`CH_4` is formed when

Options

A. Sodium acetate is heated with sodalime
B. Iodomethane is reduced
C. Aluminium carbide reacts with water
D. All of these

Explanation

To determine how `CH₄` (methane) is formed, we can analyze the three methods mentioned in the question:

### Step-by-Step Solution:

1. **Sodium Acetate with Soda Lime**:
- When sodium acetate (CH₃COONa) is heated with soda lime (a mixture of NaOH and CaO), a decarboxylation reaction occurs.
- The reaction can be represented as:
\[ \text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\Delta} \text{CH}_4 + \text{Na}_2\text{CO}_3 \]
- In this reaction, sodium acetate loses a carbon dioxide molecule, resulting in the formation of methane and sodium carbonate.

2. **Reduction of Iodo-methane**:
- Iodo-methane (CH₃I) can be reduced to methane in the presence of red phosphorus and hydroiodic acid (HI).
- The reaction can be represented as:
\[ \text{CH}_3\text{I} + \text{Red Phosphorus} \rightarrow \text{CH}_4 + \text{I}_2 \]
- Here, the iodine atom is replaced, resulting in the formation of methane.

3. **Hydrolysis of Aluminium Carbide**:
- Aluminium carbide (Al₄C₃) reacts with water to produce methane.
- The reaction can be represented as:
\[ \text{Al}_4\text{C}_3 + 12 \text{H}_2\text{O} \rightarrow 3 \text{CH}_4 + 4 \text{Al(OH)}_3 \]
- In this reaction, three moles of methane are produced along with aluminium hydroxide.

4. **Conclusion**:
- All three methods lead to the formation of methane. Therefore, the correct answer to the question is:
\[ \text{All of these} \]

### Final Answer:
`CH₄` (methane) is formed when:
- Sodium acetate is heated with soda lime,
- Iodo-methane is reduced,
- Aluminium carbide reacts with water,
- Hence, the answer is **all of these**.