Step 1

$\mathbf{55}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{mL}}\mathbf{}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{102}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{K}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$** = 5.61x10 ^{-3} mol K_{2}SO_{4}**

$\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{mL}}\mathbf{}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{114}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{Pb}{\left({C}_{2}{H}_{3}{O}_{2}\right)}_{\mathbf{2}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$** = 3.99x10 ^{-3} mol Pb(C_{2}H_{3}O_{2})_{2}**

A 60.5 mL sample of a 0.114 M potassium sulfate solution is mixed with 37.5 mL of a 0.116 M lead(II) acetate solution and the following precipitation reaction occurs:

K_{2}SO_{4}(aq) + Pb(C_{2}H_{3}O_{2})_{2}(aq) → 2KC_{2}H_{3}O_{2}(aq) + PbSO_{4}(s)

The solid PbSO_{4} is collected, dried, and found to have a mass of 1.00 g .

Determine the limiting reactant, the theoretical yield, and the percent yield.

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