Question

How much iron can be obtained by the reduction of 1kg of Fe2O3.

Answer 1

700 gms of iron can be obtained by the reduction of 1kg of Fe2O3.

Answer 2

The balance chemical equation for the reduction of ferric oxide is:  

$\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{Fe}+3 \mathrm{H}_{2} \mathrm{O}$

Given:  

$\begin{array}{l}{\text { Mass of } \mathrm{Fe}_{2} \mathrm{O}_{3}=1 \mathrm{kg}=1000 \mathrm{g}} \\ {\text { Molar mass of } \mathrm{Fe}_{2} \mathrm{O}_{3}=160 \mathrm{g} / \mathrm{mol}} \\ {\text { Moles of } \mathrm{Fe}_{2} \mathrm{O}_{3}=\frac{\text { Mass }}{\text { Molar mass }}}\end{array}$

$\begin{array}{c}{=\frac{1000 g}{160 g} m o l^{-1}} \\ {=6.25 \text { moles }}\end{array}$

$\begin{array}{l}{1 \text { mole of } \mathrm{Fe}_{2} \mathrm{O}_{3}=2 \text { moles of Fe }} \\ {\text { Hence } 6.25 \text { moles of } \mathrm{Fe}_{2} \mathrm{O}_{3}=6.25 \times 2=12.5 \text { moles }} \\ {\text { of Fe }} \\ {1 \text { mole Fe }=55.8 \mathrm{g} / \mathrm{mol}} \\ {12.5 \text { moles of Fe }=12.5 \times 55.8} \\ {=697.5 \mathrm{g} \text { of Fe }}\end{array}$