Math  /  Numbers & Operations

Question(83. Lead ions can be precipitated from solution with KCl according to the reaction: Pb2+(aq)+2KCl(aq)PbCl2(s)+2 K+(aq)\mathrm{Pb}^{2+}(a q)+2 \mathrm{KCl}(a q) \longrightarrow \mathrm{PbCl}_{2}(s)+2 \mathrm{~K}^{+}(a q)
When 28.5 g KCl is added to a solution containing 25.7 g Pb2+25.7 \mathrm{~g} \mathrm{~Pb}^{2+}, a PbCl2\mathrm{PbCl}_{2} 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\mathrm{PbCl}_{2}, and percent yield for the reaction.

Studdy Solution
The **limiting reactant** is Pb2+.
The **theoretical yield** of PbCl2 is 34.5 g\text{34.5 g}.
The **percent yield** of the reaction is \text{85.2%}.

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