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the number of moles of Cr(2)O(7)^(2) nee...

the number of moles of `Cr_(2)O_(7)^(2)` needed to oxidise `0.136` equivalent of `N_(2)H_(5)` through the reaction:
`N_(2)H_(5) + Cr_(2)O_(7)^(2-) rarr N_(2) + Cr^(3) + H_(2)O` is:

A

`0.0226`

B

`0.087`

C

`0.136`

D

`0.488`

Text Solution

Verified by Experts

The correct Answer is:
A
Equivalents of `Cr_(2)O_(7)^(2-) = 0.136`
`n_(b) xx n = 0.136 rArr n = (0.136)/(6) = 0.226`
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