Under similar conditions of pressure and temperature, 40 ml of slightly moist hydrogen chloride gas is mixed with 20 ml of ammonia gas, the final volume of gas at the same temperature and pressure will be

The correct option is "20 ml".

Let us understand the chemical reaction and calculate the final volume of the gas step-by-step.

When hydrogen chloride gas ($H\mathrm{Cl}$) reacts with ammonia gas ($N{H}_3$), they combine to form ammonium chloride ($N{H}_4\mathrm{Cl}$). The chemical equation for this reaction is:
$N{H}_3\left(g\right)+H\mathrm{Cl}\left(g\right)\to N{H}_4\mathrm{Cl}\left(s\right)$

According to Gay-Lussac's Law of Combining Volumes, gases react with each other in simple integer ratios by volume under similar conditions of temperature and pressure.
From the balanced chemical equation, 1 volume of $N{H}_3$ gas reacts completely with 1 volume of $H\mathrm{Cl}$ gas to form solid ammonium chloride ($N{H}_4\mathrm{Cl}$). Since ammonium chloride is a solid, its volume is negligible compared to the volumes of the gases and does not contribute to the final gaseous volume.

Now, let's analyze the given quantities:
Initial volume of ammonia gas ($N{H}_3$) = 20 ml
Initial volume of hydrogen chloride gas ($H\mathrm{Cl}$) = 40 ml

Since the reaction ratio is 1:1, 20 ml of $N{H}_3$ will react completely with exactly 20 ml of $H\mathrm{Cl}$.
Here, ammonia ($N{H}_3$) acts as the limiting reactant because it is present in a lesser volume than required to react with all of the $H\mathrm{Cl}$.

After the reaction goes to completion:
Volume of ammonia gas remaining = $20\text{ ml}-20\text{ ml}=0\text{ ml}$
Volume of hydrogen chloride gas remaining = $40\text{ ml}-20\text{ ml}=20\text{ ml}$

Therefore, the final volume of the gas remaining in the mixture is due only to the unreacted hydrogen chloride gas, which is 20 ml.