Lead ions can be precipitated from solution with KCl according to the reaction: Pb2+(aq)
Chapter 4, Problem 49(choose chapter or problem)
Lead ions can be precipitated from solution with KCl according to the reaction: Pb2+(aq) + 2 KCl(aq) S PbCl2(s) + 2 K+(aq) When 28.5 g KCl is added to a solution containing 25.7 g Pb2+ , a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction.
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