(a) Propane burns in air according to the following equation.

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

What volume of propane is consumed on using 1000 cm³ of air, considering only 40% of air contains oxygen?

(b) Calculate the gram molecular mass of N₂, if 360 cm³ at S.T.P. weighs 0.45 g.

(a) Given, 40% of air contains oxygen

∴ 40% of 1000 cm³ = $\dfrac{40}{100}$ x 1000 = 400 cm³

[By Lussac's law]

$\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.} & : & 4\text{ vol.} \end{matrix}

To calculate the volume of propane consumed :

$\begin{matrix}\text{O}$2 & : & \text{C}3\text{H}_8 \ 5 \text{ vol.} & : & 1 \text{ vol.} \ 400 \text{ cm.}^3 & : & \text{x} \end{matrix}

$\therefore x = \dfrac{1}{5} \times 400 = 80 \text{ cm.}^3$

Hence, volume of propane consumed is 80 cm³

(b) The mass of 22.4 L of a gas at S.T.P. is equal to its gram molecular mass.

360 cm³ of N₂ at S.T.P. weighs 0.45 g

∴ 22,400 cm³ of N₂ will weigh $\dfrac{0.45}{360} \times 22,400$ = 28 g

Hence, Gram Molecular Mass of N₂ = 28 g.