# Question #284f8

Sep 20, 2017

4 moles of Aluminum

#### Explanation:

If we assume that we are creating Aluminum Chloride from the
metal-acid reaction between Aluminum and hydrochloric acid we can begin with the balanced chemical reaction below.

$2 A l + 6 H C l \to 2 A l C {l}_{3} + 3 {H}_{2}$

We can use the mole ratio $\frac{2 m o l A l}{2 m o l A l C {l}_{3}}$

The stoichiometry equation becomes

$4 m o l A l C {l}_{3} \frac{2 m o l A l}{2 m o l A l C {l}_{3}}$

$4 \cancel{m o l A l C {l}_{3}} \frac{\cancel{2} m o l A l}{\cancel{2} \cancel{m o l A l C {l}_{3}}}$

$= 4 m o l A l$