What volume of propane is burnt for every 500 cm³ of air used in the reaction under the same conditions? (assuming oxygen is 1/5th of air)

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

Given, oxygen is 1⁄5^th of air = $\dfrac{1}{5}$ of 500 = 100 cm³

$\begin{matrix} \text{C}$3\text{H}8 & + & 5\text{O}2 & \longrightarrow & 3\text{CO}2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3 \text{ vol.} & \\ \end{matrix}

[By Gay Lussac's law]

5 Vol. of O₂ requires 1 Vol. of propane

∴ 100 cm³ of O₂ will require = $\dfrac{1}{5}$ x 100 = 20 cm³

Hence, propane burnt = 20 cm³ or 20 cc