Balanced equation: Pb2+ (aq) + 2 KCl (aq) → PbCl2 (s) + 2 K+ (aq)
Mole to mole comparison:
limiting reactant: forms the less amount of product
The mass of PbCl2 formed by KCl is:
Lead ions can be precipitated from solution with KCl according to the following reaction:
Pb2+ (aq) + 2 KCl (aq) → PbCl2 (s) + 2 K+ (aq)
When 28.6 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.
Determine the limiting reactant.
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