Question

A 61.0mL sample of a 0.112M potassium sulfate solution is mixed with 35.0mL of a 0.104M 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 0.997g .

Determine the limiting reactant, the theoretical yield, and the percent yield.

Part A.

Identify the limiting reactant.

KC2H3O2
Pb(C2H3O2)2
K2SO4
PbSO4

Answer 1

Answer:  the limiting reactant is the lead(II) acetate solution

The theoretical yield is 1.092 g PbSO4

The percent yield is 91,3 %

Explanation:

We verify  which reagent is the limiting one by comparing the amount of product  formed with each reactant, and the one with the lowest result  is the limiting reactant.

K2SO4(aq)+Pb(C2H3O2)2(aq) → 2KC2H3O2(aq)+PbSO4(s)

we multiply the giving volumen and the concentration, to find out the moles available for reacting and then we use the stoichiometric relation to find out the amount of the PbSO4 generated

• with 61.0mL of 0.112M K2SO4:  

$0.061L x \frac{0.112 moles}{1L} x \frac{1mol PbSO4}{1mol K2SO4} x \frac{303.6g PbSO4}{1mol PbSO4} =2.06g PbSO4$

• with 35.0mL of 0.104M lead(II) acetate solution

$0.035L x \frac{0.104moles}{1L} x \frac{1mol PbSO4}{1 mol. lead(II) acetate} x\frac{303.26g PbSO4}{1 mol PbSO4} = 1.092 g PbSO4$

By comparing , the between 2.06g and 1.092g , the lower one is the limiting reactant.

So the lead(II) acetate solution is the limiting reactant. and after knowning this, we continue calculating the theoretical yield, and the percent yield.

The theorical yield is the amount of product that will be produced when all the limiting reactant is consumed (100%);  the answer is already calculated above : 1.092g PbSO4 will be produced theorically.

Now the percent yield is the actual product / theorical product x 100

Percent yield = 0.997g / 1.092g X100

Percent yield = 91.3 %