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Question

2Pb(s)+O2(g)⟶2PbO(s)2Pb(s)+O2(g)⟶2PbO(s)

An excess of oxygen reacts with 451.4 g of lead, forming 372.2 g372.2 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

% yield = actual yield / theoretical yield (x100%)

We are given the actual yield of 372.2 g of PbO. Now we must find the theoretical yield

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

Atomic mass Pb = 207.2 g / mol

Molar mass PbO = 223.2 g / mol

moles Pb used = 451.4 g x 1 mol / 207.2 g = 2.179 mols

theoretical yield of PbO = 2.179 mol Pb x 2 mol PbO / 2 mol Pb x 223.2 g / mol = 486.3 g PbO

% yield = 372.2 g / 486.3 g (x100%) = 76.54%