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H(2)O(2) reduces K(4)Fe(CN)(6)...

`H_(2)O_(2)` reduces `K_(4)Fe(CN)_(6)`

A

`6.1`

B

`12.2`

C

`3.0`

D

`5.0`

Text Solution

Verified by Experts

The correct Answer is:
A
n-factor of `Ba(MnO_(4))_(2) = 10`
n-factor of `K_(4)[Fe(CN)_(6)] = 61`
`:' ("mole of" Ba(MnO_(4))_(2))/(1 "mole of" K_(4)Fe(CN)_(6)) = (61)/(10)`
`:.` mole of `Ba(MnO_(4))_(2) = (61)/(10) xx 1 = 6.1`
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