# Problem: 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:K2SO4(aq) + Pb(C2H3O2)2(aq) → 2KC2H3O2(aq) + PbSO4(s)The solid PbSO4 is collected, dried, and found to have a mass of 1.00 g .Determine the limiting reactant, the theoretical yield, and the percent yield.

###### FREE Expert Solution

Step 1

= 5.61x10-3 mol K2SO4

= 3.99x10-3 mol Pb(C2H3O2)2

84% (84 ratings) ###### Problem Details

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:

K2SO4(aq) + Pb(C2H3O2)2(aq) → 2KC2H3O2(aq) + PbSO4(s)

The solid PbSO4 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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