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65: Lead ions can be precipitated from solution with NaCl according to the reaction: Pb2 (aq) + 2NaCl(ag) -> PbCl2(s) + 2Na (aq), When 135.8 g of NaCI are added to a solution containing 195.7 g of Pb2*, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 252,4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction,

65: Lead ions can be precipitated from solution with NaCl according to the reaction: Pb2 (aq) + 2NaCl(ag) -> PbCl2(s) + 2Na (aq), When 135.8 g of NaCI are added to a solution containing 195.7 g of Pb2*, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 252,4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction,

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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65: Lead ions can be precipitated from solution with NaCl according to the reaction: Pb2 (aq) + 2NaCl(ag) -> PbCl2(s) + 2Na (aq), When 135.8 g of NaCI are added to a solution containing 195.7 g of Pb2*, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 252,4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction,
Transcribed Image Text:65: Lead ions can be precipitated from solution with NaCl according to the reaction: Pb2 (aq) + 2NaCl(ag) -> PbCl2(s) + 2Na (aq), When 135.8 g of NaCI are added to a solution containing 195.7 g of Pb2*, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 252,4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction,
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