Question

A 62.0 mL sample of a 0.122 M potassium sulfate solution is mixed with 36.5 mL of a 0.108 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 0.992 g . Determine the limiting reactant, the theoretical yield, and the percent yield.

Answer 1

no. of mole = molarity volume of solution in liter.

no. of mole of potassium sulphate = 0.122 0.062 = 0.007564 mole

no. of mole of lead(ll) acetate = 0.108 0.0365 = 0.003942 mole

According to reaction potassium sulphate and lead(ll) acetate react with equimolar proportion (1:1) that mean

1 mole of potassium sulphate react with 1 mole of  lead(ll) acetate then 0.007564 mole potassium sulphate require 0.007564 mole of  lead(ll) acetate for complete reaction but lead(ll) acetate given 0.003942 mole therefore

lead(ll) acetate is limiting reactant.

According to reaction 1mole lead(ll) acetate produce 1 mole of lead sulphate then 0.003942 mole of lead(ll) acetate produce 0.003942 mole of lead sulphate.

molar mass of lead sulphate = 303.26 gm/mol

1 mole of lead sulphate = 303.26 gm then 0.003942 mole of lead sulphate = 0.003942 303.26 = 1.195 gm

1.195 gm lead sulphate is therotical yield.

1.195 gm yield = 100% yield then

0.992 gm = 0.992 100/1.195 = 83.0125 %

83.0125% is percetage yield.