Chemistry - Redox Reactions Question with Solution | TestHub

Equivalent volume 11.2 L at STP means one mole of gas carries $n$ = 22.4/11.2 = 2 g-eq. (a) $\mathrm{N}{\mathrm{H}}_3$ $\to$ ${\mathrm{N}}_2$: N goes -3 $\to$ 0, $n$ = 3, so 7.47 L. (b) ${\mathrm{O}}_3$ as oxidant (${\mathrm{O}}_3$ $\to$ ${\mathrm{O}}_2$ + ${\mathrm{O}}^{2-}$): accepts 2 ${\mathrm{e}}^{-}$ per molecule, $n$ = 2, so 11.2 L. (c) $\mathrm{C}{\mathrm{H}}_4$ combustion: C goes -4 $\to$ +4, $n$ = 8, so 2.8 L. (d) C $\to$ $\mathrm{C}{\mathrm{O}}_2$: $n$ = 4, so 5.6 L. Only (b).