1. Identify the reaction involved: Kolbe electrolysis

On electrolysis of sodium salts of carboxylic acids, the anion gets oxidized at the anode:

$\mathrm{RCOO^- \rightarrow R^. + CO_2 + e^-}$

Then two radicals combine:

$\mathrm{R^. + R^. \rightarrow R-R}$

So, each carboxylate gives an alkyl radical after loss of $\mathrm{CO_2}$.

2. Write radicals formed from each salt

• From sodium acetate, $\mathrm{CH_3COONa}$:

$\mathrm{CH_3COO^- \rightarrow CH_3^. + CO_2}$

So radical formed is $\mathrm{CH_3^.}$.

• From sodium propionate, $\mathrm{C_2H_5COONa}$:

$\mathrm{C_2H_5COO^- \rightarrow C_2H_5^. + CO_2}$

So radical formed is $\mathrm{C_2H_5^.}$.

3. List all possible radical combinations

Since both radicals are present, they can couple in all possible ways:

(i) $\mathrm{CH_3^. + CH_3^.}$

$\mathrm{CH_3-CH_3}$

This gives ethane.

(ii) $\mathrm{C_2H_5^. + C_2H_5^.}$

$\mathrm{C_2H_5-C_2H_5}$

This gives butane.

(iii) $\mathrm{CH_3^. + C_2H_5^.}$

$\mathrm{CH_3-C_2H_5}$

This gives propane.

4. Count the number of alkanes formed

The distinct alkanes are:

1. Ethane
2. Propane
3. Butane

Therefore, the number of alkanes obtained is

$\boxed{3}$

5. Compare with stored correct answer

Stored correct answer = $3$

My derived answer = $3$

So, the answer agrees with the stored correct answer.