So, let's write an chemical equation:
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%.