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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_2S`

B

S

C

`K_2SO_3`

D

`MnO_2`

Text Solution

Verified by Experts

The correct Answer is:
B
`KMnO_4//H^(+)` oxidises `H_2S` to S
`({:(2KMnO_4+3H_2SO_4to,K_2SO_4+2MnSO_4+3H_2O+50),(H_2S+O to, H_2O +S"]"xx5):})/ul(5H_2S + 2KMnO_4+3H_2SO_4 to 8H_2O + K_2SO_4 +2MnSO_4+5S)`
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