Question

Write the balanced chemical equation for the following:

dilute Sulphuric acid is poured over Sodium sulphite.

Answer 1

Reaction between dilute Sulphuric acid and Sodium sulphite:

When dilute Sulphuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ is poured over Sodium sulphite $\left({\mathrm{Na}}_2{\mathrm{SO}}_3\right)$, it will give Sodium sulphate$\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)$, Sulphur dioxide gas $\left({\mathrm{SO}}_2\right)$ and water $\left({\mathrm{H}}_2\mathrm{O}\right)$ as a product.

Balanced chemical equation :

$\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}+\underset{\mathrm{Sodium}\mathrm{sulphite}}{{\mathrm{Na}}_2{\mathrm{SO}}_3\left(\mathrm{s}\right)}\to \underset{\mathrm{Sodium}\mathrm{sulphate}}{{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Sulphur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}$