Question

Solid lead sulfate can be produced by the following reaction. If 10.0 g of Pb and 100. mL of 1.50 M H2SO4 are mixed, a) what is the theoretical yield of lead sulfate? b) If the actual yield was 19.8 g, what was the percent yield? Pb(s) + PbO2 + 2H2SO4 → 2PbSO4(s) + 2H2O(l)

Answer 1

moles of Pb = 10.0 / 207.2 = 0.0483

moles of H2SO4 = 100 x1.50 / 1000 = 0.15

Pb(s) + PbO2 + 2H2SO4 ------------> 2PbSO4(s) + 2H2O(l)

1                            2                               2

0.0483                0.15                            ?

limiting reagent is Pb

moles of PbSO4 fomred = 2 x 0.0483 = 0.0966

a)

theoretical yield of lead sulfate = 0.0966 x 303.3

                                                 = 29.3 g

b)

percent yield = actual yield x 100 / theoretical yield

                     = 19.8 x 100 / 29.3

                    = 67.6 %