Question

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

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

$\underset{\mathrm{Ammonia}}{{\mathrm{NH}}_3\left(\mathrm{l}\right)}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Ammonium}\mathrm{Sulphate}}{{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}$

• It is a redox synthesis process that leads to the formation of Ammonium Sulphate.
• The following table shows the number of atoms on each side:

Reactant side	Product side
1 Nitrogen atom from ${\mathrm{NH}}_3$	2 Nitrogen atoms from ${\left({\mathrm{NH}}_4\right)}_2\mathrm{SO}$
5 Hydrogen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$ and ${\mathrm{NH}}_3$	8 Hydrogen atoms from ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$
1 Sulphur atom from ${\mathrm{H}}_2{\mathrm{SO}}_4$	1 Sulphur atom from ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$
4 Oxygen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$	4 Oxygen atoms from ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

• As there are two Nitrogen atoms on the right side, we must put a $2$ before the Ammonia on the left side.

$\underset{\mathrm{Ammonia}}{2{\mathrm{NH}}_3\left(\mathrm{l}\right)}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Ammonium}\mathrm{Sulphate}}{{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}$

• Finally, the other atoms must be balanced, but now, in this case, there are already an equal number of Sulphur, Hydrogen, and Oxygen atoms on each side.

Reactant side	Product side
2 Nitrogen atoms from ${\mathrm{NH}}_3$	2 Nitrogen atoms from ${\left({\mathrm{NH}}_4\right)}_2\mathrm{SO}$
8 Hydrogen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$ and ${\mathrm{NH}}_3$	8 Hydrogen atoms from ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$
1 Sulphur atom from ${\mathrm{H}}_2{\mathrm{SO}}_4$	1 Sulphur atom from ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$
4 Oxygen atoms from ${\mathrm{H}}_2{\mathrm{SO}}_4$	4 Oxygen atoms from ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

• Thus, the balanced chemical reaction is:

$\underset{\mathrm{Ammonia}}{2{\mathrm{NH}}_3\left(\mathrm{l}\right)}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Ammonium}\mathrm{Sulphate}}{{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}$

Hence, the reaction is balanced.