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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 + 4SO2
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%

Responses

gr. 12 chemistry - drwls, Saturday, May 24, 2008 at 5:08pm
That equation does not look balanced to me. Check the oxygen on both sides.

-this was the balanced equation out of the textbook...i can't believe the textbook's wrong (again)...

you should get half the moles you started with.

expected yield=.133 moles or .133*molmassFe2O3

percentyield=molesproduct/expectedmoles

= (16.5g/159.7)/(.133)

check that thinking. In your second step you solve for xO2. Why?

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