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Lead ions can be precipitated from solution with KCI according to the reaction: When 28.5 g KCI (MM = 74.5 g/mol) is added to a solution %3D containing 25.7 g Pb2*(At. wt. = 207.2 g/mol) %3D a 29.4 g PbC12 (MM = 278.2 g/mol) precipitate is obtained. Determine the limiting reactant and the percent yield of PbClz. P+ (ag) + 2 KCI (aq) + PBC1, (s) + 2 K (aq) O 2 KCI is L.R and 85.3 % O KCI is LR and 55.6 % O Pb?*is LR and 55.6 % O Pb2*is L.R and 85.3 %

Lead ions can be precipitated from solution with KCI according to the reaction: When 28.5 g KCI (MM = 74.5 g/mol) is added to a solution %3D containing 25.7 g Pb2*(At. wt. = 207.2 g/mol) %3D a 29.4 g PbC12 (MM = 278.2 g/mol) precipitate is obtained. Determine the limiting reactant and the percent yield of PbClz. P+ (ag) + 2 KCI (aq) + PBC1, (s) + 2 K (aq) O 2 KCI is L.R and 85.3 % O KCI is LR and 55.6 % O Pb?*is LR and 55.6 % O Pb2*is L.R and 85.3 %

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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Lead ions can be precipitated from solution with KCI according to the reaction: When 28.5 g KCI (MM = 74.5 g/mol) is added to a solution containing 25.7 g Pb2*(At. wt. = 207.2 g/mol) %3D a 29.4 g PbC12 (MM = 278.2 g/mol) precipitate is obtained. Determine the limiting reactant and the percent yield of PbClz. P (ag) + 2 KCI (aq) - PbCl, (s) + 2 K' (aq) O 2 KCI is L.R and 85.3 % O KCI is LR and 55.6 % O Pb2*is L.R and 55.6 % O Pb2*is L.R and 85.3 % O KCI is LR and 85.3 %
Transcribed Image Text:Lead ions can be precipitated from solution with KCI according to the reaction: When 28.5 g KCI (MM = 74.5 g/mol) is added to a solution containing 25.7 g Pb2*(At. wt. = 207.2 g/mol) %3D a 29.4 g PbC12 (MM = 278.2 g/mol) precipitate is obtained. Determine the limiting reactant and the percent yield of PbClz. P (ag) + 2 KCI (aq) - PbCl, (s) + 2 K' (aq) O 2 KCI is L.R and 85.3 % O KCI is LR and 55.6 % O Pb2*is L.R and 55.6 % O Pb2*is L.R and 85.3 % O KCI is LR and 85.3 %
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