# How would we represent the reaction between hydrochloric acid and aluminum hydroxide?

Sep 22, 2016

We need the stoichiometric equation:

$A l {\left(O H\right)}_{3} \left(s\right) + 3 H C l \left(a q\right) \rightarrow A l C {l}_{3} \left(a q\right) + 3 {H}_{2} O \left(l\right)$

#### Explanation:

Perhaps more realistically, we could write:

$A {l}_{2} {O}_{3} \left(s\right) + 6 H C l \left(a q\right) \rightarrow 2 A {l}^{3 +} + 3 {H}_{2} O \left(l\right)$

In each equation the stoichiometry is identical: 1 equiv aluminum reacts with 3 equiv hydrochloric acid.

$\text{Moles of aluminum}$ $=$ $\frac{0.80 \cdot g}{78.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.0 \times {10}^{-} 2 \cdot m o l$.

And thus we need 3 equiv of the given acid:

$=$ $\frac{3 \times 1.0 \times {10}^{-} 2 \cdot \cancel{m o l}}{6.0 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1}} \times {10}^{3} \cdot m L \cdot \cancel{{L}^{-} 1}$ $\cong$ $5 \cdot m L$