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

Nov 30, 2015

You balance it stoichiometrically.

#### Explanation:

This is NOT the stoichiometric equation:

$A l \left(s\right) + 3 H C l \left(a q\right) \rightarrow A l C {l}_{3} \left(a q\right) + 2 {H}_{2} \left(g\right)$

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

Nov 30, 2015

$\text{2Al(s)" + "6HCl(aq)}$$\rightarrow$$\text{2AlCl"_3("aq") + "3H"_2("g")}$

#### Explanation:

$\text{Al(s)" + "HCl(aq)}$$\rightarrow$$\text{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 $\text{HCl}$.

$\text{Al(s)" + "3HCl(aq)}$$\rightarrow$$\text{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 $\text{H"_2}$.

$\text{Al(s)" + "6HCl(aq)}$$\rightarrow$$\text{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 $\text{AlCl"_3}$.

$\text{Al(s)" + "6HCl(aq)}$$\rightarrow$$\text{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.

$\text{2Al(s)" + "6HCl(aq)}$$\rightarrow$$\text{2AlCl"_3("aq") + "3H"_2("g")}$

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

Reactants: $\text{2 Al atoms} ,$$\text{6 H atoms} ,$$\text{6 Cl atoms}$
Products: $\text{2 Al atoms} ,$$\text{6 H atoms} ,$$\text{6 Cl atoms}$