Question

A 61.0 mL sample of a 0.122 M potassium sulfate solution is mixed with 34.5 mL of a 0.120 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 0.999 g . Determine the theoretical yield (mass of PbSO4) , and the percent yield.

Answer 1

Number of moles of K2SO4 = Volume of solution (in L) * molarity = 61/1000 * 0.122 = 0.007442 moles

Number of Pb(C2H3O2)2 = Volume of solution (in L) * molarity = 34.5/1000 * 0.120 = 0.00414 moles

Since Pb(C2H3O2)2 is a limiting reagent, hence number of moles of PbSO4 formed for theoritical yield = 0.00414 moles

Molar mass of PbSO4 = 303.26 gm/mol

Theoritical yield of PbSO4 = 0.00414 moles * 303.26 gm/mol = 1.255 gms

Percent Yield = 0.999/1.255 * 100 = 79.57%