Question

How do you balance the equation:
$Al + HCl \to AlC{l_3} + {H_2}$

Answer 1

Hint : The main thing to keep in mind while balancing the equation is that the number of atoms of each element on the reactant side must be equal to the number of atoms of each element on the product side.

Complete Step By Step Answer:
In order to balance this equation, we check the number of atoms on both sides.
The number of chlorine atoms on the product side is $3$ , while that in reactant only $1$ chlorine atom is present. We’ll add $3$ in front of $HCl$ .
$Al + 3HCl \to AlC{l_3} + {H_2}$
Now, we can see that the number of hydrogens on both sides are unequal. We can add a coefficient of $3$ for ${H_2}$ and a coefficient of $6$ for $HCl$ .
$Al + 6HCl \to AlC{l_3} + 3{H_2}$
This will make chlorine atoms on both sides unequal, we can add $2$ in front of $AlC{l_3}$ on the product side to make the number of chlorine atoms equal.
$Al + 6HCl \to 2AlC{l_3} + 3{H_2}$
Now, everything is equal except the aluminium atoms, we add a coefficient of $2$ in front of $Al$ in the reactant side.
$2Al + 6HCl \to 2AlC{l_3} + 3{H_2}$
The number of atoms of elements are equal on reactant and product side. Thus, our balanced equation is obtained.

Additional Information:
Aluminium reacts with dilute hydrochloric acid producing aluminium chloride and colourless hydrogen gas, this reaction is irreversible.

Note :
For balancing any equation, we have to use hit and trial. In this example as well we have added a coefficient $3$ in front of $HCl$ and then changed it to $6$ . Thus by adding coefficients we have to make the number of atoms of elements on reactant and product side equal.