1.

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

Answer»

`2.4 g`
`2.8 g`
`3.0` g
`2.0 g`

Solution :`N_(2)^(4-) to OVERSET(o) N_(2) + 4e^(-)`
` Cr^(6+) + 3eto Cr^(3+)`
Eq. of `K_(2)CrO_(4) = " Eq. of " N_(2)H_(4)`
` :.(19.4)/(194//3) = W/(32//4) rArrW = 2.4g`


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