What mass of ${\mathrm{N}}_2{\mathrm{H}}_4$ can be oxidized to ${\mathrm{N}}_2$ by 24 g of ${\mathrm{K}}_2{\mathrm{CrO}}_4$ which is reduced to $\mathrm{Cr}{\left(\mathrm{OH}\right)}_4^{-}$ ?

$$\begin{gathered} {\stackrel{-2}{\mathrm{N}}}_2{\mathrm{H}}_4\to {\stackrel{0}{\mathrm{N}}}_2 ; \mathrm{change} \mathrm{in} \mathrm{oxidation} \mathrm{state}=4 \\ {\mathrm{K}}_2\stackrel{+6}{\mathrm{Cr}}{\mathrm{O}}_4\to \stackrel{+3}{\mathrm{Cr}}{\left(\mathrm{OH}\right)}_4^{-} ; \mathrm{change} \mathrm{in} \mathrm{oxidation} \mathrm{state}=3 \end{gathered}$$

According to oxidation no. method, $3{\mathrm{N}}_2{\mathrm{H}}_4+4{\mathrm{K}}_2{\mathrm{CrO}}_4$

$$\begin{gathered} \mathrm{Moles} {\mathrm{ofK}}_2{\mathrm{CrO}}_4 \mathrm{reacted} = \frac{24}{194} \mathrm{moles}. \\ 4 \mathrm{moles} \mathrm{of} {\mathrm{K}}_2{\mathrm{CrO}}_4 \mathrm{reacts} \mathrm{with} 3 \mathrm{moles} \mathrm{of} {\mathrm{N}}_2{\mathrm{H}}_4 \\ \therefore \frac{24}{194} \mathrm{moles} \mathrm{of} {\mathrm{K}}_2{\mathrm{CrO}}_4 \mathrm{reacts} \mathrm{with} \frac{3}{4} \times \frac{24}{194} \mathrm{moles} {\mathrm{N}}_2{\mathrm{H}}_4 \\ \therefore \mathrm{Amount} \mathrm{of} {\mathrm{N}}_2{\mathrm{H}}_4 \mathrm{reacted} = \frac{3}{4} \times \frac{24}{194} \mathrm{moles} \\ = \frac{3}{4} \times \frac{24}{194} \times 32 \mathrm{g} =2.969 \mathrm{g} \end{gathered}$$