The gas produced when a mixture of sodium propanoate and soda lime is heated is

The correct answer is E. ethyne (also known as acetylene).

When a mixture of sodium propanoate (CH₃CH₂COONa) and soda lime (NaOH and CaO) is heated, the sodium propanoate decomposes to form ethyne (C₂H₂) gas. This reaction is an example of a decarboxylation reaction, where the carboxyl group (-COOH) is removed from the sodium propanoate, leaving behind the ethyne molecule.

Here's a simplified equation for the reaction:

CH₃CH₂COONa + NaOH → C₂H₂ + Na₂CO₃ + H₂O

The other options are not correct because:

- Methane (A) is a different hydrocarbon gas (CH₄).
- Pentane (B) is a larger hydrocarbon molecule (C₅H₁₂).
- Ethane (C) is a smaller hydrocarbon molecule (C₂H₆).
- Butene (D) is a hydrocarbon molecule with a double bond (C₄H₈).

Ethyne (E) is the correct answer because it's the gas produced by the decomposition of sodium propanoate with soda lime.