In the reaction, `CH_(3)COONa+NaOHrarr` the gas obtained is

Options

A. `C_(2)H_(6)`
B. `C_(2)H_(2)`
C. `CH_(4)`
D. `C_(3)H_(8)`

Explanation

To determine the gas obtained from the reaction between sodium acetate (CH₃COONa) and sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in the reaction are:
- Sodium acetate (CH₃COONa)
- Sodium hydroxide (NaOH)

### Step 2: Understand the Reaction
When sodium acetate reacts with sodium hydroxide, a de-carboxylation reaction can occur. This means that a carboxyl group (-COOH) is removed from the acetate ion.

### Step 3: Write the Reaction
The reaction can be represented as follows:
\[ \text{CH}_3\text{COONa} + \text{NaOH} \rightarrow \text{CH}_4 + \text{Na}_2\text{CO}_3 \]

### Step 4: Identify the Products
From the reaction, we can identify the products:
- Methane (CH₄) is produced as a gas.
- Sodium carbonate (Na₂CO₃) is formed as a solid byproduct.

### Step 5: Conclusion
The gas obtained from the reaction is methane (CH₄).

### Final Answer
The gas obtained in the reaction between sodium acetate and sodium hydroxide is **methane (CH₄)**.

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