Answer to Question #125761 in Inorganic Chemistry for Anelisa

Question

If 300g of iron pyrites is burned in 200g of O2,143 of ferric oxide is produced.What is the percentage yield of ferric oxide?

Answer 1

So, let's write an chemical equation:

4FeS2+11O2=2Fe2O3+8SO2

Let's find the amounts of substances of oxygen and iron pyrite.

n(O2)=m(O2)/M(O2)=200 g/32 g/mol = 6.25 mol

n(FeS2)=m(FeS2)/M(FeS2)=300 g/120 g/mol = 2.5 mol

Iron pyrite is in the excess in this task, because for burning 2.5 mol of iron pyrite, 6.875 mol of oxygen is needed. 

So, the theoretical amount of Fe2O3 forming during this reaction can be found from next ratio:

n(O2)/n(Fe2O3)=11/2 it makes the n(Fe2O3)=2*n(O2)/11 (all n are for theoretical reaction)

n theor. (Fe2O3)=2*6.25/11=1.136 mol

m theor. (Fe2O3)=n(Fe2O3)*M(Fe2O3)=1.136 mol * 160 g/mol = 181.76 g

η=m(pract.)/m(theor.)∗100%=143g/181.76 g *100%=78.68 %

So, the yield is 78.68%.