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Unformatted text preview: Chapter 5: Chemical Reactions 171 End-of-Chapter Solutions for Chapter 5 Summary Problem Answer: (a) Au is oxidized. (b) O 2 is reduced. (c) O 2 is the oxidizing agent. (d) Au is the reducing agent. (e) 3.7 g NaCN (f) 190 g Au (g) 95 g NaCN (h) 26 L NaCN solution Strategy and Explanation: Given a reaction, identify what substance is oxidized, what substance is reduced, the oxidizing agent and the reducing agent. Given a volume and molarity of a solution, determine the mass of solute needed to prepare it. Given the percent mass of a reactant by weight in an ore, determine how many grams of another reaction would be needed to react. Given the concentration of a solution, determine the volume of solution needed to react with a specific mass of reactant. (a) Find the oxidation numbers (Ox. #s) that change during the reaction: 2 Au + 4 CN + O 2 + 2 H 2 O 4 Au(CN) 2 + 4 OH The CN remains unaltered, though it does bind with the metal in the products. Using the rules described on page 185 in Section 5.4, Ox. # H = +1 when combined with nonmetals, and Ox. # O = 2, so these atoms do not change oxidation state. In the reactant, Ox. # Au = 0, since it is in elemental form. In the products, the metal is combined with CN . Ox. # Au + 2(charge of CN ) = 1 Ox. # Au + 2(1) = 1 Ox. # Au = +1 In the reactant, Ox. # O = 0, since it is in elemental form, O 2 . In the products, the O atom is part of OH , so Ox. # Au = 1. The reactant that is losing electrons is oxidized. Because the Au oxidation number increases (0 to +1), it is losing electrons. Au is oxidized. (b) The reactant that is gaining electrons is reduced. Because the O oxidation number decreases (0 to 1), it is losing electrons. O 2 is reduced. (c) Oxidizing agents assist in the oxidation of another species. The reactant reduced is the oxidizing agent. O 2 is reduced, so O 2 is the oxidizing agent. (d) Reducing agents assist in the reduction of another species. The reactant oxidized is the reducing agent. Au is oxidized, so Au is the reducing agent. (e) Use the molarity as a conversion factor to determine moles of solute. Then use the molar mass to find grams of solute. 1.0 L solution " 0.075mol NaCN 1 L solution " 49.0072 g NaCN 1 mol NaCN = 3.7 g NaCN (f) First, determine the number of grams, using a metric conversion. Then use the percent by mass as a conversion factor to determine the mass of Au (see Section 3.9). If an ore has 0.019% Au by weight, then a 100.0-g sample of ore contains 0.019 g Au). 1000 kg ore (exact) " 1000 g ore 1 kg ore " 0.019 g Au 100 g ore = 1.9 " 10 2 g Au (g) First, determine the moles of Au, using the molar mass as a conversion factor. Then use the stoichiometry of the equation to determine the moles of CN . Use the stoichiometry of the formula to determine the moles of NaCN. Then, use the molar mass of NaCN to find the mass. NaCN....
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