masses of SiO<sub>2<\/sub> and SiC. Then calculate the number of moles of SiC required, and through the mole relation of SiO<sub>2<\/sub> to SiC, find the mass of SiO<sub>2<\/sub> required. The conversions required are: [latex]\\text{g SiC}\\rightarrow\\text{mol SiC}\\rightarrow\\text{mol}{\\text{SiO}}_{2}\\rightarrow\\text{g}{\\text{SiO}}_{2}[\/latex]<br \/>\nMolar masses: SiO<sub>2<\/sub> = 60.0843 g mol<sup>\u20131<\/sup>; SiC = 40.0955 g mol<sup>\u20131<\/sup><br \/>\n[latex]\\text{mass SiO}2=3000\\cancel{\\text{g SiC}}\\times \\frac{1\\cancel{\\text{mol SiC}}}{40.955\\cancel{\\text{g SiC}}}\\times \\frac{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}{1\\cancel{\\text{mol SiC}}}\\times \\frac{\\text{60.843 g}{\\text{SiO}}_{2}}{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}=\\text{4496 g}{\\text{SiO}}_{2}=\\text{4.50 kg}{\\text{SiO}}_{2}[\/latex]<\/p>\n<p>12. Molar mass urea = 12.011 + 15.9994 + 2(14.0067) + 4(1.0079) = 60.054 g mol<sup>\u20131<\/sup><br \/>\n[latex]\\text{1 mol C}\\rightarrow 1{\\text{mol CO}}_{2}\\rightarrow 1\\text{mol urea}[\/latex]<br \/>\n[latex]\\begin{array}{ll}\\hfill \\text{mass urea}& =1.00\\times {10}^{3}\\cancel{\\text{kg}}\\times \\frac{1000\\cancel{\\text{g}}}{\\cancel{\\text{kg}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.0\\cancel{\\text{g C}}}\\times \\frac{1\\cancel{\\text{mol urea}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{60.054 g urea}}{1\\cancel{\\text{mol urea}}}\\\\ & =5.00\\times {10}^{6}\\text{g or}5.00\\times {10}^{3}\\text{kg}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>stoichiometric factor: <\/strong>ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products<\/p>\n<p><strong>stoichiometry: <\/strong>relationships between the amounts of reactants and products of a chemical reaction<\/p>\n","protected":false},"author":6181,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1154","chapter","type-chapter","status-publish","hentry"],"part":114,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1154","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":19,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1154\/revisions"}],"predecessor-version":[{"id":1953,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1154\/revisions\/1953"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/114"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1154\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/media?parent=1154"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1154"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/contributor?post=1154"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/license?post=1154"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}