In the reaction ${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3\mathrm{C}\left(\mathrm{s}\right)\to 2\mathrm{Fe}\left(\mathrm{s}\right)+3\mathrm{CO}\left(\mathrm{g}\right),453\mathrm{kg}$ of iron was obtained from 752 kg of a sample of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ The perentage of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ in the sample is (Given:  Molar mass of $\left.\mathrm{Fe}=56\mathrm{g}{\mathrm{mol}}^{-1}\right)$

We have ${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3\mathrm{C}\left(\mathrm{s}\right)\to 2\mathrm{Fe}\left(\mathrm{s}\right)+3\mathrm{CO}\left(\mathrm{g}\right)$

Molar mass $160\mathrm{g}{\mathrm{mol}}^{-1}56\mathrm{g}{\mathrm{mol}}^{-1}$

$2\times 56\mathrm{g}$ of Fe(s)  will be obtained from 160g of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ Mass of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ from which 453kg of Fe is obtained will be

$m=\frac{160\mathrm{g}}{2\times 56\mathrm{g}}\times 453\mathrm{kg}=647.14\mathrm{kg}$

Hence, Per cent purity of sample $=\frac{647.14\mathrm{kg}}{752\mathrm{kg}}\times 100=86\mathrm{\%}$