N₂​(g) reacts with H_2​(g) in either of the following ways depending upon the supply of H_2​(g).

N₂​(g)+H₂​(g)→N_2​H₂​(l)

N₂​(g)+2H₂​(g)→N₂​H₄​(g)

If 5L N_2​(g) and 3L H₂​(g) are taken initially (at same temperature and pressure). Choose the correct statement.

The reaction follows as,

$\begin{array}{l}{\mathrm{N}}_2+{\mathrm{H}}_2\to {\mathrm{N}}_2{\mathrm{H}}_2\\ {\mathrm{N}}_2+2{\mathrm{H}}_2\to {\mathrm{N}}_2{\mathrm{H}}_4\end{array}$

Considering the above reactions, three moles of H₂ reacts with two moles of N₂. Given 5 L of N₂ and 3 L of H₂.

H₂ is the limiting reagent. So, the first reaction will follow.

According to the first reaction, 3 L of H₂​ will react and form 3 L of the product. So, the final volume of solution 3+(5−3)=5 L.
Contraction in volume is V_i​−V_f​=5+3−5=3 L