Question

What mass of $N_{2}H_{4}$ can be oxidised to $N_{2}$ by 24.0 g $K_{2}CrO_{4}$ , which is reduced to $\left(C r \left(\right. O H \left.\right)_{4}^{- \, \, }$ (Given: Molar mass of $\text{K}_{\text{2}} \text{CrO}_{\text{4}} \, \text{=} \, \text{194} \text{.2}$ )

• A. 9.97 g
• B. 2.97 g
• C. 3.97 g
• D. 4.97 g

Answer 1

Answer: (B)

$N_{2}^{2 -} \rightarrow N_{2}^{0}+ \, 4e$
$C r^{6 +} + \, 3 e \, \, \rightarrow C r^{3 +}$
Meq. of $N_{2}H_{4}$ = Meq. of $K_{2}C r O_{4}$
$\frac{w \, \, \, }{32 / 4}\times \, 1000 \, =\frac{24}{194.2 / 3}\times 1000$ $\left(E . \, w t \, = \frac{M \, w t .}{v a l e n c e \, f a c t o r} \, \right)$
$\text{w}_{\text{N}_{\text{2}} \text{H}_{\text{4}}} \text{= 2} \text{.97} \, \text{g}$