`KMnO_(4)` can be prepared from `K_(2)MnO_(4)` as per reaction
`3MnO_(4)^(2-)+2H_(2)OhArr2MnO_(4)^(-)+MnO_(2)+4OH^(-)`

Options

A. HCl
B. KOH
C. `CO_(2)` is formed as the product
D. `sO_(2)`

Answer

D

Explanation

Since,`OH^(-)` are generated from weak acid`(H_(2)O)`. A weak acid (like `CO_(2)`) should be used to remove it because strong acid (HCl) reverse the reaction . KOH increases the concentration of `OH^(-)`,thus again sgifts the reaction in backward side . `Co_(2)` combines with `OH^(-)` to give carbonate which is easily removed. `SO_(2)`react with water to give strong acid , so it cannot be used.