Question

Forty litres of a mixture of methane and nitrogen requires 24 L of pure oxygen for complete combustion. The percentage of methane in the mixture is (a) 40 (b) 25 (c) 30 (d) 60

Answer 1

Answer:

$\boxed{30 \, \%}$

Explanation:

We can use Gay-Lussac's Law of Combining Volumes to solve this problem.

Gases at the same temperature and pressure react in the same ratios as their coefficients in the balanced equation.

1. Write the chemical equation.

Ratio:  1 L     2 L

           CH₄ + 2O₂ → CO₂ + 2H₂O

V/L:                24

2. Calculate the volume of O₂.

According to Gay-Lussac, 1 L of CH₄ reacts with 2 L of O₂.

Then, the conversion factor is (1 L CH₄/2 L O₂).

$\text{Volume of CH}_{4} = \text{24 L O}_{2} \times \dfrac{ \text{1 L CH}_{4} }{\text{ 2 L O}_{2}}= \text{12 L CH}_{4}$

3. Calculate the percentage of methane

$\text{\% Methane } = \dfrac{\text{12 L}}{\text{40 L}} \times 100 \%= \mathbf{30 \, \%} \\\\\text{The percentage of methane in the mixture is }\boxed{\mathbf{30 \, \%}}}$