# Question #0a95e

Jul 1, 2015

#### Explanation:

Relative atomic mass of $A l = 27$ and $C l = 35.5$
To form $A l C {l}_{3}$ you need these in the ratio:

$\frac{m \left(C l\right)}{m \left(A l\right)} = \frac{3 \cdot 35.5}{1 \cdot 27} = \frac{106.5}{27} \to$

$m \left(C l\right) = \frac{106.5}{27} \cdot m \left(A l\right) = \frac{106.5}{27} \cdot 13.5 = 53.2 g$

The mass of the $A l C {l}_{3}$
$m \left(A l C {l}_{3}\right) = 13.5 + 53.2 = 66.7 g$