`CH_3COOH+NaOH to A+B`
`X+NaOH to C(gas)+D`
C(gas) is:

Options

A. `CH_4`
B. `CO_2`
C. `O_2`
D. `H_2`

Answer

To solve the problem step by step, we will analyze the reactions given in the question:

1. **Identify the Reactants and Products in the First Reaction:**
- The first reaction involves ethanoic acid (CH₃COOH) reacting with sodium hydroxide (NaOH).
- This is a neutralization reaction where an acid reacts with a base to form a salt and water.
- The products of this reaction are sodium ethanoate (CH₃COONa) and water (H₂O).

**Reaction:**
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

- Here, A = CH₃COONa (sodium ethanoate) and B = H₂O (water).

2. **Identify the Reactants and Products in the Second Reaction:**
- The second reaction involves sodium ethanoate (A) reacting with sodium hydroxide (NaOH).
- In this reaction, sodium ethanoate can undergo a process called saponification or can be subjected to a reaction that produces a gas.
- The products of this reaction are methane gas (CH₄) and sodium carbonate (Na₂CO₃).

**Reaction:**
\[ \text{CH}_3\text{COONa} + \text{NaOH} \rightarrow \text{CH}_4 + \text{Na}_2\text{CO}_3 \]

- Here, C (the gas) = CH₄ (methane) and D = Na₂CO₃ (sodium carbonate).

3. **Conclusion:**
- The gas produced in the second reaction (C) is methane (CH₄).

Thus, the final answer is:
C(gas) is methane gas (CH₄).

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