Question

A 65.0 mL sample of a 0.122 M potassium sulfate solution is mixed with 34.0 mL of a 0.100 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 0.994 g .

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

Answer 1

1)

volume of K2SO4, V = 65.0 mL

= 6.5*10^-2 L

we have below equation to be used:

number of mol in K2SO4,

n = Molarity * Volume

= 0.122*0.065

= 7.93*10^-3 mol

volume of Pb(C2H3O2)2, V = 34.0 mL

= 3.4*10^-2 L

we have below equation to be used:

number of mol in Pb(C2H3O2)2,

n = Molarity * Volume

= 0.1*0.034

= 3.4*10^-3 mol

we have the Balanced chemical equation as:

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

1 mol of K2SO4 reacts with 1 mol of Pb(C2H3O2)2

for 7.93*10^-3 mol of K2SO4, 7.93*10^-3 mol of Pb(C2H3O2)2 is required

But we have 3.4*10^-3 mol of Pb(C2H3O2)2

so, Pb(C2H3O2)2 is limiting reagent

b)

we will use Pb(C2H3O2)2 in further calculation

Molar mass of PbSO4 = 1*MM(Pb) + 1*MM(S) + 4*MM(O)

= 1*207.2 + 1*32.07 + 4*16.0

= 303.27 g/mol

From balanced chemical reaction, we see that

when 1 mol of Pb(C2H3O2)2 reacts, 1 mol of PbSO4 is formed

mol of PbSO4 formed = (1/1)* moles of Pb(C2H3O2)2

= (1/1)*3.4*10^-3

= 3.4*10^-3 mol

we have below equation to be used:

mass of PbSO4 = number of mol * molar mass

= 3.4*10^-3*3.033*10^2

= 1.03 g

Answer: 1.03 g

% yield = actual mass*100/theoretical mass

= 0.994*100/1.03

= 96.4 %

Answer: 96.4 %