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%
Your calculations are correct, but there seems to be a mistake in your final calculation for the percentage yield. Let's go through the calculations step by step to ensure accuracy:
1. Calculate the moles of FeS:
nFeS = mass / molar mass
nFeS = 23.4 g / 87.92 g/mol
nFeS = 0.266 mol
2. Use the stoichiometry of the balanced equation to calculate the moles of O2:
4 mol FeS = 7 mol O2
0.266 mol FeS = (0.266 mol FeS)*(7 mol O2 / 4 mol FeS) = 0.466 mol O2
3. Calculate the theoretical yield of Fe2O3:
mFe2O3 = nFe2O3 * molar mass
mFe2O3 = 0.466 mol * 159.7 g/mol
mFe2O3 = 74.4 g
4. Calculate the percentage yield:
% yield = (actual yield / theoretical yield) * 100%
% yield = (16.5 g / 74.4 g) * 100%
% yield = 22.2%
Therefore, the correct answer is a percentage yield of 22.2%.