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102 Chapter Three Stoichiometry Before doing any calculations involving a chemical reaction, be sure the equation for the reaction is balanced. 3 ~ Because all the molecules in this equation are of about equal complexity, where we start in balancing it is rather arbitrary. Let's begin by balancing the hydrogen. A coefficient of 2 for NH3 and a coefficient of 3 for H 2 0 give six atoms of hydrogen on both sides: 2NH 3 (g) + 02(g) -+ NO(g) + 3H 2 0(g) The nitrogen can be balanced with a coefficient of 2 for NO: 2NH 3 (g) + 02(g) -+ 2NO(g) + 3H 2 0(g) Finally, note that there are two atoms of oxygen on the left and five on the right. The OJ(Y- gen can be balanced with a coefficient of ~for 02: 5 2NH 3 (g) + '2 02(g) -+ 2NO(g) + 3H 2 0(g) However, the usual custom is to have whole-number coefficients. We simply multiplv the entire equation by 2. 4NH 3 (g) + 50 2 (g) -+ 4NO(g) + 6H 2 0(g) Reality Check: There are 4N, 12H, and 10 ° on both sides, so the equation isbalanced We can represent this balanced equation visually as + ---. + SEE EXERCISES 3.93 THROUGH 3. 3.10 ~ Stoichiometric Calculations: Amounts
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Unformatted text preview: Reactants and Products As we have seen in previous sections of this chapter, the coefficients in chemical ~ represent numbers of molecules, not masses of molecules. However, when a reaction is ill run in a laboratory or chemical plant, the amounts of substances needed cannot be determizef by counting molecules directly. Counting is always done by weighing. In this section we see how chemical equations can be used to determine the masses of reacting chemicals. To develop the principles for dealing with the stoichiometry of reactions, we will sider the reaction of propane with oxygen to produce carbon dioxide and water. We consider the question: "What mass of oxygen will react with 96.1 grams of propane?" doing stoichiometry, the first thing we must do is write the balanced chemical equ.-for the reaction. In this case the balanced equation is C3H8(g) + 50 2 (g) ~ 3C0 2 (g) + 4H 2 0(g)...
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