Question

How would you balance: $\mathrm{Al}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{H}}_2$

Answer 1

Balancing the Chemical equation:

• When we balance chemical equations, we want the same number of each sort of atom on both sides of the equation.
• Only the coefficients are changed. The numbers in front of the molecule are called coefficients.
• Subscripts should never be changed.
• Metals must be balanced first, followed by nonmetals such as oxygen and hydrogen.
• The unbalanced chemical reaction provided is

$\underset{\mathrm{Aluminium}}{\mathrm{Al}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Aluminium}\mathrm{sulphate}}{{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{l}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{{\mathrm{H}}_2\uparrow \left(\mathrm{g}\right)}$

• It is a single displacement redox process in which Aluminium ($\mathrm{Al}$) and Iron ($\mathrm{H}$) swap places.
• The following table shows the number of atoms on each side:

Reactant side	Product side
1 Aluminium atom	2 Aluminium atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$
2 Hydrogen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$	2 Hydrogen atoms from ${\mathrm{H}}_2$
1 Sulphur atom from ${\mathrm{H}}_2{\mathrm{SO}}_4$	3 Sulphur atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$
4 Oxygen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$	12 Oxygen atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$

• As there are two Aluminium atoms on the right side, we must put a $2$ before the Aluminium on the left side.
• Second, the Sulphate must be balanced. As shown on the right, there are three Sulphate atoms in ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$, thus, a $3$ must be inserted on the left side to balance them.

Reactant side	Product side
2 Aluminium atom	2 Aluminium atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$
6 Hydrogen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$	2 Hydrogen atoms from ${\mathrm{H}}_2$
3 Sulphur atom from ${\mathrm{H}}_2{\mathrm{SO}}_4$	3 Sulphur atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$
12 Oxygen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$	12 Oxygen atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$

• Finally, the oxygens must be balanced, but now in this case, they are already at $12$ on each side.
• Now, at last, we will balance Hydrogen; there are $6$ hydrogen atoms on the left, so we must put $3$ on the right side.

Reactant side	Product side
2 Aluminium atom	2 Aluminium atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$
6 Hydrogen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$	6 Hydrogen atoms from ${\mathrm{H}}_2$
3 Sulphur atom from ${\mathrm{H}}_2{\mathrm{SO}}_4$	3 Sulphur atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$
12 Oxygen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$	12 Oxygen atoms from ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$

• Thus, the balanced chemical reaction is:

$\underset{\mathrm{Aluminium}}{2\mathrm{Al}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{3{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Aluminium}\mathrm{sulphate}}{{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{l}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{3{\mathrm{H}}_2\uparrow \left(\mathrm{g}\right)}$