Question

27. Complete and balance the following equation:

2. ${\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to$

Answer 1

Step 1: Write the skeletal equation for the given reaction

Sulfur trioxide $\left({\mathrm{SO}}_3\right)$reacts with water $\left({\mathrm{H}}_2\mathrm{O}\right)$ and forms Sulphuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$. The skeletal chemical equation is as follows:

$\underset{\mathrm{Sulphur}\mathrm{trioxide}}{{\mathrm{SO}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}$

Step 2: Count the number of atoms of each element on both sides of the equation

For balancing the equation, both reactants and products should be balanced on both the sides. The number of atoms of each element present on both sides of the equation are listed below in the table:

Reactant	Product
$\mathrm{S}=1$	$\mathrm{S}=1$
$\mathrm{O}=4$	$\mathrm{O}=4$
$\mathrm{H}=2$	$\mathrm{H}=2$

Since the number of atoms is equal on both sides thus, the equation is said to be balanced.

Conclusion:

Therefore, the balanced chemical equation for the given reaction is as follows:

$\underset{\mathrm{Sulphur}\mathrm{trioxide}}{{\mathrm{SO}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}$