Question #60368

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?

Expert's answer

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

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

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