8 L of H2 and 6 L Cl2 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
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B
14 L
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C
2 L
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D
None of these
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Solution
The correct option is B 14 L H2+Cl2→2HClVolume before reaction8L6L0Volume after reaction 2012 As per stoichiometry, 1 vol. of H2 reacts to form 1 vol. of Cl2 to form 2 vol. of HCl Limiting reagent will be Cl2 since it is present in less quantity and will be consumed Volume of mixture after reaction = Volume of H2 left + Volume of HCl formed Volume of mixture after reaction = 2 + 12 = 14 L.