A 0.50 g sample containing only anhydrous FeCl3 ( molar mass: 162.5gmol−1) and AlCl3 (molar mass: 133.5 g mol−1 yielded 1.435g of  AgCl (molar mass: 143.5gmol−1).  The mass of FeCl3 in the sample is

# 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: ${\mathrm{mol}}^{-1}$ 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

1. A

0.3126 g

2. B

0.4157 g

3. C

0.2345 g

4. D

0.1567 g

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### Solution:

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

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