Question

If a chemist mixes 0.250 L of 0.150 M potassium carbonate with 0.300 L of 0.125 M lead (II) nitrate, and obtains 4.35 g of lead carbonate, what is the percent yield of the reaction?

Answer 1

moles of K2CO3 = 0.250 x 0.150

                           = 0.0375 mol

moles of Pb(NO3)2 = 0.300 x 0.125

                               = 0.0375 mol

Pb(NO3)2 (aq) +   K2CO3 (aq)   ------------> PbCO3 (s) + 2 KNO3 (aq)

         1                          1                                      1

0.0375                       0.0375

moles of PbCO3 formed = 0.0375

mass of PbCO3 formed = 0.0375 x 267.21

                                        = 10.02 g

% yield = (actual / theoretical) x 100

            = (4.35 / 10.02) x 100

percent yield = 43.4 %