Lead ions can be precipitated from solution with NaCl according to the reaction: Pb²+ (aq) + 2 NaCl(aq) → PbCl₂(s) + 2 Na+ (aq) When 135.8 g of NaCl 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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65. Lead ions can be precipitated from solution with NaCl
according to the reaction:
Pb²+ (aq) + 2 NaCl(aq)
PbCl₂(s) + 2 Na+ (aq)
When 135.8 g of NaCl 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) + 2 NaCl(aq) PbCl₂(s) + 2 Na+ (aq) When 135.8 g of NaCl 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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