Yes, that's where you're getting screwed up. The mole ratio you want is
0.266 mol FeS x (2 mol Fe2O3/4 mol FeS) = 0.133 as Bob Pursley wrote. You don't need to worry about a ratio of oxygen since it's the percent yield of Fe2O3 you want.
Posted by Sarah H on Friday, May 23, 2008 at 8:10pm.
Hi, please check my work:
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.
My Answer:
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
Chemistry - please help - bobpursley, Friday, May 23, 2008 at 8:50pm
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.
----BOB: WHERE DID U GET THE .133 MOLE FROM? VERY CONFUSED. IKNOW IT'S HARD TO SEE ON HERE, BUT I SET IT UP AS A "MOLE RATIO" TO FIGURE OUT WHAT X AMOUNT OF O2 MOLES WERE (I MULTIPLIED 7 MOL 02 x 0.266 MOLES AND THEN DIVIDED THAT BY 4 MOL FeS TO GET MY 0.466 MOL of 02 end product- but not sure if this is where im getting screwed up.
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