i.(A)`underset(Delta)overset(NaOH)rarr(B)(g) overset(HCI)rarr` While fumes.
ii. After (B) is expelled completely, resultant alkline solution again gives gas (B) on heating with zine
iii.(A) `overset(Delta)rarr N_(2)O +H_(2)O`
What is the formula of white fumes?

Options

A. `NH_(4)NO_(3)`
B. `NH_(4)CI`
C. `NH_(4)NO_(2)`
D. `NH_(3)`

Explanation

To solve the given problem, we will analyze the steps provided in the question and derive the necessary conclusions step by step.

### Step 1: Identify Compound A
The question states that compound A, when heated, produces \( N_2O \) (dinitrogen oxide) and \( H_2O \) (water). The only known compound that decomposes to give one mole of \( N_2O \) and one mole of \( H_2O \) upon heating is **ammonium nitrate** \((NH_4NO_3)\).

**Conclusion**:
- **Compound A** is \( NH_4NO_3 \).

### Step 2: Reaction of A with NaOH
Next, we consider the reaction of compound A with sodium hydroxide (NaOH). When ammonium nitrate is heated with NaOH, it decomposes to produce ammonia gas (B), sodium nitrate, and water.

The reaction can be summarized as:
\[ NH_4NO_3 + NaOH \xrightarrow{\Delta} NH_3(g) + NaNO_3 + H_2O \]

**Conclusion**:
- **Gas B** is \( NH_3 \) (ammonia).

### Step 3: Reaction of Gas B with HCl
The question states that gas B (ammonia) reacts with hydrochloric acid (HCl) to form white fumes. The reaction between ammonia and hydrochloric acid produces ammonium chloride \((NH_4Cl)\), which is a white solid that appears as white fumes when formed in the air.

The reaction can be summarized as:
\[ NH_3(g) + HCl(g) \rightarrow NH_4Cl(s) \]

**Conclusion**:
- The white fumes are due to the formation of **ammonium chloride** \((NH_4Cl)\).

### Step 4: Resultant Alkaline Solution
After gas B (ammonia) is expelled completely, the resultant alkaline solution (which contains sodium nitrate) can also produce gas B (ammonia) upon heating with zinc. This is consistent with the behavior of ammonium compounds in alkaline solutions.

### Final Answer
The formula of the white fumes formed is:
\[ \text{NH}_4\text{Cl} \]

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