1 mole of conc. HCl requires X moles of dilute NaOH for neutralisation and 1 mole of concentrate `H_(2)SO_(4)` requires Y moles of small dilute NaOH then which of the following reaction is true

Options

A. `Y=(1)/(2)X`
B. `X=(1)/(2)Y`
C. `X=2Y`
D. None of these

Explanation

To determine the values of X and Y for the neutralization reactions of HCl and H2SO4 with NaOH, we need to consider the number of hydrogen ions (H⁺) released by each acid and the stoichiometry of the neutralization reactions.

### Step-by-Step Solution:

1. **Identify the Acids and Their Properties:**
- **HCl (Hydrochloric Acid):** It is a strong acid that dissociates completely in solution to release 1 mole of H⁺ ions per mole of HCl.
- **H2SO4 (Sulfuric Acid):** It is also a strong acid, but it is diprotic, meaning it can release 2 moles of H⁺ ions per mole of H2SO4.

2. **Neutralization Reaction of HCl:**
- The neutralization reaction for HCl with NaOH can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- Since 1 mole of HCl releases 1 mole of H⁺, it will require 1 mole of NaOH to neutralize it.
- Therefore, for 1 mole of HCl, we have:
\[ X = 1 \]

3. **Neutralization Reaction of H2SO4:**
- The neutralization reaction for H2SO4 with NaOH can be represented as:
\[ \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \]
- Since 1 mole of H2SO4 releases 2 moles of H⁺, it will require 2 moles of NaOH to neutralize it.
- Therefore, for 1 mole of H2SO4, we have:
\[ Y = 2 \]

4. **Conclusion:**
- From the above analysis, we conclude that:
- \( X = 1 \) (for HCl)
- \( Y = 2 \) (for H2SO4)

5. **Final Statement:**
- The relationship between X and Y can be summarized as:
\[ Y = 2X \]
- Therefore, the correct statement regarding the neutralization of HCl and H2SO4 with NaOH is that 1 mole of HCl requires 1 mole of NaOH (X = 1) and 1 mole of H2SO4 requires 2 moles of NaOH (Y = 2).