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

The percent yield of PbSO4 is 90.6 %.

The equation of the reaction is;

K2SO4(aq) + Pb(C2H3O2)2(aq)----------> 2KC2H3O2(aq) + PbSO4(s)

Number of moles K2SO4 = 61.0/1000 L × 0.112M = 0.0068 moles

Number of moles of Pb(C2H3O2)2 = 35/1000 L × 0.104 = 0.0036 moles

Given that the reaction is 1:1, we can see that Pb(C2H3O2)2 is the limiting reactant.

The theoretical yield of PbSO4 is 0.0036 moles × 303.26 g/mol

= 1.1 g of PbSO4

The percent yield is obtained from;

actual yield/theoretical yield × 100/1

% yield = 0.997g/1.1 g × 100/1

% yield = 90.6 %

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