# How many moles of solid aluminum, Al(s) will react completely with 6 moles of HCI (aq)?

Sep 16, 2017

#### Answer:

Approx. $2 \cdot m o l$ metal, i.e. approx. $54 \cdot g$.....

#### Explanation:

We write the oxidation reaction......

$A l \left(s\right) \rightarrow A {l}^{3 +} + 3 {e}^{-}$ ${E}^{\circ} = 1.662 \cdot V$

And we also write the reduction reaction.....

${H}^{+} + {e}^{-} \rightarrow \frac{1}{2} {H}_{2} \left(g\right)$ ${E}^{\circ} = 0 \cdot V$

And thus......

$A l \left(s\right) + 3 {H}^{+} \rightarrow A {l}^{3 +} + \frac{3}{2} {H}_{2} \left(g\right)$ ;E_"cell"^@=1.662*V

And because ${E}_{\text{cell}}^{\circ}$ is positive, the oxidation reaction as written is spontaneous. Given the stoichiometry if we got 6 equiv. hydrochloric acid, two equivs of aluminum should be oxidized.