{"id":215,"date":"2016-01-11T19:59:45","date_gmt":"2016-01-11T19:59:45","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/the-mole-in-chemical-reactions-2\/"},"modified":"2020-05-06T20:20:14","modified_gmt":"2020-05-06T20:20:14","slug":"the-mole-in-chemical-reactions","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/the-mole-in-chemical-reactions\/","title":{"raw":"The Mole in Chemical Reactions","rendered":"The Mole in Chemical Reactions"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Balance a chemical equation in terms of moles.<\/li>\r\n \t
  2. Use the balanced equation to construct conversion factors in terms of moles.<\/li>\r\n \t
  3. Calculate moles of one substance from moles of another substance using a balanced chemical equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nConsider this balanced chemical equation:\r\n

    2H2<\/sub> +\u00a0O2<\/sub> \u2192\u00a02H2<\/sub>O<\/p>\r\nWe interpret this as \u201ctwo molecules of hydrogen react with one molecule of oxygen to make two molecules of water.\u201d The chemical equation is balanced as long as the coefficients are in the ratio 2:1:2. For instance, this chemical equation is also balanced:\r\n

    100H2<\/sub> +\u00a050O2<\/sub> \u2192\u00a0100H2<\/sub>O<\/p>\r\nThis equation is not conventional\u2014because convention says that we use the lowest ratio of coefficients \u2014 but it is balanced. So is this chemical equation:\r\n

    5,000H2<\/sub> +\u00a02,500O2<\/sub> \u2192\u00a05,000H2<\/sub>O<\/p>\r\nAgain, this is not conventional, but it is still balanced. Suppose we use a much larger number:\r\n

    12.044 \u00d7 1023<\/sup> H2<\/sub> +\u00a06.022 \u00d7 1023<\/sup> O2<\/sub> \u2192\u00a012.044 \u00d7 1023<\/sup> H2<\/sub>O<\/p>\r\nThese coefficients are also in the ratio of 2:1:2. But these numbers are related to the number of things in a mole: the first and last numbers are two times Avogadro\u2019s number, while the second number is Avogadro\u2019s number. That means that the first and last numbers represent 2 mol, while the middle number is just 1 mol. Well, why not just use the number of moles in balancing the chemical equation?\r\n

    2H2<\/sub> +\u00a0O2<\/sub> \u2192\u00a02H2<\/sub>O<\/p>\r\nThis is the same balanced chemical equation we started with! What this means is that chemical equations are not just balanced in terms of molecules; they are also balanced in terms of moles<\/i>. We can just as easily read this chemical equation as \u201ctwo moles of hydrogen react with one mole of oxygen to make two moles of water.\u201d All balanced chemical reactions are balanced in terms of moles.\r\n

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    Example 5.12<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nInterpret this balanced chemical equation in terms of moles.\r\n

    P4<\/sub> + 5O2<\/sub> \u2192 P4<\/sub>O10<\/sub><\/p>\r\n\r\n

    Solution<\/h2>\r\nThe coefficients represent the number of moles that react, not just molecules. We would speak of this equation as \u201cone mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide.\u201d\r\n

    Test Yourself<\/h1>\r\nInterpret this balanced chemical equation in terms of moles.\r\n

    N2<\/sub> + 3H2<\/sub> \u2192 2NH3<\/sub><\/p>\r\n\r\n

    Answer<\/h2>\r\nOne mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.\r\n\r\n<\/div>\r\n<\/div>\r\nIn Chapter 4 \"Chemical Reactions and Equations\"<\/a>, in the section called \"The Chemical Equation\"<\/a>, we stated that a chemical equation is simply a recipe for a chemical reaction. As such, chemical equations also give us equivalences\u2014equivalences between the reactants and the products. However, now we understand that these equivalences are expressed in terms of moles<\/em>. Consider the following chemical equation:\r\n

    2H2<\/sub> +\u00a0O2<\/sub> \u2192\u00a02H2<\/sub>O<\/p>\r\nThis chemical reaction gives us the following equivalences:\r\n

    2 mol H2<\/sub> \u21d4 1 mol O2<\/sub> \u21d4 2 mol H2<\/sub>O<\/p>\r\nAny two of these quantities can be used to construct a conversion factor that lets us relate the number of moles of one substance to an equivalent number of moles of another substance. If, for example, we want to know how many moles of oxygen will react with 17.6 mol of hydrogen, we construct a conversion factor between 2 mol of H2<\/sub> and 1 mol of O2<\/sub> and use it to convert from moles of one substance to moles of another:\r\n\r\n\\[17.6\\cancel{\\text{ mol }\\ce{H2}}\\times \\dfrac{1\\text{ mol }\\ce{O2}}{2\\cancel{\\text{ mol }\\ce{H2}}}=8.80\\text{ mol }\\ce{O2}\\]\r\n\r\nNote how the mol H2<\/sub> unit cancels, and mol O2<\/sub> is the new unit introduced. This is an example of a [pb_glossary id=\"1530\"]mole-mole calculation[\/pb_glossary], when you start with moles of one substance and convert to moles of another substance by using the balanced chemical equation. The example may seem simple because the numbers are small, but numbers won\u2019t always be so simple!\r\n

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    Example 5.13<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nConsider the following balanced chemical equation:\r\n

    2C4<\/sub>H10<\/sub>(g) + 13O2<\/sub> \u2192 8CO2<\/sub>(g) + 10H2<\/sub>O(\u2113)<\/p>\r\nIf 154 mol of O2<\/sub> are reacted, how many moles of CO2<\/sub> are produced?\r\n

    Solution<\/h2>\r\nWe are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is:\r\n

    13 mol O2<\/sub> \u21d4 8 mol CO2<\/sub><\/p>\r\nWe can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:\r\n\r\n\\[154\\cancel{\\text{ mol }\\ce{O2}}\\times \\dfrac{8\\text{ mol }\\ce{CO2}}{13\\cancel{\\text{ mol }\\ce{O2}}}=94.8\\text{ mol }\\ce{CO2}\\]\r\n\r\nThe mol O2<\/sub> unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they don\u2019t contribute to the number of significant figures in the final answer.\r\n

    Test Yourself<\/h1>\r\nUsing the above equation, how many moles of H2<\/sub>O are produced when 154 mol of O2<\/sub> react?\r\n

    Answer<\/h2>\r\n118 mol\r\n\r\n<\/div>\r\n<\/div>\r\nIt is important to reiterate that balanced chemical equations are balanced in terms of moles<\/i>. Not grams, kilograms, or litres \u2014 but moles. Any stoichiometry problem will likely need to work through the mole unit at some point, especially if you are working with a balanced chemical reaction.\r\n
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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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