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By passing H(2)S in acidified KMnO(4) so...

By passing `H_(2)S` in acidified `KMnO_(4)` solution we get

A

`K_2 S`

B

`S`

C

`K_2 SO_3`

D

`MnO_2`

Text Solution

Verified by Experts

The correct Answer is:
B
`H_2 S ` decolourlses acidified `KMnO_4` (purpie) solution by reducing it to colourless `Mn^(2+)` ions.
`2KMnO_4 + 3H_2 SO_4 + 5H_2 S to K_2 SO_4 + underset("Colourless")(2MnSO_4) + 8H_2 O + underset("sulphur")underset("Colloidal")(5S)`
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