Calculate the weight of `N_2H_4` (hydrazine) oxidised to `N_2` by `19.4 g K_2CrO_4`, which is reduced to `Cr(OH)_4^(ɵ)` in basic medium.
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Mw of `K_2CrO_4=2xx39+52+4xx16=194g`
Mw of `N_2H_4=2xx14+4=32g`
`3N_2H_4+4H_2O+4CrO_4^(2-)to3N_2+4Cr(OH)_4^(ɵ)+4overset(ɵ)(O)H`
`4xx194g of K_2CrO_4-=3xx32 g of N_2H_4`
`19.4 g K_2CrO_4=(3xx32xx19.4)/(4xx194)=2.4g`