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 + 4SO2
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
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)
= about 77 percent
check that thinking. In your second step you solve for xO2. Why?
Your calculations are correct, but it seems like there is an error in the balanced equation you provided. As pointed out by drwls, the equation doesn't appear to be balanced.
To double-check, let's balance the equation:
4FeS + 11O2 -> 2Fe2O3 + 4SO2
Now that we have the correctly balanced equation, we can proceed with the calculations:
1. Calculate the number of moles of FeS:
nFeS = mFeS / MFeS
nFeS = 23.4 g / 87.92 g/mol
nFeS = 0.266 mol
2. Use the mole ratio from the balanced equation to find the moles of O2 required:
4 mol FeS = 11 mol O2
0.266 mol FeS = 11 mol O2 / 4 mol FeS * 0.266 mol FeS
xO2 = 0.732 mol
3. Calculate the theoretical yield of Fe2O3:
mFe2O3 = nFe2O3 * MFe2O3
mFe2O3 = 0.732 mol * 159.7 g/mol
mFe2O3 = 116.8 g
4. Calculate the percent yield:
% yield = (actual yield / theoretical yield) x 100%
% yield = (16.5 g / 116.8 g) x 100%
% yield = 14.1%
So the corrected percent yield is 14.1%.