An unbalanced equation is shown. in this reaction, 200.0 g of fes2 is burned in 100.0 g of oxygen, and 55.00 g of fe2o3 is produced.
4fes2 + 11o2 mc021-1.jpg fe2o3 + so2
what is the percent yield of fe2o3?

4FeS2 + 11O2 = 2Fe2O3 + 8SO2

Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.

Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3

To find for the theoretical yield, we first determine the limiting reactant.

100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2

Therefore, the limiting reactant is O2.

Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3

Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%

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