# What are the products of the following reaction? 2Al(s) + 6H^+ (aq) ->

Jul 2, 2016

Aluminium cations and hydrogen gas.

#### Explanation:

You were given the reactants in the net ionic equation that describes the reaction between aluminium metal, $\text{Al}$, and hydrochloric acid, $\text{HCl}$.

Hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to form hydrogen ions, ${\text{H}}^{+}$, which you'll very often see referred to hydronium cations, ${\text{H"_3"O}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$.

${\text{HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

You can thus rewrite the equation as

2"Al"_ ((s)) + 6"H"_ ((aq))^(+) + 6"Cl"_ ((aq))^(-) -> ?

Remember, the stoichiometric coefficient of the hydrogen ions must also be distributed to the chloride anions, since

$6 {\text{HCl"_ ((aq)) -> 6"H"_ ((aq))^(+) + 6"Cl}}_{\left(a q\right)}^{-}$

Now, when aluminium reacts with hydrochloric acid, it gets oxidized to aluminium cations, ${\text{Al}}^{3 +}$. At the same time, the hydrogen ions get reduced to hydrogen gas, ${\text{H}}_{2}$.

$2 {\text{Al"_ ((s)) + 6"H"_ ((aq))^(+) + color(red)(cancel(color(black)(6"Cl"_ ((aq))^(-)))) -> "Al"_ ((aq))^(3+) + color(red)(cancel(color(black)(6"Cl"_ ((aq))^(-)))) + "H}}_{2 \left(g\right)} \uparrow$

As you can see ,the chloride anions are spectator ions, which is why the initial equation didn't include them

$2 {\text{Al"_ ((s)) + 6"H"_ ((aq))^(+) -> "Al"_ ((aq))^(3+) + "H}}_{2 \left(g\right)} \uparrow$

Now all you have to do is balance the aluminium and hydrogen atoms

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{2 {\text{Al"_ ((s)) + 6"H"_ ((aq))^(+) -> 2"Al"_ ((aq))^(3+) + 3"H}}_{2 \left(g\right)} \uparrow} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The products of the reaction will thus be aqueous aluminium cations, ${\text{Al}}^{3 +}$, and hydrogen gas, ${\text{H}}_{2}$.

If you want, you can add in the chloride anions to get

$2 {\text{Al"_ ((s)) + 6"H"_ ((aq))^(+) + 6"Cl"_ ((aq))^(-) -> 2"Al"_ ((aq))^(3+) + 6"Cl"_ ((aq))^(-) + 3"H}}_{2 \left(g\right)} \uparrow$

This is equivalent to

$2 {\text{Al"_ ((s)) + 6"HCl"_ ((aq)) -> 2"AlCl"_ (3(aq)) + 3"H}}_{2 \left(g\right)} \uparrow$

The single replacement reaction reaction between aluminium metal and hydrochloric acid produces aqueous aluminium chloride and hydrogen gas.