Question

Air contains 79% N₂ and 21% O₂ on a molar basis. Methane (CH₄) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N₂ in the products is) ________.

Answer 1

Concept:

One litre of air contains 21% O₂ and 79% N₂, so 1 litre of air = [O₂ + 3.76 N₂]

Now stoichiometric reaction

CH₄ + 2[O₂ + 3.76 N₂] → 2H₂O + 2 × 3.76 N₂ + CO₂

50% excess air is supplied

CH₄ + 1.5 × 2 [O₂ + 3.76 N₂] → 2H₂₀ + CO₂ + 3 × 3.76 N₂ + O₂

\(\% \;of\;{N_2} = \frac{{3 \times 3.76}}{{2 + 1 + 1 + 3 \times 3.76}} = 73.82\% \)