`A + NaNO_2 to N_2O + NaCl + 2H_2O` in this reaction A can be

Options

A. `H_2SO_4`.dil
B. `H_3PO_4`
C. `NH_2OH.HCl`
D. `dil.HCl`

Answer

To determine what compound A can be in the reaction \( A + \text{NaNO}_2 \rightarrow \text{N}_2\text{O} + \text{NaCl} + 2\text{H}_2\text{O} \), we can follow these steps:

### Step 1: Analyze the Reaction
The reaction shows that compound A reacts with sodium nitrite (\( \text{NaNO}_2 \)) to produce nitrous oxide (\( \text{N}_2\text{O} \)), sodium chloride (\( \text{NaCl} \)), and water (\( \text{H}_2\text{O} \)). Notably, there are two nitrogen atoms in the product \( \text{N}_2\text{O} \).

### Step 2: Identify Possible Compounds for A
Since \( \text{N}_2\text{O} \) contains nitrogen, compound A must also contain nitrogen. Possible candidates could include:
- Hydroxylamine (\( \text{NH}_2\text{OH} \))
- Ammonium salts
- Other nitrogen-containing compounds

### Step 3: Exclude Non-Nitrogen Compounds
We can eliminate certain compounds based on the products formed:
- \( \text{H}_2\text{SO}_4 \) cannot be A because no sulfur product is formed.
- \( \text{H}_3\text{PO}_4 \) cannot be A because no phosphorus product is formed.

### Step 4: Test Hydroxylamine Hydrochloride
Let’s consider \( \text{NH}_2\text{OH} \cdot \text{HCl} \) (hydroxylamine hydrochloride):
- When \( \text{NH}_2\text{OH} \cdot \text{HCl} \) reacts with \( \text{NaNO}_2 \), it can produce \( \text{N}_2\text{O} \), \( \text{NaCl} \), and \( \text{H}_2\text{O} \).

### Step 5: Confirm the Reaction
The balanced reaction would be:
\[ \text{NH}_2\text{OH} \cdot \text{HCl} + \text{NaNO}_2 \rightarrow \text{N}_2\text{O} + \text{NaCl} + 2\text{H}_2\text{O} \]
This confirms that \( \text{NH}_2\text{OH} \cdot \text{HCl} \) fits as compound A.

### Conclusion
Thus, the compound A in the reaction is \( \text{NH}_2\text{OH} \cdot \text{HCl} \).