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$KMnO_4$ can be prepared from $K_2MnO_4$ as per the reaction: $3MnO_4^{2-} + 2H_2O \rightleftharpoons 2MnO_4^- + MnO_2 + 4OH^-$. The reaction can go to completion by removing $OH^-$ ions by adding:
A
$SO_2$
B
HCl
C
KOH
D
$CO_2$
Detailed Solution
$CO_2$ is acidic and removes $OH^-$ ions: $2OH^- + CO_2 \rightarrow CO_3^{2-} + H_2O$, which drives the reaction to completion.
HCl cannot be used because it is oxidised by $KMnO_4$, and $SO_2$ is a reducing agent.
HCl cannot be used because it is oxidised by $KMnO_4$, and $SO_2$ is a reducing agent.