Stoichiometric methane combustion with air (all gaseous products):\nFor CH4 + 2 O2 → CO2 + 2 H2O with stoichiometric air (21% O2, 79% N2 by volume), what is the water vapour mole fraction in the product mixture?

Introduction / Context:
Combustion stoichiometry is central to furnace design, flue-gas analysis, and emissions calculations. Here, we burn methane with exactly the stoichiometric amount of air and treat all species as gases. The task is to determine the mole fraction of water vapour in the product mixture at the chemical-equilibrium stoichiometry (no excess air or fuel).

Given Data / Assumptions:

• Reaction basis: 1 mol CH4.
• Stoichiometric oxygen requirement: 2 mol O2 per mol CH4.
• Air composition: 21% O2, 79% N2 by volume (mole basis).
• Products considered: CO2, H2O, and N2 only (complete combustion).

Concept / Approach:
Perform a molar balance. Determine the moles of N2 accompanying the necessary O2 in stoichiometric air, then sum product moles and compute the water fraction yH2O = nH2O / ntotal. No enthalpy data are needed for this composition calculation.

Step-by-Step Solution:

Verification / Alternative check:
Because stoichiometric air introduces approximately 3.76 mol N2 per mol O2, the N2 content dominates the product moles, placing yH2O near 2 / (1 + 2 + 2*3.76) ≈ 0.19. This quick check matches the detailed calculation.

Why Other Options Are Wrong:

• 0.33, 0.40, 0.67: these overestimate water fraction by ignoring the large N2 ballast.
• 0.10: underestimates by roughly a factor of two relative to the correct stoichiometric outcome.

Common Pitfalls:
Forgetting to include nitrogen from air; using mass rather than mole fractions; assuming dry products when the question asks for total product mixture including steam.

Final Answer:
0.19