`Impure Al_2O_3.xH_2O overset(NaOH) to A overset(B(gas)) to C`
Then A,B,C are respectively

Options

A. `Na[Al(OH)_4],CO_2,Al_2O_3`
B. `Al(OH)3,CO_2,Al_2O_3.XH_2O`
C. `Al(OH)3,SO_2,Al_2O_3.XH_2O`
D. `Na[Al(OH)_4],HCl,Al_2O_3`

Answer

To solve the problem step by step, we need to identify the compounds A, B, and C based on the reactions described.

### Step 1: Identify the reaction of impure Al₂O₃·xH₂O with NaOH

Impure Al₂O₃·xH₂O (bauxite) is amphoteric, meaning it can react with both acids and bases. When it reacts with sodium hydroxide (NaOH), it forms a soluble complex.

**Reaction:**
\[ \text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O} + \text{NaOH} \rightarrow \text{NaAl(OH)}_4 \]

**Compound A:**
Thus, A is Sodium Aluminate, which is represented as:
\[ \text{A} = \text{NaAl(OH)}_4 \]

### Step 2: Identify the gas B that reacts with A

The compound A (Sodium Aluminate) can react with an acidic oxide. The most common acidic oxide that reacts with sodium aluminate is carbon dioxide (CO₂).

**Reaction:**
\[ \text{NaAl(OH)}_4 + \text{CO}_2 \rightarrow \text{Al(OH)}_3 + \text{NaHCO}_3 \]

**Compound B:**
Thus, B is Carbon Dioxide, represented as:
\[ \text{B} = \text{CO}_2 \]

### Step 3: Identify the compound C formed after the reaction with B

When Sodium Aluminate reacts with Carbon Dioxide, it forms Aluminum Hydroxide (Al(OH)₃), which can further dehydrate to form Aluminum Oxide (Al₂O₃).

**Final Reaction:**
\[ \text{Al(OH)}_3 \rightarrow \text{Al}_2\text{O}_3 + \text{H}_2\text{O} \]

**Compound C:**
Thus, C is Aluminum Oxide, represented as:
\[ \text{C} = \text{Al}_2\text{O}_3 \]

### Summary of Compounds:
- A = NaAl(OH)₄ (Sodium Aluminate)
- B = CO₂ (Carbon Dioxide)
- C = Al₂O₃ (Aluminum Oxide)

### Final Answer:
A, B, C are respectively:
- A: Sodium Aluminate (NaAl(OH)₄)
- B: Carbon Dioxide (CO₂)
- C: Aluminum Oxide (Al₂O₃)

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