A 0.50 g sample containing only anhydrous ${\mathrm{FeCl}}_3$ ( molar mass: $162.5\mathrm{g}{\mathrm{mol}}^{-1}$) and ${\mathrm{AlCl}}_3$ (molar mass: $133.5 \mathrm{g}$${\mathrm{mol}}^{-1}$ yielded $1.435\mathrm{g}$ of  $Ag\mathrm{Cl}$ (molar mass: $143.5\mathrm{g}{\mathrm{mol}}^{-1}$).  The mass of ${\mathrm{FeCl}}_3$ in the sample is

Amount of AgCl obtained, $\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}=\frac{1.435\mathrm{g}}{143.5{\mathrm{gmol}}^{-1}}=0.01\mathrm{mol}$

Let x be the mass of ${\mathrm{FeCl}}_3$, then

Amount of ${\mathrm{FeCl}}_3$ $=\frac{x}{162\mathrm{g}{\mathrm{mol}}^{-1}}$ and  Amount of ${\mathrm{AlCl}}_3=\frac{0.50\mathrm{g}-x}{133.5\mathrm{g}{\mathrm{mol}}^{-1}}$

Hence,  $3\left[\frac{x}{162\mathrm{g}{\mathrm{mol}}^{-1}}+\frac{0.50\mathrm{g}-x}{133.5\mathrm{g}{\mathrm{mol}}^{-1}}\right]=0.01\mathrm{mol}$

Solving for x, we get

$\mathrm{x}=\frac{1}{(28.5)}\left[\frac{0.01\times 162\times 133.5}{3}-81\right]\mathrm{g}=0.3126\mathrm{g}$