Question

A sample of iron ore haematite contains 80% of pure Fe2O3.Calculate the mass of iron in 40 tonnes of the iron ore.

Answer 1

Answer:

22.3 tonnes of iron is present 40 tonnes of an ore.

Explanation:

Percentage of iron(III) oxide in an ore = 80%

Mass of iron(III) oxide in an 40 tonnes ore = $40 tonnes\times \frac{80}{100}=32 tonnes$

1 tonnes = 1,000,000 g

Mass of iron(III) oxide = 32,000,000 g

Moles of iron(III) oxide =$\frac{32,000,000 g}{159.69 g/mol}=200,388.2 mol$

In 1 mol of iron(III) oxide there are two mol of iron atoms.

Then in 200,388.2 mol of iron(III) oxide there will be:

$2\times 200,388.2 mol=400,776.5 mol$

Mass of iron: $55.85 g/mol\times 400,776.5 mol=2.23\times 10^{7}g=22.3 tonnes$

22.3 tonnes of iron is present 40 tonnes of an ore.