NEET Solved Paper 2014 Question 8
Question: When 22.4 L of $ H _2(g) $ is mixed with 11.2 L of $ Cl _2 $ each at STP, the moles of $ HCl(g) $ formed is equal to [AIPMT 2014]
Options:
A) 1 mole of $ HCl(g) $
B) 2 moles of $ HCl(g) $
C) 0.5 mole of $ HCl(g) $
D) 1.5moleof $ HCl(g) $
Show Answer
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].
