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%
That equation does not look balanced to me. Check the oxygen on both sides.
To check your work, let's go through the calculations step by step:
1. Calculate the number of moles of FeS:
nFeS = mass of FeS / molar mass of FeS
= 23.4 g / 87.92 g/mol
≈ 0.266 mol
This step is correct.
2. Use the balanced equation to calculate the number of moles of O2:
4 mol FeS : 7 mol O2 :: 0.266 mol FeS : x mol O2
Cross-multiplying gives:
4 mol FeS * x mol O2 = 0.266 mol FeS * 7 mol O2
x mol O2 ≈ 0.466 mol
This step is correct.
3. Calculate the mass of Fe2O3 produced:
mFe2O3 = nFe2O3 * Molar mass of Fe2O3
= 0.466 mol * 159.7 g/mol
≈ 74.4 g
This step is correct.
4. Calculate the percent yield:
% yield = (actual yield / theoretical yield) * 100%
= (16.5 g / 74.4 g) * 100%
≈ 22.2%
Your calculations are correct, and the percent yield is approximately 22.2%. Well done!