Decarboxylation of CH₃COONa with NaOH and CaO

The given reaction is between sodium acetate (CH₃COONa) and sodium hydroxide (NaOH) in the presence of calcium oxide (CaO) and heat. This reaction is a classic example of the decarboxylation of a sodium salt of a carboxylic acid using soda lime (a mixture of NaOH and CaO).

The purpose of CaO is to keep the NaOH dry and prevent the formation of sodium carbonate as the reaction proceeds. It also helps in making the fusion mixture less fusible.

The overall chemical equation for the reaction is:

\( \text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{CH}_4 + \text{Na}_2\text{CO}_3 \)

In this reaction:

• Sodium acetate (CH₃COONa) is the reactant containing the carboxylate group.
• Sodium hydroxide (NaOH) is the strong base used for removing the carboxyl group.
• Calcium oxide (CaO) is the catalyst and drying agent (part of soda lime).
• Heat (\( \Delta \)) is required to drive the reaction.

The products formed are:

• An alkane, which is methane (CH₄) in this case, formed by the removal of the \( \text{-COO}^- \) group and its replacement with a hydrogen atom from NaOH.
• Sodium carbonate (Na₂CO₃), formed from the sodium ions, the carboxyl carbon, and oxygen atoms.

Therefore, in the reaction \( \text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{A} + \text{B} \), the products A and B are methane (CH₄) and sodium carbonate (Na₂CO₃) respectively.

Comparing this with the given options:

• Option 1 suggests CH₃COOH and Na₂CO₃. CH₃COOH is the original acid, not a product of decarboxylation.
• Option 2 suggests C₂H₅OH and Na₂CO₃. Ethanol is not formed in this reaction.
• Option 3 suggests C₂H₆ and Na₂CO₃. Ethane would be formed from sodium propionate (CH₃CH₂COONa), not sodium acetate.
• Option 4 suggests CH₄ and Na₂CO₃. This matches the products of the decarboxylation of sodium acetate.

Thus, A is CH₄ and B is Na₂CO₃.