(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.
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(i) Given, 20% of air contains oxygen
∴ 20% of 1000 cm3 = $\frac{20}{100}$ 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 :
$$\begin{matrix}\text{O}_2 & : & \text{C}_3\text{H}_8 \\ 5 \text{ vol.} & : & 1 \text{ vol.} \\ 200 \text{ cm.}^3 & : & \text{x} \end{matrix}$$
$\therefore x = \frac{1}{5} \times 200 = 40 \text{ cm.}^3$
Hence, volume of propane is consumed is 40 cm3
(ii) 11.2 lit. weighs 24 g
∴ 22.4 lit. will weigh = $\frac{24}{11.2}$ x 22.4 = 48 g
Hence, gram molecular mass of the gas = 48 g.