Consider the reaction. 2Pb(s)+O2(g)⟶2PbO(s) An excess of oxygen reacts with 451.4 g of lead, forming 339.4 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

2Pb(s) + O₂(g) ⟶ 2PbO(s) ... balanced equation

Since there is an excess of O2, we need not worry about limiting reactants, and yield will depend only on Pb.

atomic mass Pb = 207.2 g/mol

atomic mass PbO = 223.2 g/mol

451.4 g Pb x 1 mol Pb / 207.2 g x 2 mol PbO / 2 mol Pb x 223.2 g PbO/mol = 486.3 g PbO = theoretical yield

% yield = actual yield / theoretical yield (x100%) = 339.4 g / 486.3 g (x100%) = 69.8% yield