Question

Write the balanced chemical equations for the following reactions

Sulphur trioxide reacts with water to give sulphuric acid

Answer 1

Balanced chemical equation:

A chemical equation in which the number of atoms of each element on both sides of the equation is equal is called a balanced chemical equation.

We will use the hit-trial method to balance the equation on both sides.

Step 1: Representation of the chemical reaction:

When Sulphur trioxide$\left({\mathrm{SO}}_3\right)$ reacts with Water$\left({\mathrm{H}}_2\mathrm{O}\right)$, Sulphuric acid$\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ is formed.

$\underset{\left(\text{Sulphur trioxide}\right)}{S{O}_3}+\underset{\left(\text{Water}\right)}{H_{2}O}\to \underset{\text{(Sulphuric acid)}}{H_{2}S{O}_4}$

Step 2: Calculating the number of atoms on both sides of reaction.

Elements	The number of atoms on the reactant side	The number of atoms on the product side
$\mathrm{S}$	$1$	$1$
$\mathrm{O}$	$4$	$4$
$\mathrm{H}$	$2$	$2$

Here, the number of atoms on the reactant side is equal to the number of atoms on the product side.

Hence, it is a balanced chemical equation.

$\underset{\left(\text{Sulphur trioxide}\right)}{S{O}_3\left(\mathrm{g}\right)}+\underset{\left(\text{Water}\right)}{H_{2}O\mathit{\left(}l\mathit{\right)}}\to \underset{\text{(Sulphuric acid)}}{H_{2}S{O}_4\mathit{\left(}l\mathit{\right)}}$