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