Question

Iron ore, which is composed of mostly Fe2O3, is converted to cast iron (Fe) in a blast furnace.

The reaction is given below:

Fe2O3 + 3CO
⟶
2Fe + 3CO2

If 159.2 kg of Fe is obtained from 267.5 kg of ore and the percent yield of the reaction is 95.0% , what is the percent of Fe2O3 in the iron ore? Assume that there is enough CO to react with all the Fe2O3.

Molar mass of Fe2O3 = 159.69 g/mol

Molar mass of Fe = 55.85 g/mol

Answer 1

First we must find the theoretical yield of iron in order to find the exact amount of iron in the original sample of iron ore
Percent yield = actual yield/theoretical yield ×100


95%=159.2/theoretical yield ×100


Rearrange theoretical yield =159.2/0.95

Theoretical yield Fe= 167.5789474 kg

Therefore the amount of iron ore in original sample is 267.5kg iron ore -167.5789474kg Iron =99.9210526 kg÷total amount of Fe2O3(267.5)

0.37 x 100 =37%