When ethyl iodide and n-propyl iodide are allowed to react with sodium in the presence of dry ether, the number of alkanes that would be produced is

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When ethyl iodide and n-propyl iodide are allowed to react with sodium in the presence of dry ether, the number of alkanes that would be three as follows
$2 C H_{3} - C H_{2} - I \overset{N a}{\to} C H_{3} - C H_{2} - C H_{2} - C H_{3}$
$2 C H_{3} - C H_{2} - C H_{2} - I \overset{N a}{\to} \underset{n - h e x a n e}{C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{3}}$
$C H_{3} - C H_{2} - I + C H_{3} - C H_{2} - C H_{2} - I \overset{N a}{\to} \underset{n - p e n t a n e}{C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H_{3}}$

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