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. Determine the percent yield.

FREE Expert Solution

Recall molarity (M):

molarity(M)=moles soluteL solution


Calculate the mass of PbSO4 produced from each reactant:

PbSO4 from potassium sulfate (K2SO4)

molar mass PbSO4 = 303.26 g/mol

mass PbSO4=55.0 mL×10-3 L1 mL×0.102 mol K2SO41 L                                 ×1 mol PbSO41 mol K2SO4×303.26 g PbSO41 mol PbSO4

mass PbSO4 = 1.70 g


PbSO4 from lead(II) acetate (Pb(C2H3O2)2)

molar mass PbSO4 = 303.26 g/mol

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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. Determine the percent yield.

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