Lead ions can be precipitated from solution with KCI according to the following reaction: Pb (ag) + 2 KOI(aq) PbCla (a) + 2 K (aq) When 28.7g KCI is added to a solution containing 25.8 g Pba PbCla precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.6 g.

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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Part A
Determine the limiting reactant.
O KCI
O Pb2+
Part B
Determine the theoretical yield of PbCl2.
m%3=
Transcribed Image Text:Part A Determine the limiting reactant. O KCI O Pb2+ Part B Determine the theoretical yield of PbCl2. m%3=
Lead ions can be precipitated from solution with KCI
according to the following reaction:
Pb (aq) + 2 KCI(aq) PbCl2 (s) + 2 K" (aq)
When 28.7 g KCI is added to a solution containing 25.8 g
Pb, a PbCla precipitate forms. The precipitate is
filtered and dried and found to have a mass of 29.6 g
Transcribed Image Text:Lead ions can be precipitated from solution with KCI according to the following reaction: Pb (aq) + 2 KCI(aq) PbCl2 (s) + 2 K" (aq) When 28.7 g KCI is added to a solution containing 25.8 g Pb, a PbCla precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.6 g
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