The volume of oxygen gas ($ ext{O}_2$) needed to completely burn $1 ext{ L}$ of propane gas ($ ext{C}_3 e

Question

The volume of oxygen gas ($	ext{O}_2$) needed to completely burn $1 	ext{ L}$ of propane gas ($	ext{C}_3	ext{H}_8$) (both $	ext{O}_2$ & propane measured at $0^\circ	ext{C}$ and $1 	ext{ atm}$) will be:

Accepted Answer

The balanced chemical equation for the complete combustion of propane ($	ext{C}_3	ext{H}_8$) is:
$	ext{C}_3	ext{H}_8(g) + 5	ext{O}_2(g) ightarrow 3	ext{CO}_2(g) + 4	ext{H}_2	ext{O}(l)$ 
According to Gay Lussac's law of gaseous volumes, gases combine or are produced in a chemical reaction in a simple ratio by volume, provided all gases are at the same temperature and pressure .
From the stoichiometry of the balanced equation, $1 	ext{ volume}$ of propane gas requires $5 	ext{ volumes}$ of oxygen gas for complete combustion.
Therefore, $1 	ext{ L}$ of propane gas will require $5 	imes 1 	ext{ L} = 5 	ext{ L}$ of oxygen gas.

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