Question

How do you balance the equation$\mathrm{Al}\left(\mathrm{s}\right)+\mathrm{HCl}\left(\mathrm{l}\right)\to {\mathrm{AlCl}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$?

Answer 1

Step 1: Note the unbalanced reaction

Write the unbalanced reaction as it is.

$\underset{\mathrm{Aluminium}}{\mathrm{Al}\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{l}\right)}\to \underset{\mathrm{Aluminium}\mathrm{chloride}}{{\mathrm{AlCl}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrogen}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$

Step 2: Determination of the oxygen state of each atom

Write the oxidation numbers of each atom.

On the reactant side or in RHS, the aluminium is in metallic form, hence its oxidation state is zero. When it forms Aluminium trichloride oxidation state changes to $+3$ as each Chlorine accepts one electron from Aluminium.

The oxidation state of Hydrogen in Hydrochloric acid is $+1$ as it donates one electron to the Chlorine atom. On the product side, it liberates as Hydrogen gas which has an oxidation state of zero.

Chlorine is more electronegative than Hydrogen and Aluminium, so it accepts an electron from both and has an oxidation state $-1$ on both sides.

$$\begin{gathered} \mathrm{On}\mathrm{RHS},\mathrm{Al}=0,\mathrm{H}=+1,\mathrm{Cl}=-1 \\ \mathrm{On}\mathrm{LHS},\mathrm{Al}=+3,\mathrm{H}=0,\mathrm{Cl}=-1 \end{gathered}$$

Step 3: Separate the oxidation and reduction reactions.

Write the oxidation and reduction couplets.

Reduction: In Hydrochloric acid, the oxidation state of Hydrogen is $+1$ and in Hydrogen gas, it changes to $0$, therefore it is reduction half-reaction.

$\mathrm{HCl}\to {\mathrm{H}}_2$

Oxidation: In the Aluminium metal state, the oxidation state of Aluminium is $0$ and in Aluminium trichloride, it changes to $+3$, therefore it is oxidation half-reaction.

$\mathrm{Al}\to {\mathrm{AlCl}}_3$

Step 4: Balancing of all atoms except Oxygen and Hydrogen.

Balance all other elements except Oxygen and Hydrogen. To balance the number of Chlorine atoms on both sides multiply $\mathrm{HCl}$ by $3$.

The reaction becomes,

$\underset{\mathrm{Aluminium}}{\mathrm{Al}\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{3\mathrm{HCl}\left(\mathrm{l}\right)}\to \underset{\mathrm{Aluminium}\mathrm{chloride}}{{\mathrm{AlCl}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrogen}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$

Step 5: Balancing of Hydrogen and Oxygen.

Balance the Hydrogen atoms.

To balance the Hydrogen atom multiply ${\mathrm{H}}_2$$\mathrm{by}$$3$, $\mathrm{Al}\mathrm{and}{\mathrm{AlCl}}_3\mathrm{by}2$, and $\mathrm{HCl}\mathrm{by}2$

Therefore, the balanced chemical reaction is

$\underset{\mathrm{Aluminium}}{2\mathrm{Al}\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{aicd}}{6\mathrm{HCl}\left(\mathrm{l}\right)}\to \underset{\mathrm{Aluminium}\mathrm{chloride}}{2{\mathrm{AlCl}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrogen}}{3{\mathrm{H}}_2\left(\mathrm{g}\right)}$