Question

8 L of $H_2$ and 6 L $C{l}_2$ are allowed to react to their maximum possible extent. Find out the final volume of reaction mixture. Suppose P and T remains constant throughout the course of reaction.

• A. 7 L
• B. 14 L
• C. 2 L
• D. None of these

Answer 1

The correct option is B 14 L

$\begin{array}{ccccccc}& H_2& +& C{l}_2& \to & 2HCl& \\ \text{Volume before reaction}& 8\text{L}& & 6\text{L}& & 0& \\ \text{Volume after reaction}& 2& & 0& & 12& \end{array}$
As per stoichiometry,
1 vol. of $H_2$ reacts to form 1 vol. of $C{l}_2$ to form 2 vol. of $HCl$
Limiting reagent will be $C{l}_2$ since it is present in less quantity and will be consumed
Volume of mixture after reaction = Volume of $H_2$ left + Volume of $HCl$ formed
Volume of mixture after reaction = 2 + 12 = 14 L.