An excess of oxygen reacts with 451.4 g of lead, forming 367.3 g of lead(II) oxide. Calculate the percent yield of the reaction.

Question

An excess of oxygen reacts with 451.4 g of lead, forming 367.3 g367.3 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

Lead(II) oxide = PbO

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

atomic mass Pb = 207.2 g / mol

molar mass PbO = 223.2 g / mol

Theoretical yield: 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

Percent yield = actual yield / theoretical yield (x100%) = 367.3 g / 486.3 g (x100%) = 75.48% yield