Sulphuryl chloride `SO_(2)Cl_(2)` reacts with water to give a mixture of `H_(2)SO_(4)` and `HCl`. Moles of `NaOH` requried to neutralise the solution formed by adding 1 " mol of "`SO_(2)Cl_(2)` to excess water is are

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the reaction of sulfuryl chloride (SO₂Cl₂) with water and determine how many moles of sodium hydroxide (NaOH) are required to neutralize the resulting acids.

### Step-by-Step Solution:

1. **Identify the Reaction**:
Sulfuryl chloride reacts with water to produce sulfuric acid (H₂SO₄) and hydrochloric acid (HCl).
\[ \text{SO}_2\text{Cl}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 + \text{HCl} \]

2. **Determine the Moles of Products**:
From the reaction, we see that 1 mole of SO₂Cl₂ produces:
- 1 mole of H₂SO₄
- 1 mole of HCl

3. **Neutralization of H₂SO₄**:
Sulfuric acid (H₂SO₄) is a diprotic acid, meaning it can donate two protons (H⁺ ions). Therefore, to neutralize 1 mole of H₂SO₄, we need 2 moles of NaOH:
\[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow 2 \text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \]

4. **Neutralization of HCl**:
Hydrochloric acid (HCl) is a monoprotic acid, meaning it can donate one proton (H⁺ ion). Therefore, to neutralize 1 mole of HCl, we need 1 mole of NaOH:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{H}_2\text{O} + \text{NaCl} \]

5. **Calculate Total Moles of NaOH Required**:
- For 1 mole of H₂SO₄: 2 moles of NaOH
- For 1 mole of HCl: 1 mole of NaOH
- Total NaOH required = 2 (for H₂SO₄) + 1 (for HCl) = 3 moles of NaOH

6. **Conclusion**:
Therefore, the total moles of NaOH required to neutralize the solution formed by adding 1 mole of SO₂Cl₂ to excess water is **3 moles**.

### Final Answer:
3 moles of NaOH are required to neutralize the solution.

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