Mole Concept and Stoichiometry — Question 14

$\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1\text{ vol.} \\ 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

1 Vol. CH₄ requires 2 Vol. of O₂

∴ 10 cm³ CH₄ will require 2 x 10

= 20 cm³ of O₂

Given, air contains 20% O₂ by volume.

Let $x$ volume of air contain 20 cm³ of O₂

$\Rightarrow \dfrac{20}{100} \times x = 20 \\[1em] \Rightarrow x = \dfrac{100}{20} \times 20 \\[1em] \Rightarrow x = 100 \text{ cm}^3$

∴ 20 cm³ O₂ is present in 100 cm³ of air.

Similarly, 2 Vol C₂H₂ requires 5 Vol. of oxygen

∴ 10 cm³ C₂H₂ will require $\dfrac{5}{2}$ x 10

= 25 cm³ of oxygen

Given, air contains 20% O₂ by volume

Let $x$ volume of air contain 25 cm³ of O₂

$\Rightarrow \dfrac{20}{100} \times x = 25 \\[1em] \Rightarrow x = \dfrac{100}{20} \times 25 \\[1em] \Rightarrow x = 125 \text{ cm}^3$

∴ 25 cm³ O₂ is present in 125 cm³ of air.

Hence, total volume of air required is 100 + 125 = 225 cm³