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68 Chapter 4 Chemical Quantities and Aqueous Reactions 4.43 Given: 4.2 mol ZnS; 6.8 mol O₂ Find: mole amount of excess reactant left Conceptual Plan: mol ZnS mol ZnO 2 mol ZnO smaller mol amount determines limiting reactant 2 mol ZnS mol O₂ mol ZnO 2 mol ZnO 3 mol O₂ mol limiting reactant mol excess reactant required mol excess reactant left 3 mol O₂ 2 Solution: 4.2 mol X 2 2 mol ZnO = 4.2 mol ZnO 6.8 mol X 2 3 = 4.5 mol ZnO mol ZnO ZnS is the limiting reactant; therefore, O₂ is the excess reactant. 4.2 X 2 3 molZnS mol O₂ = 6.3 mol O₂ required 6.8 mol O₂ - 6.3 mol O₂ = 0.5 mol O₂ left Check: The units of the answer (mol O₂) are correct. The magnitude is reasonable because it is less than the original amount of O₂. 4.45 (a) Given: 2.0 g Al; 2.0 g Cl₂ Find: theoretical yield in g Conceptual Plan: g Al mol Al mol 1 mol Al 2 mol smaller mol amount determines limiting reactant 26.98 g Al 2 mol Al g mol mol 1 mol Cl₂ 2 mol 70.90 g Cl₂ 3 mol Cl₂ then mol g 133.33 g mol Solution: 2.0 26.98 1 X 2 2 mol mol = 0.074 mol 2.0 X 70.90 1 mol X 2 3 mol = 0.0188 mol 0.0188 mol X 133.33 = 2.5 g mol Check: The units of the answer (g are correct. The answer is reasonable because Cl₂ produced the smaller amount of product and is the limiting reactant. (b) Given: 7.5 g Al; 24.8 g Cl₂ Find: theoretical yield in g Conceptual Plan: g Al mol Al mol 1 mol Al 2 mol smaller mol amount determines limiting reactant 26.98 g Al 2 mol Al g Cl₂ mol Cl₂ mol 1 mol Cl₂ 2 mol 70.90 g Cl₂ 3 mol Cl₂ then mol g 133.33 g mol Copyright © 2017 Pearson Education, Inc.