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A hydrocarbon (C_nHₘ) is burnt in air (O₂ + 3.78N₂). The stoichiometric fuel to air mass ratio for this process is Note: Atomic Wt. C (12), H (1) Effective Molecular Wt. Air (28.8) Ignore any conversion of N₂ in air to the oxides of nitrogen (NOₓ)

1. 0.0291 ((4n + m)/(12n + m))
2. 34.42 ((12n + m)/(4n + m))
3. 34.42 ((4n + m)/(12n + m))
4. 0.0291 ((12n + m)/(4n + m))

Correct answer: 0.0291 ((12n + m)/(4n + m))

Solution

The correct option is derived from the stoichiometric calculations of the combustion process, where the mass of the hydrocarbon is compared to the mass of the air required for complete combustion. The formula reflects the relationship between the molecular weights of the components involved, ensuring that the fuel-to-air ratio is accurately represented.

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