Question

ron metal is produced in a blast furnace by the reaction of iron (III) oxide and coke (pure carbon). If 25.0 moles of pure Fe2O3 is used, how many grams of iron can be produced?

Answer 1

Solution:
n(Fe2O3)=25.0mol
n(Fe0)-?

1. For calculation n(Fe) we need to calculate m(Fe2O3):
Mr(Fe2O3)=55.85*2+16*3=159.6(g/mol)
n=m/Mr; hereof m(Fe2O3)=n*Mr=159.6(g/mol)*25mol=3990g

2. Compose proportion:
3990(g) it is 159.6
X(g) it is 55.85*2
Where X is the total Iron weight in Fe2O3
Hereof X=3990*55.85*2/159.6=2792.5(g)

3. m(Fe0)=2792.5/55.85=50(mol)
Answer: 50 mol of Iron can be produced from 25mol of Fe2O3