What are the products of the following reaction? #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))^(-)#
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"HCl"_ ((aq)) -> 6"H"_ ((aq))^(+) + 6"Cl"_((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.