Answer:

Correct Answer: A

Solution:

The given problem is related to the concept of stoichiometry of chemical equations.

Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction].

The limiting reagent gives the moles of product formed in the reaction.

$ \underset{lnitial,vol\text{.}}{\overset{{}}{\mathop{{}}}}\ \underset{22\text{.4L}}{\mathop{\ {H_2}\text{(g)}}},\text{+}\underset{11\text{.2L}}{\mathop{C{l_2}\text{(g)}}},\to \underset{2mol}{\mathop{\text{2HCl(g)}}}, $ $ \because $ 22.4 L volume at STP is occupied by $ Cl _2 $ = 1 mole,

$ \therefore $ 11.2 L volume will be occupied by, $ Cl _2=\frac{1\times 11.2}{22.4}mole $ = 0.5 mol Thus, $ \underset{1,mol}{\mathop{H _2(g)}},+\underset{0.5mol}{\mathop{Cl _2(g)}},\to 2HCl(g) $

Since, $ Cl _2 $ possesses minimum number of moles, thus it is the limiting reagent.

As per equation, $ 1,mol,C{l_2},\equiv ,2,mol,HCl $

$ \therefore $ $ 0.5molCl _2=2\times 0.5molHCl $ $ =1.0molHCl $

Hence, 1.0 mole of HCI (g) is produced by 0.5 mole of $ Cl _2 $ [or 11.2 L].