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%
1 answer
That equation does not look balanced to me. Check the oxygen on both sides.
0 answers