What are the products of the following reaction? 2Al(s) + 6H^+ (aq) ->2Al(s)+6H+(aq)→
1 Answer
Aluminium cations and hydrogen gas.
Explanation:
You were given the reactants in the net ionic equation that describes the reaction between aluminium metal,
Hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to form hydrogen ions,
"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)HCl(aq)→H+(aq)+Cl−(aq)
You can thus rewrite the equation as
2"Al"_ ((s)) + 6"H"_ ((aq))^(+) + 6"Cl"_ ((aq))^(-) -> ?2Al(s)+6H+(aq)+6Cl−(aq)→?
Remember, the stoichiometric coefficient of the hydrogen ions must also be distributed to the chloride anions, since
6"HCl"_ ((aq)) -> 6"H"_ ((aq))^(+) + 6"Cl"_((aq))^(-)6HCl(aq)→6H+(aq)+6Cl−(aq)
Now, when aluminium reacts with hydrochloric acid, it gets oxidized to aluminium cations,
2"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(g)) uarr
As you can see ,the chloride anions are spectator ions, which is why the initial equation didn't include them
2"Al"_ ((s)) + 6"H"_ ((aq))^(+) -> "Al"_ ((aq))^(3+) + "H"_ (2(g)) uarr
Now all you have to do is balance the aluminium and hydrogen atoms
color(green)(|bar(ul(color(white)(a/a)color(black)(2"Al"_ ((s)) + 6"H"_ ((aq))^(+) -> 2"Al"_ ((aq))^(3+) + 3"H"_ (2(g)) uarr)color(white)(a/a)|)))
The products of the reaction will thus be aqueous aluminium cations,
If you want, you can add in the chloride anions to get
2"Al"_ ((s)) + 6"H"_ ((aq))^(+) + 6"Cl"_ ((aq))^(-) -> 2"Al"_ ((aq))^(3+) + 6"Cl"_ ((aq))^(-) + 3"H"_ (2(g)) uarr
This is equivalent to
2"Al"_ ((s)) + 6"HCl"_ ((aq)) -> 2"AlCl"_ (3(aq)) + 3"H"_ (2(g)) uarr
The single replacement reaction reaction between aluminium metal and hydrochloric acid produces aqueous aluminium chloride and hydrogen gas.
