Pure methane can be prepared by

soda-lime de-carboxylation

Pure methane can be prepared using sodium acetate (CH3COONa) in the presence of a strong base like sodium hydroxide (NaOH) and a dehydrating agent such as calcium oxide (CaO). This reaction is known as the Decarboxylation Reaction, where a carboxylic acid (or its salt) loses a carbon dioxide molecule to form a hydrocarbon.

Reaction:

${\mathrm{CH}}_3-{\mathrm{CO}}_2\mathrm{H}\underset{\mathrm{\Delta }}{\overset{\mathrm{NaOH},\mathrm{CaO}}{\to }}{\mathrm{CH}}_4+{\mathrm{CO}}_2$

Explanation:

1. Sodium acetate (CH3COONa): Acts as the source of methane.
2. Sodium hydroxide (NaOH): Acts as a strong base to facilitate the reaction.
3. Calcium oxide (CaO): Used as a dehydrating agent to ensure water does not interfere with the reaction.
4. Heating (Δ): Provides the necessary energy to break the carboxylate bond and release carbon dioxide.

In this process:

• The sodium acetate undergoes decarboxylation, where the carboxyl group (-COO) is removed in the form of carbon dioxide (CO2CO2CO2).
• This results in the formation of methane (CH4CH4CH4), which is a simple and pure hydrocarbon.

Concept: Decarboxylation Reaction

The decarboxylation reaction is widely used in organic chemistry to reduce a compound by removing a carbon dioxide molecule. When carboxylic acids or their salts are treated with strong bases like NaOH and heated, they lose their -COO group as CO2CO2CO2 and form a lower hydrocarbon.

For example:

R-COONa + NaOH → Δ, CaO R-H + Na₂CO₃

Here, RRR represents the alkyl group. In this specific case, RRR is CH3CH3CH3, leading to the formation of methane.