# A mass of 45.0 grams of aluminum will react with how many grams of zinc chloride?

Apr 12, 2016

$328.6 \frac{g}{m o l}$

#### Explanation:

In order to solve this problem we need to begin with a balanced chemical equation.

Zinc Chloride $\left(Z n C {l}_{2} - Z {n}^{+} 2 \mathmr{and} C {l}^{-} 1\right)$
Reacts with Aluminum in solution to form
Aluminum Chloride $\left(A l C {l}_{3} - A {l}^{+} 3 \mathmr{and} C {l}^{-} 1\right)$

$2 A l + 3 Z n C {l}_{2} \to 2 A l C {l}_{3} + 3 Z n$

Now we must use stoichiometry to convert grams of Al to grams of ZnCl_2

molar mass of aluminum is $27.99 \frac{g}{m o l}$

molar mass of zinc chloride is $136.29 \frac{g}{m o l}$
Zn $65.39$
Cl $2 \left(35.45\right) = 70.90$
$65.39 + 70.90 = 136.329$

the mole ratio of $Z n C {l}_{2}$ to $A l$ is $3 : 2$

grams $\to$ moles $\to$ moles $\to$ grams

$45 g A l$ x $\frac{1 \cancel{m o l A l}}{27.99 \cancel{g A l}}$ x $\frac{\cancel{3 \left(m o l Z n C {l}_{2}\right)}}{2 \cancel{m o l A l}}$ x $\frac{136.29 g Z n C {l}_{2}}{1 \cancel{m o l Z n C {l}_{2}}}$ =

$\frac{45 \left(3\right) \left(136.29\right)}{\left(27.99\right) \left(2\right)}$ = $328.6 \frac{g}{m o l}$