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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. |
Answer» 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` |
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