When H(2)O(2) react with N(2)H(4) then H...

When `H_(2)O_(2)` react with `N_(2)H_(4)` then `H_(2)O_(2)` oxidised `N_(2)H_(4)`. The no. of mole of `H_(2)O_(2)` required to oxidised 64gm `N_(2)H_(4)`.

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The correct Answer is:

4

`2H_(2)O_(2)rarr2H_(2)O+2[O]`
`N_(2)H_(4)+2[O]rarr N_(2)uarr +2H_(2)O`
`ul(bar(N_(2)H_(4)+2H_(2)O rarrN_(2)uarr+4H_(2)O))`

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