Number of alkanes obtained on electrolysis of a mixture of $$CH_3COONa$$ and $$C_2H_5COONa$$ is _____.

This problem involves Kolbe's electrolysis of a mixture of two sodium carboxylates.

Key Concept: Kolbe's Electrolysis

In Kolbe's electrolysis, sodium salts of carboxylic acids are electrolysed. At the anode, the carboxylate ions lose electrons and undergo decarboxylation to form free radicals:

$$RCOO^- \rightarrow RCOO \cdot + e^- \rightarrow R \cdot + CO_2$$

These free radicals then combine (couple) to form alkanes.

From $$CH_3COONa$$: the methyl radical $$CH_3 \cdot$$ is produced.

From $$C_2H_5COONa$$: the ethyl radical $$C_2H_5 \cdot$$ is produced.

When two types of radicals are present in the reaction mixture, they can combine in all possible pairwise ways:

1. $$CH_3 \cdot + CH_3 \cdot \rightarrow CH_3CH_3$$ (ethane, $$C_2H_6$$)

2. $$CH_3 \cdot + C_2H_5 \cdot \rightarrow CH_3C_2H_5$$ (propane, $$C_3H_8$$)

3. $$C_2H_5 \cdot + C_2H_5 \cdot \rightarrow C_2H_5C_2H_5$$ (butane, $$C_4H_{10}$$)

We obtain three distinct alkanes: ethane, propane, and butane.

The answer is 3.