{"id":683,"date":"2018-03-20T15:21:54","date_gmt":"2018-03-20T15:21:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&p=683"},"modified":"2018-08-07T18:12:36","modified_gmt":"2018-08-07T18:12:36","slug":"6-4-mole-mole-relationships-in-chemical-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/chapter\/6-4-mole-mole-relationships-in-chemical-reactions\/","title":{"raw":"6.4 Mole-Mole Relationships in Chemical Reactions","rendered":"6.4 Mole-Mole Relationships in Chemical Reactions"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  1. Use a balanced chemical reaction to determine molar relationships between the substances.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    In Chapter 5 \"Introduction to Chemical Reactions\"<\/a>, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). Here, we will extend the meaning of the coefficients in a chemical equation.<\/p>\r\n

    Consider the simple chemical equation<\/p>\r\n2H2<\/sub> + O2<\/sub> \u2192 2H2<\/sub>O<\/span><\/span>\r\n

    The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as<\/p>\r\n4H2<\/sub> + 2O2<\/sub> \u2192 4H2<\/sub>O<\/span><\/span>\r\n

    The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as<\/p>\r\n22H2<\/sub> + 11O2<\/sub> \u2192 22H2<\/sub>O<\/span><\/span>\r\n

    because 22:11:22 also reduces to 2:1:2.<\/p>\r\n

    Suppose we want to use larger numbers. Consider the following coefficients:<\/p>\r\n12.044 \u00d7 1023<\/sup> H2<\/sub> + 6.022 \u00d7 1023<\/sup> O2<\/sub> \u2192 12.044 \u00d7 1023<\/sup> H2<\/sub>O<\/span><\/span>\r\n

    These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 \u00d7 1023<\/sup> is 1 mol, while 12.044 \u00d7 1023<\/sup> is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as<\/p>\r\n2 mol H2<\/sub> + 1 mol O2<\/sub> \u2192 2 mol H2<\/sub>O<\/span><\/span>\r\n

    We can leave out the word mol<\/em> and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is<\/p>\r\n2H2<\/sub> + O2<\/sub> \u2192 2H2<\/sub>O<\/span><\/span>\r\n

    Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level but also in terms of molar amounts of reactants and products.<\/em> Thus, we can read this reaction as \u201ctwo moles of hydrogen react with one mole of oxygen to produce two moles of water.\u201d<\/p>\r\n

    By the same token, the ratios we constructed in Chapter 5 \"Introduction to Chemical Reactions\"<\/a> can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios:<\/p>\r\n [latex]\\frac{2\\,mol\\,H_2}{1\\,mol\\,O_2}[\/latex]\u00a0 or\u00a0 [latex]\\frac{1\\,mol\\,O_2}{2\\,mol\\,H_2}[\/latex]\u00a0<\/span>\r\n[latex]\\frac{2\\,mol\\,H_2O}{1\\,mol\\,O_2}[\/latex]\u00a0 or [latex]\\frac{1\\,mol\\,O_2}{2\\,mol\\,H_2O}[\/latex]<\/span>\r\n [latex]\\frac{2\\,mol\\,H_2}{2\\,mol\\,H_2O}[\/latex]\u00a0 or\u00a0 [latex]\\frac{2\\,mol\\,H_2O}{2\\,mol\\,H_2}[\/latex]\u00a0<\/span>\r\n

    We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry<\/em>.<\/p>\r\n\r\n

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    Example 7<\/h3>\r\n

    How many moles of oxygen react with hydrogen to produce 27.6 mol of H2<\/sub>O? The balanced equation is as follows:<\/p>\r\n2H2<\/sub> + O2<\/sub> \u2192 2H2<\/sub>O<\/span><\/span>\r\n

    Solution<\/p>\r\n

    Because we are dealing with quantities of H2<\/sub>O and O2<\/sub>, we will use a ratio that relates those two substances. Because we are given an amount of H2<\/sub>O and want to determine an amount of O2<\/sub>, we will use the ratio that has H2<\/sub>O in the denominator (so it cancels) and O2<\/sub> in the numerator (so it is introduced in the answer). Thus,<\/p>\r\n [latex]27.6\\,\\rlap{{-----------}}mol\\,H_2O\\,\u00d7\\,\\frac{1\\,mol\\,O_2}{2\\,\\rlap{{-----------}}mol\\,H_2O}\\,=\\,13.8\\,mol\\,O_2[\/latex] <\/span>\r\n

    To produce 27.6 mol of H2<\/sub>O, 13.8 mol of O2<\/sub> react.<\/p>\r\n\r\n<\/div>\r\n

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    Skill-Building Exercise<\/h3>\r\n
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      Using 2H2<\/sub> + O2<\/sub> \u2192 2H2<\/sub>O, how many moles of hydrogen react with 3.07 mol of oxygen to produce H2<\/sub>O?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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      Concept Review Exercise<\/h3>\r\n
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        How do we relate molar amounts of substances in chemical reactions?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

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        Answer<\/h3>\r\n[reveal-answer q=\"190743\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"190743\"]\r\n\r\n1. Amounts of substances in chemical reactions are related by their coefficients in the balanced chemical equation.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n
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        Key Takeaway<\/h3>\r\n<\/div>\r\n
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