Question

calculate the mass of iron (iii) oxide from which ten kilogram of iron was produced.

Answer 1

Answer: The mass of iron (III) oxide used will be 14.324 kg.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

We are given:

Given mass of iron = 10 kg = 10000 g   (Conversion Factor: 1 kg = 1000 g)

Molar mass of iron = 55.845 g/mol

Putting values in above equation:

$\text{Moles of iron}=\frac{10000g}{55.845g/mol}=179.06mol$

The chemical equation for the decomposition of iron (iii) oxide follows:

$2Fe_2O_3\rightarrow 4Fe+3O_2$

By Stoichiometry of the reaction:

4 moles of iron are produced from 2 moles of iron (iii) oxide.

So, 179.06 moles of iron are produced from = $\frac{2}{4}\times 179.06=89.53mol$ of iron (iii) oxide.

Now, to calculate the mass of iron (iii) oxide, we use equation 1, we get:

Molar mass of iron (iii) oxide = 160 g/mol

Moles of iron (iii) oxide = 89.53 mol

Putting values in equation 1, we get:

$89.53mol=\frac{\text{Mass of iron (iii) oxide}}{160g/mol}\\\\\text{Mass of }Fe_2O_3=14324.8g=14.324kg$

Hence, the mass of iron (III) oxide used will be 14.324 kg.