Question

You may want to reference (Page) Section 4.4 while completing this problem.

A 61.5 mL sample of a 0.122 M potassium sulfate solution is mixed with 37.0 mL of a 0.122 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.

Answer 1

Moles of K2SO4 = 0.122*61.5 / 1000 = 0.007503

Moles of lead acetate = 0.122*37 / 1000 = 0.004514

From balanced reaction,

1 mole K2SO4 requires 1 mole of lead acetate

However, actual moles of K2SO4 > actual moles of lead acetate

Hence, lead acetate is limiting reagent and will drive the yield of PbSO4. .... Answer

Again from reaction,

1 mole lead acetate produces 1 mole PbSO4

So, moles of PbSO4 = moles of lead acetate = 0.004514

Mass of PbSO4 (theoretical yield )= moles * molar mass = 0.004514 * 303 = 1.368 grams ..... Answer

% yield = 1*100/1.368 = 73.1 % .... Answer