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On heating sodium propanoate with sodium...

On heating sodium propanoate with sodium hydroxide and quick lime the gas evolved is

A

`C_(2)H_(2)`

B

`C_(2)H_(6)`

C

`CH_(4)`

D

`C_(2)H_(4)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the gas evolved when sodium propanoate is heated with sodium hydroxide and quick lime, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reactants**: - Sodium propanoate (C2H5COONa) - Sodium hydroxide (NaOH) - Quick lime (Calcium oxide, CaO) 2. **Understand the Reaction**: - When sodium propanoate is heated with sodium hydroxide and quick lime, it undergoes a process known as decarboxylation. This reaction typically involves the removal of a carboxyl group (-COOH) from the sodium propanoate. 3. **Write the Reaction**: - The reaction can be represented as: \[ \text{C}_2\text{H}_5\text{COONa} + \text{NaOH} + \text{CaO} \rightarrow \text{C}_2\text{H}_6 + \text{Na}_2\text{CO}_3 \] - Here, ethane (C2H6) is formed as a gas, and sodium carbonate (Na2CO3) is produced as a solid byproduct. 4. **Identify the Gas Evolved**: - The gas evolved in this reaction is ethane (C2H6). 5. **Conclusion**: - Therefore, the answer to the question is that the gas evolved is ethane (C2H6). ### Final Answer: The gas evolved is **ethane (C2H6)**. ---
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