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(C) The reaction is: $NH_3(g) + HCl(g) 	o NH_4Cl(s)$.Given volumes are $V_{NH_3} = 4 \ litre$ and $V_{HCl} = 1.5 \ litre$.Since $NH_4Cl$ is formed as

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$2.5$

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(C) The reaction is: $NH_3(g) + HCl(g) 	o NH_4Cl(s)$.Given volumes are $V_{NH_3} = 4 \ litre$ and $V_{HCl} = 1.5 \ litre$.Since $NH_4Cl$ is formed as a solid,it does not contribute to the gaseous volume.$HCl$ is the limiting reagent because it is present in a smaller amount compared to $NH_3$.$1.5 \ litre$ of $HCl$ will react with $1.5 \ litre$ of $NH_3$ to form $NH_4Cl(s)$.Remaining volume of $NH_3 = 4 \ litre - 1.5 \ litre = 2.5 \ litre$.Therefore,the volume of the mixture after the reaction is $2.5 \ litre$.

What will be the volume of the mixture after the reaction? $litre$
$NH_3(4 \ litre) + HCl(1.5 \ litre) \to NH_4Cl(s)$

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What will be the volume of the mixture after the reaction? $litre$$NH_3(4 \ litre) + HCl(1.5 \ litre) \to NH_4Cl(s)$