Question

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

Answer 1

Approx. `2*mol` metal, i.e. approx. `54*g`.....

Explanation:

We write the oxidation reaction......

`Al(s) rarr Al^(3+) + 3e^(-)` `E^@=1.662*V`

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

`H^(+) +e^(-) rarr 1/2H_2(g)` `E^@=0*V`

And thus......

`Al(s) +3H^+ rarrAl^(3+) + 3/2H_2(g)` `;E_"cell"^@=1.662*V`

And because `E_"cell"^@` 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.