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Using Mole-Mole Conversions from Balanced Equations
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Mass and energy are conserved It s the Law Unfortunately, this means that there s no such thing as a free lunch, or any other type of free meal Ever On the other hand, the conservation of mass makes it possible to predict how chemical reactions will turn out 8 describes why chemical reaction equations should be balanced for equal mass in reactants and products You balance an equation by adjusting the coefficients that precede reactant and product compounds within the equation Balancing equations can seem like a chore, like taking out the trash But a balanced equation is far better than any collection of coffee grounds and orange peels because such an equation is a useful tool After you ve got a balanced equation, you can use the coefficients to build mole-mole conversion factors These kinds of conversion factors tell you how much of any given product you get by reacting any given amount of reactant This is one of those calculations that makes chemists particularly useful, so they needn t get by on looks and charm alone
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Consider the following balanced equation for generating ammonia from nitrogen and hydrogen gases: N2(g) + 3H2(g) 2NH3(g) Industrial chemists around the globe perform this reaction, humorlessly fixating on how much ammonia product they ll end up with at the end of the day (In fact, clever methods for improving the rate and yield of this reaction garnered Nobel Prizes for two German gentlemen, die Herren Haber und Bosch) In any event, how are chemists to judge how closely their reactions have approached completion The heart of the answer lies in a balanced equation and the mole-mole conversion factors that spring from it For every mole of dinitrogen reactant, a chemist expects 2 moles of ammonia product Similarly, for every 3 moles of dihydrogen reactant, the chemist expects 2 moles of ammonia product These expectations are based on the coefficients of the balanced equation and are expressed as mole-mole conversion factors as shown in Figure 9-1
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2 mol NH3 1 mol N2
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Figure 9-1: Building mole-mole conversion factors from a balanced equation
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N2 + 3 H2
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2 NH3
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2 mol NH3 3 mol H2
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Q A
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How many moles of ammonia can be expected from the reaction of 278 moles of N2 gas 556 moles of ammonia Begin with your known quantity, the 278 moles of dinitrogen that is to be reacted Multiply that quantity by the mole-mole conversion factor that relates moles of dinitrogen to
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moles of ammonia Use the orientation of the conversion factor that places mol NH3 on top and mol N2 on the bottom This way, the mol N2 units cancel, leaving you with the desired units, mol NH3: 278 mol N 2 2 mol NH 3 = 556 mol NH 3 1 1 mol N 2
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9: Putting Stoichiometry to Work
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One source of dihydrogen gas is from the electrolysis of water, in which electricity is passed through water to break hydrogenoxygen bonds, yielding dihydrogen and dioxygen gases: 2H2O(l) 2H2(g) + O2(g) a How many moles of dihydrogen gas result from the electrolysis of 784 moles of water b How many moles of water are required to produce 905 moles of dihydrogen c Running the electrolysis reaction in reverse constitutes the combustion of hydrogen How many moles of dioxygen are required to combust 846 moles of dihydrogen
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Aluminum reacts with copper (II) sulfate to produce aluminum sulfate and copper, as summarized by the skeleton equation: Al(s) + CuSO4(aq) Al2(SO4)3(aq) + Cu(s) a Balance the equation b How many moles of aluminum are needed to react with 1038 moles of copper (II) sulfate c How many moles of copper are produced if 208 moles of copper (II) sulfate react with aluminum d How many moles of copper (II) sulfate are needed to produce 096 moles of aluminum sulfate e How many moles of aluminum are needed to produce 2001 moles of copper