Question

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

Answer 1

You balance it stoichiometrically.

Explanation:

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`!

Answer 2

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

Explanation:

`"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"`