Quantidades Químicas e Reações

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Chapter 4 Chemical Quantities and Aqueous Reactions 69 Solution: 7.5 1 mol X 2 2 mol = 0.2780 mol 24.8 70.90 1 X 2 3 mol = 0.2332 mol 0.2332 mol X 133.33 = 31.1 g mol Check: The units of the answer (g are correct. The answer is reasonable Cl₂ produced the smaller amount of product and is the limiting reactant. (c) Given: 0.235 g Al; 1.15 g Cl₂ Find: theoretical yield in g AICl₃ Conceptual Plan: g A mol Al mol AICI₃ 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 AICl₃ g 133.33 g AICI₃ mol Solution: 0.235 X 26.98 1 X 2 2 mol = 0.008710 mol 1.15 X 70.90 1 X 2 3 mol = 0.01081 mol 0.008710 mol X 133.33 mol = 1.16 g Check: The units of the answer (g are correct. The answer is reasonable because Al produced the smaller amount of product and is the limiting reactant. 4.47 Given: 22.55 g Fe₂O₃; 14.78 g CO Find: mole amount of excess reactant left Conceptual Plan: g Fe₂O₃ mol Fe₂O₃ mol Fe 1 mol Fe₂O₃ 2 mol Fe smaller mol amount determines limiting reactant 159.7 g Fe₂O₃ 1 mol Fe₂O₃ g CO mol CO mol Fe 1 mol CO 2 mol Fe 28.01 g CO 3 mol CO then mol limiting reactant mol excess reactant required mol excess reactant left g excess 3 1 molFe₂O₃ 28.01 g CO reactant left or 1 159.70 1 mol CO Solution: 22.55 g X 1 mol Fe₂O₃ X 1 2 mol Fe = 0.2824 mol Fe 14.78 28.01 1 g X 3 2 mol mol Fe = 0.3518 mol Fe Fe₂O₃ is the limiting reactant; therefore, CO is the excess reactant. 22.55 g X 159.70 1 X 1 3 X 28.01 1 mol CO = 11.865 g CO required 14.78 g CO - 11.87 g CO = 2.91 g CO left Check: The units of the answer (g CO) are correct. The magnitude is reasonable because it is less than the original amount of CO. 4.49 Given: 28.5 g 25.7 g Pb²⁺; 29.4 g PbCl₂ Find: limiting reactant; theoretical yield PbCl₂; % yield Conceptual Plan: g KCI mol KCI mol PbCl₂ 1 mol KCI 1 mol PbCl₂ smaller mol amount determines limiting reactant 74.55 g KCI 2 mol Copyright © 2017 Pearson Education, Inc.