1.

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

Answer»

HCl
KOH
`CO_(2)` is formed as the product
`sO_(2)`

Solution :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.


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