The given chemical equation represents the combustion reaction of propanoic acid:
On balancing we get,
[tex]2CH_{3}CH_{2}COOH (l)+ 7O_{2}(g)--->6CO_{2}(g) + 6H_{2}O(g)[/tex]
The above balanced chemical equation tells us that, 2 mol [tex]CH_{3}CH_{2}COOH[/tex] require 7 mol [tex]O_{2}[/tex] for complete combustion to produce [tex]6 mol H_{2}O[/tex] and [tex]6 mol CO_{2}[/tex]
Given moles of [tex]CH_{3}CH_{2}COOH[/tex] = 1 mol
Calculating moles of [tex]O_{2}[/tex] required for combustion:
[tex]1 mol CH_{3}CH_{2}COOH * \frac{7 mol O_{2} }{2 mol CH_{3}CH_{2}COOH } =3.5 mol O_{2}[/tex]
Therefore, 3.5 mol Oxygen is required for complete combustion of 1 mol propanoic acid.