Identify the gas that readily decolourises acidified KMnO_4...

Acidified $\text{KMnO}_4$ acts as a strong oxidizing agent. $\text{SO}_2$ is a good reducing agent and readily reduces the purple-coloured acidified $\text{KMnO}_4$ solution to colourless $\text{Mn}^{2+}$ ions.

The chemical reaction is given by:
$2\text{KMnO}_4 + 5\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 2\text{H}_2\text{SO}_4$

$\text{CO}_2$ does not react with $\text{KMnO}_4$. While $\text{H}_2\text{S}$ can also reduce $\text{KMnO}_4$, it produces a milky suspension of colloidal sulphur, whereas $\text{SO}_2$ yields a clear, colourless solution. Thus, $\text{SO}_2$ is the gas that readily decolourises the solution.

Answer: $\text{SO}_2$