# Problem: A 55.0-mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and this precipitation reaction occurs: K2SO4(aq) + Pb(C2H3O2)2(aq) → 2 KC2H3O2(aq) + PbSO4(s)The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Identify the limiting reactant.

###### FREE Expert Solution

= 5.61x10-3 mol K2SO4

= 3.99x10-3 mol Pb(C2H3O2)2

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###### Problem Details

A 55.0-mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and this precipitation reaction occurs: K2SO4(aq) + Pb(C2H3O2)2(aq) → 2 KC2H3O2(aq) + PbSO4(s)

The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Identify the limiting reactant.

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