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70 Chapter 4 Chemical Quantities and Aqueous Reactions g Pb²⁺ mol Pb²⁺ mol PbCl₂ 1 mol Pb²⁺ 1 mol PbCl₂ 207.2 g Pb²⁺ 1 mol Pb²⁺ then mol PbCl₂ g PbCl₂ then determine % yield 278.1 g PbCl₂ actual yield g PbCl₂ 100% mol PbCl₂ theoretical yield g PbCl₂ Solution: 28.5 g KCI X 74.55 1 KCI 1 2 mol PbCl₂ = 0.1911 mol PbCl₂ 1 mol 25.7 207.2 1 1 mol PbCl₂ = 0.1240 mol PbCl₂ Pb²⁺ is the limiting reactant. 0.1240 278.1 1 g PbCl₂ = 34.5 g PbCl₂ 29.4 X 100% II 85.2% Check: The theoretical yield has the correct units (g PbCl₂) and has a reasonable magnitude compared to the mass of Pb²⁺, the limiting reactant. The % yield is reasonable, under 100%. 4.51 Given: 136.4 kg NH₃; 211.4 kg CO₂; 168.4 kg Find: limiting reactant; theoretical yield % yield Conceptual Plan: kg NH₃ g mol mol 1000 g 1 mol NH₃ 1 mol smaller amount determines limiting reactant 1 kg 17.04 g NH₃ 2 mol NH₃ kg CO₂ g mol mol 1000 g 1 mol 1 mol 1 kg 44.01 g CO₂ 1 mol then mol g kg then determine % yield 60.06 g actual yield kg 100% 1 mol 1000 g theoretical yield kg CH₄N₂O Solution: 136.4 X 1000 1kg X 1 1 mol 2 = 4002.3 mol 211.4 1 X 1 mol 1 = 4803.5 mol CH₄N₂O NH₃ is the limiting reactant. 4002.3 X 60.06 1 1 kg = 240.38 CH₄N₂O 100% = 70.06% 240.38 Check: The theoretical yield has the correct units (kg and has a reasonable magnitude compared to the mass of NH₃, the limiting reactant. The % yield is reasonable, under 100%. Solution Concentration and Solution Stoichiometry 4.53 (a) Given: 3.25 mol LiCl; 2.78 L solution Find: molarity LiCl Conceptual Plan: mol LiCl, L solution molarity amount of solute (in moles) molarity (M) = volume of solution (in L) 3.25 mol LiCl Solution: = 1.169 M = 1.17 M 2.78 L solution Check: The units of the answer (M) are correct. The magnitude of the answer is reasonable. Concentrations are usually between 0 M and 18 M. Copyright © 2017 Pearson Education, Inc.