`CH_4` is formed when

Options

A. sodium acetate is heated with soda lime
B. iodomethane is reduced
C. aluminium carbide reacts with water
D. all of the above.

Answer

To determine when CH₄ (methane) is formed, we can analyze the three options provided in the question:

### Step 1: Decarboxylation of Sodium Acetate
- Sodium acetate (CH₃COONa) can undergo decarboxylation when treated with soda lime (a mixture of NaOH and CaO).
- The reaction is as follows:
\[ \text{CH}_3\text{COONa} \xrightarrow{\text{NaOH, CaO} } \text{CH}_4 + \text{Na}_2\text{CO}_3 \]
- This reaction produces methane (CH₄) and sodium carbonate (Na₂CO₃).

### Step 2: Reduction of Iodomethane
- Iodomethane (CH₃I) can be reduced using hydrogen (H₂) in the presence of a catalyst.
- The reaction is:
\[ \text{CH}_3\text{I} + \text{H}_2 \rightarrow \text{CH}_4 + \text{HI} \]
- This reaction also yields methane (CH₄) and hydrogen iodide (HI).

### Step 3: Hydrolysis of Aluminum Carbide
- Aluminum carbide (Al₄C₃) reacts with water (H₂O) to produce methane.
- The reaction is:
\[ \text{Al}_4\text{C}_3 + 12\text{H}_2\text{O} \rightarrow 3\text{CH}_4 + 4\text{Al(OH)}_3 \]
- This reaction results in the formation of methane (CH₄) and aluminum hydroxide (Al(OH)₃).

### Conclusion
Since all three reactions lead to the formation of methane (CH₄), we conclude that CH₄ is formed when:
1. Sodium acetate undergoes decarboxylation.
2. Iodomethane is reduced.
3. Aluminum carbide is hydrolyzed.

Thus, the correct answer is that CH₄ is formed in all of the above scenarios.

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