$\mathrm{Al}\left(\mathrm{s}\right)+{\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)\to {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+\mathrm{Fe}\left(\mathrm{s}\right)$ (atomic masses  of $\mathrm{Al}=27\mathrm{U},\mathrm{Fe}=56\mathrm{U},\text{ and }\mathrm{O}=16\mathrm{U}$) the amount of aluminium required to produced 1120 Kg of iron is

The chemical reaction between Al and Fe₂O₃ is

$2\mathrm{Al}\left(\mathrm{s}\right)+{\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)\to {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+2\mathrm{Fe}\left(\mathrm{s}\right)$

$$\begin{gathered} 2\times 27\mathrm{g}\text{ of }\mathrm{Al}\to 2\times 56\mathrm{g}\text{ of }\mathrm{Fe} \\ \mathrm{xg}\text{ of }\mathrm{Al}\to 1120\times 1000\mathrm{g}\text{ of }\mathrm{Fe};(1\mathrm{Kg}=1000\mathrm{g}) \\ \begin{array}{l}\mathrm{x}=\frac{2\times 27\mathrm{gAl}}{2\times 56\mathrm{Fe}}\times 1120\times 1000\mathrm{g}\text{ of }\mathrm{Fe}\\ \therefore \mathrm{x}=540\mathrm{kg}\end{array} \end{gathered}$$