How do you balance the equation #Al + HCl -> AlCl_3 + H2#?

This is NOT the stoichiometric equation:

#Al(s) + 3HCl(aq) rarr AlCl_3(aq) + 2H_2(g)#

How would you modify it so that the left had side is equivalent to the right hand side? You can certainly use half-coeffcients, #1/2# or #3/2#!

#"Al(s)" + "HCl(aq)"##rarr##"AlCl"_3("aq") + "H"_2("g")"# is unbalanced.

Balance Cl.
There are 3 Cl atoms on the product side and 1 on the reactant side. Add a coefficient of 3 in front of #"HCl"#.

#"Al(s)" + "3HCl(aq)"##rarr##"AlCl"_3("aq") + "H"_2("g")"#

There are now 3 Cl atoms on both sides.

Balance H.
There are 3 H atoms on the reactant side and 2 on the product side. 3 and 2 are multiples of 6. So change the coefficient in front of HCl from 3 to 6, and add a coefficient of 3 in front of the #"H"_2"#.

#"Al(s)" + "6HCl(aq)"##rarr##"AlCl"_3("aq") + "3H"_2("g")"#

There are now 6 H atoms on both sides.

Go back to the Cl. There are now 6 Cl on the reactant side and 3 on the product side. Add a coefficient of 2 in front of #"AlCl"_3"#.

#"Al(s)" + "6HCl(aq)"##rarr##"2AlCl"_3("aq") + "3H"_2("g")"#

There are now 6 Cl atoms on both sides.

Balance the Al
There is 1 Al on the reactant side and 2 Al on the product side. Add a coefficient in front of Al on the reactant side.

#"2Al(s)" + "6HCl(aq)"##rarr##"2AlCl"_3("aq") + "3H"_2("g")"#

There are now 2 Al on both sides and the equation is balanced.

Reactants: #"2 Al atoms",##"6 H atoms",##"6 Cl atoms"#
Products: #"2 Al atoms",##"6 H atoms",##"6 Cl atoms"#