Chemistry - Mole Concept Question with Solution | TestHub

$\stackrel{-2}{{\mathrm{N}}_2{\mathrm{H}}_4}+\stackrel{+6}{K_{2}Cr{O}_4}\to \stackrel{0}{N_2}+\stackrel{+3}{Cr{\left(OH\right)}_4^{-}}$

n-factor 4 3

Moles of K₂CrO₄ reacted = $\frac{24}{194}$

Equivalents of K₂CrO₄ reacted = $\frac{3\times 24}{194}=\frac{72}{194}$

Equivalents of N₂H₄ oxidized = $\frac{72}{194}$

Moles of N₂H₄ oxidized = $\frac{72}{4\times 194}=\frac{18}{194}$

Mass of N₂H₄ oxidized = $\frac{18\times 32}{194}=2.97g$