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Lead ions can be precipitated from solution with NaCl according to the reaction:Pb2+
Chapter 8, Problem 65P(choose chapter or problem)
Lead ions can be precipitated from solution with NaCl according to the reaction:
\(\mathrm{Pb}^{2+}(a q)+2 \mathrm{NaCl}(a q) \longrightarrow \mathrm{PbCl}_{2}(s)+2 \mathrm{Na}^{+}(a q)\)
When 135.8 g of NaCl are added to a solution containing 195.7 g of \(\mathrm{Pb}^{2+}\), a \(\mathrm{PbCl}_{2}\) 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 \(\mathrm{PbCl}_{2}\), and percent yield for the reaction.
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QUESTION:
Lead ions can be precipitated from solution with NaCl according to the reaction:
\(\mathrm{Pb}^{2+}(a q)+2 \mathrm{NaCl}(a q) \longrightarrow \mathrm{PbCl}_{2}(s)+2 \mathrm{Na}^{+}(a q)\)
When 135.8 g of NaCl are added to a solution containing 195.7 g of \(\mathrm{Pb}^{2+}\), a \(\mathrm{PbCl}_{2}\) 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 \(\mathrm{PbCl}_{2}\), and percent yield for the reaction.
ANSWER:Step 1 of 5
Theoretical yield is the amount of product that can be made in a chemical reaction based on the amount of limiting reactant.
Actual yield is the amount of product actually produced by a chemical reaction.
\(\text { Percent yield }=\frac{\text { actual yield }}{\text { theoretical yield }} \times 100 \%\)
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