`H_(2)SO_(4)+NaCl(s)rarrNaHSO_(4)+HCl` . Hydrochloric acid is liberated because

Options

A. `H_(2)SO_(4)` is a reducing agent
B. `HCl` is a smaller molecule than `H_(2)SO_(4)`
C. `HCl` is more volatile than `H_(2)SO_(4)`
D. `(2)` and `(3)` both

Explanation

To understand why hydrochloric acid (HCl) is liberated in the reaction \( H_2SO_4 + NaCl(s) \rightarrow NaHSO_4 + HCl \), we can break down the reasoning into several steps.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactants are sulfuric acid (\( H_2SO_4 \)) and sodium chloride (\( NaCl \)).
- The products formed are sodium hydrogen sulfate (\( NaHSO_4 \)) and hydrochloric acid (\( HCl \)).

2. **Understand the Nature of \( H_2SO_4 \)**:
- Sulfuric acid is a strong acid and has strong hydrogen bonding due to its molecular structure. This results in it being less volatile compared to other acids.

3. **Understand the Nature of \( HCl \)**:
- Hydrochloric acid is a gas at room temperature and is more volatile than sulfuric acid. This means that \( HCl \) can easily escape into the atmosphere when formed.

4. **Displacement Reaction**:
- In the reaction, \( H_2SO_4 \) reacts with \( NaCl \) to form \( NaHSO_4 \) and \( HCl \). The formation of \( HCl \) occurs because \( H_2SO_4 \) displaces \( HCl \) from sodium chloride due to the difference in volatility.

5. **Conclusion**:
- The liberation of hydrochloric acid occurs because \( HCl \) is more volatile than \( H_2SO_4 \). The strong hydrogen bonding in \( H_2SO_4 \) makes it less likely to escape, allowing \( HCl \) to be released.

### Final Answer:
Hydrochloric acid is liberated because \( HCl \) is more volatile than \( H_2SO_4 \), allowing it to be displaced from sodium chloride.