In which one of the following reactions, the maximum quantity of H₂ gas is produced by the decomposition of 1 g of compound by H₂O/O₂?
- ACH₄ + H₂O → CO + 3H₂
- BCO + H₂O → CO₂ + H₂
- CCH₄ + ½O₂ → CO + 2H₂
- DC₁₂H₂₄ + 6O₂ → 12CO₂ + 12H₂
Explanation
Computing H2 yield per gram of the named compound: (A) CH4 + H2O -> CO + 3H2 gives 3 mol H2 per 16 g, i.e. 0.1875 mol/g; (B) CO + H2O -> CO2 + H2 gives 1 mol per 28 g, i.e. about 0.036 mol/g; (C) CH4 + 1/2 O2 -> CO + 2H2 gives 2 mol per 16 g, i.e. 0.125 mol/g; (D) C12H24 + 6O2 -> 12CO2 + 12H2 gives 12 mol per 168 g, i.e. about 0.071 mol/g. The steam reforming of methane (reaction A) wins because it produces three light hydrogen molecules from the lightest hydrocarbon feed. None of the heavier carbon monoxide or C12H24 feeds can match this hydrogen-to-mass ratio. Reaction A therefore produces the maximum quantity of H2 per gram of compound.