(i) 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 20% of air contains oxygen.

(ii) The mass of 11.2 litres of a certain gas at s.t.p. is 24 g. Find the gram molecular mass of the gas.

(i) Given, 20% of air contains oxygen

∴ 20% of 1000 cm³ = $\dfrac{20}{100}$ x 1000 = 200 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.} \ 200 \text{ cm.}^3 & : & \text{x} \end{matrix}

$\therefore x = \dfrac{1}{5} \times 200 = 40 \text{ cm.}^3$

Hence, volume of propane is consumed is 40 cm³

(ii) 11.2 lit. weighs 24 g

∴ 22.4 lit. will weigh = $\dfrac{24}{11.2}$ x 22.4 = 48 g

Hence, gram molecular mass of the gas = 48 g.