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The amount of hydrazine (N(2)H(4)) oxidi...

The amount of hydrazine `(N_(2)H_(4))` oxidized to `N_(2)` by `19.4 g K_(2)CrO_(4)` which itself reduces to `Cr(OH)_(4)^(-)` is :

A

`2.4 g`

B

`2.8 g`

C

`3.0` g

D

`2.0 g`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2)^(4-) to overset(o) N_(2) + 4e^(-)`
` Cr^(6+) + 3e to Cr^(3+)`
Eq. of `K_(2)CrO_(4) = " Eq. of " N_(2)H_(4)`
` :. (19.4)/(194//3) = W/(32//4) rArr W = 2.4 g`
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