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.
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.
It appears that your calculations may have been incorrect. Let's break down the steps to determine the correct % yield in this experiment.
1. Calculate the moles of FeS (iron sulfide):
mass of FeS = 23.4 g
molar mass of FeS = 87.92 g/mol
moles of FeS = mass of FeS / molar mass of FeS
2. Use the balanced equation to determine the theoretical moles of O2 (oxygen) required:
4 mol FeS = 7 mol O2
moles of O2 = (moles of FeS / 4) * 7
3. Convert the moles of O2 to grams (theoretical yield):
molar mass of O2 = 32 g/mol (approximately)
mass of O2 = moles of O2 * molar mass of O2
4. Calculate the % yield:
% yield = (actual yield / theoretical yield) * 100%
Let's redo the calculations:
1. moles of FeS = 23.4 g / 87.92 g/mol ≈ 0.266 mol
2. moles of O2 = (0.266 mol / 4) * 7 ≈ 0.466 mol
3. mass of O2 = 0.466 mol * 32 g/mol ≈ 14.9 g
4. % yield = (16.5 g / 14.9 g) * 100% ≈ 110.7%
It seems like there was a mistake in your initial calculation of the theoretical yield. By using the correct formula and values, the % yield is approximately 110.7%.