The given information in the question can be summarized as follows:

- A colourless liquid is reacting with soda lime.
- The reaction produces sodium salt of carboxylic acid and ammonia gas.
- We need to identify the liquid from the given options.

Now, let's analyze each option to see which one fits the given information.

Option A: Propanamide

Propanamide is an amide compound that contains a carbonyl group bonded to a nitrogen atom. When it reacts with soda lime, it undergoes a hydrolysis reaction to form propanoic acid and ammonia gas. This reaction does not match the given information, as we are not getting the sodium salt of carboxylic acid.

Option B: Propanoic acid

Propanoic acid is a carboxylic acid that contains a carboxyl group (-COOH) attached to a three-carbon chain. When it reacts with soda lime, it produces sodium propanoate and water. This reaction also does not match the given information, as we are not getting ammonia gas.

Option C: Formamide

Formamide is an amide compound that contains a carbonyl group bonded to a nitrogen atom and a hydrogen atom. When it reacts with soda lime, it undergoes a dehydrogenation reaction to form sodium formate and ammonia gas. However, if we consider the reverse reaction, i.e., the reaction of sodium formate with soda lime, it produces formamide and calcium carbonate. This matches the given information, as we are getting the sodium salt of carboxylic acid (sodium formate) and ammonia gas.

Option D: Methyl ethanoate

Methyl ethanoate is an ester compound that contains a carbonyl group (-CO-) attached to an oxygen atom and a methyl group (-CH3). When it reacts with soda lime, it undergoes a hydrolysis reaction to form sodium ethanoate and methanol. This reaction also does not match the given information, as we are not getting ammonia gas.

Therefore, the correct answer is option C: Formamide.