In an experiment, 23.4 g of FES are added to excess oxygen and 16.5 g of FE2O3 are produced. The balanced equation is:
4FeS = 7O2 --> 2Fe2O3 + 2SO2
Calculate the % yield.

nFeS=23.4g/87.92 g/mol
=0.266 mol

4mol FeS / 0.266 mol = 7mol O2/ x02
xO2=0.466 mol

mFe2O3 = nXM
= 0.466 mol x 159.7 g/mol
= 74.4 g

% yield = actual yield / theoretical yield X 100%
=16.5 g/74.4 g x 100%
=22.2%

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1. You started with .266 mol
you should have ended with .133 mole of product.

Actual product moles: 16.5/160 = .103

Yield: .104/.133= 77 percent.

check my thinking.

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