NEET 2021 Chemistry Question 95 Solution

When sodium salts of carboxylic acids are heated strongly with soda lime, they undergo decarboxylation (loss of carbon dioxide) to form an alkane containing one less carbon atom than the parent carboxylic acid: $$R-COO^-Na^+ + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} R-H + Na_2CO_3$$

Here, sodium propanoate ($CH_3CH_2COONa$, 3 carbons) is heated to produce ethane ($CH_3CH_3$, 2 carbons) and sodium carbonate:

In soda lime ($3\text{NaOH} : 1\text{CaO}$): 1. $\text{CaO}$ keeps the reaction mixture dry by absorbing atmospheric moisture ($\text{NaOH}$ is deliquescent and would turn into a sticky liquid). 2. It raises the fusion temperature, allowing the reaction mixture to be heated to high temperatures without melting glass containers.