A metal X reacts with aqueous NaOH solution to form Y and a highly inflammalbe gas. Solution Y is heated and `CO_(2)` is poured through it. Z precipitates out and `Na_(2)CO_(3)` is formed. Z on the heating gives `AI_(2)O_(3)`. Identify X,Y and Z.
X Y Z

Options

A. `Al, NaAIO_(2), AI(OH)_(3)`
B. `AI_(2)O_(3), NaAIO_(2), AI_(2)CO_(3)`
C. `AI_(2)O_(3), [Na_(2)AIO_(2)]^(+) OH^(-), AI(OH)_(3)`
D. `AI, AI(OH)_(3) AI_(2)O_(3)`

Explanation

To solve the problem step by step, we need to identify the metal X, the compound Y formed when X reacts with NaOH, and the compound Z that precipitates when Y is treated with CO2.

### Step-by-Step Solution:

1. **Identify Metal X:**
- The problem states that a metal X reacts with aqueous NaOH to produce a compound Y and a highly flammable gas.
- A common metal that reacts with NaOH to produce hydrogen gas (which is highly flammable) is **Aluminum (Al)**.
- Thus, **X = Aluminum (Al)**.

2. **Reaction of X with NaOH:**
- When aluminum reacts with sodium hydroxide (NaOH), it forms sodium aluminate (NaAlO2) and hydrogen gas (H2).
- The balanced equation for this reaction is:
\[ 2Al + 2NaOH + 6H2O \rightarrow 2NaAl(OH)4 + 3H2 \]
- Here, **Y = Sodium Aluminate (NaAlO2)**.

3. **Heating Solution Y and Adding CO2:**
- When sodium aluminate (Y) is heated and CO2 is bubbled through it, it forms aluminum hydroxide (Al(OH)3) and sodium carbonate (Na2CO3).
- The balanced equation for this reaction is:
\[ NaAlO2 + CO2 + 2H2O \rightarrow Al(OH)3 \downarrow + Na2CO3 \]
- The precipitate formed here is **Z = Aluminum Hydroxide (Al(OH)3)**.

4. **Heating Z:**
- When aluminum hydroxide (Z) is heated, it decomposes to form aluminum oxide (Al2O3) and water (H2O).
- The balanced equation for this reaction is:
\[ 2Al(OH)3 \rightarrow Al2O3 + 3H2O \]

### Final Identification:
- **X = Aluminum (Al)**
- **Y = Sodium Aluminate (NaAlO2)**
- **Z = Aluminum Hydroxide (Al(OH)3)**

### Summary:
- X: Aluminum (Al)
- Y: Sodium Aluminate (NaAlO2)
- Z: Aluminum Hydroxide (Al(OH)3)