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GATE 2025 CE (CE1) – Question 22

Air Pollution and Municipal Solid Wastes · Air pollution: pollutants, sources, impacts, control and air quality index · 1 mark · Multiple choice

A hydrocarbon ($C_nH_m$) is burnt in air ($O_2 + 3.78N_2$). The stoichiometric fuel to air mass ratio for this process is

Note: Atomic Weight: C(12), H(1)

Effective Molecular Weight: Air(28.8)

Ignore any conversion of $N_2$ in air to the oxides of nitrogen ($NO_x$)

  1. $0.0291\,\frac{(4n + m)}{(12n + m)}$
  2. $34.42\,\frac{(12n + m)}{(4n + m)}$
  3. $34.42\,\frac{(4n + m)}{(12n + m)}$
  4. $0.0291\,\frac{(12n + m)}{(4n + m)}$

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Correct answer: (D) $0.0291\,\frac{(12n + m)}{(4n + m)}$

Explanation

The reaction is $C_nH_m + \left(n + \frac{m}{4}\right)(O_2 + 3.78N_2) \rightarrow nCO_2 + \frac{m}{2}H_2O + 3.78\left(n + \frac{m}{4}\right)N_2$. One mole of fuel has a mass of $12n + m$. The air needed is $4.78\left(n + \frac{m}{4}\right)$ moles, which has a mass of $4.78 \times 28.8 \times \frac{4n + m}{4} = 34.42(4n + m)$. So the fuel to air mass ratio is $\frac{12n + m}{34.42(4n + m)} = 0.0291\,\frac{12n + m}{4n + m}$. Option B is the air to fuel ratio.