One gram of a mixture of ${\mathrm{Fe}}_3{\mathrm{O}}_4$ and ${\mathrm{Fe}}_2{\mathrm{O}}_3$ is treated with $H_2$ when the following reaction occurs:

$3{\mathrm{Fe}}_2{\mathrm{O}}_3+{\mathrm{H}}_2\to 2{\mathrm{Fe}}_3{\mathrm{O}}_4+{\mathrm{H}}_2\mathrm{O}$

The sample weighs 0.97 g after the above reaction is completed. The per cent of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ in the mixture is about

Molar mass of ${\mathrm{Fe}}_2{\mathrm{O}}_3=(2\times 56+3\times 16)\mathrm{g}{\mathrm{mol}}^{-1}=160\mathrm{g}{\mathrm{mol}}^{-1}$

Molar mass of ${\mathrm{Fe}}_3{\mathrm{O}}_4=(3\times 56+4\times 16)\mathrm{g}{\mathrm{mol}}^{-1}=232\mathrm{g}{\mathrm{mol}}^{-1}$

Loss of mass in the reaction due to conversion of ${\mathrm{Fe}}_2{\mathrm{O}}_3\text{ to }{\mathrm{Fe}}_3{\mathrm{O}}_4\text{ is }(2\times 232-3\times 160)\mathrm{g}{\mathrm{mol}}^{-1}=-16\mathrm{g}{\mathrm{mol}}^{-1}$lf x is the mass of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ in the given mixture, then $\frac{16x}{3\times 160}=(1-0.97)g$ or $x=\frac{0.03\times 480}{16}\mathrm{g}=0.9\mathrm{g}$

Per cent of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ in the mixture $=0.9\times 100=90\mathrm{\%}$