In which one of the following reactions, the maximum quantity of H₂ ga

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?

CH₄ + H₂O → CO + 3H₂
CO + H₂O → CO₂ + H₂
CH₄ + ½O₂ → CO + 2H₂
C₁₂H₂₄ + 6O₂ → 12CO + 12H₂
This question was previously asked in
UPSC CDS-2 – 2018
The reaction in option A produces the maximum quantity of H₂ gas per gram of the reactant compound.
To find the maximum quantity of H₂ produced per gram of reactant, we need to calculate the moles of H₂ produced per gram of the principal reactant in each reaction. Let’s calculate the moles of H₂ per gram of the compound undergoing decomposition/reaction:
A) CH₄ + H₂O → CO + 3H₂: Molar mass of CH₄ is 16 g/mol. 1 mol CH₄ produces 3 mol H₂. Moles of H₂ per gram of CH₄ = (3 mol H₂) / (16 g CH₄) = 3/16 mol/g.
B) CO + H₂O → CO₂ + H₂: Molar mass of CO is 28 g/mol. 1 mol CO produces 1 mol H₂. Moles of H₂ per gram of CO = (1 mol H₂) / (28 g CO) = 1/28 mol/g.
C) CH₄ + ½O₂ → CO + 2H₂: Molar mass of CH₄ is 16 g/mol. 1 mol CH₄ produces 2 mol H₂. Moles of H₂ per gram of CH₄ = (2 mol H₂) / (16 g CH₄) = 2/16 = 1/8 mol/g.
D) C₁₂H₂₄ + 6O₂ → 12CO + 12H₂: Molar mass of C₁₂H₂₄ is 168 g/mol. 1 mol C₁₂H₂₄ produces 12 mol H₂. Moles of H₂ per gram of C₁₂H₂₄ = (12 mol H₂) / (168 g C₁₂H₂₄) = 12/168 = 1/14 mol/g.
Comparing the values: 3/16 (0.1875), 1/28 (~0.0357), 1/8 (0.125), 1/14 (~0.0714). The largest value is 3/16 mol/g, corresponding to reaction A.
These reactions represent different methods for producing hydrogen gas, often from hydrocarbon fuels or syngas (CO + H₂). Reaction A is a simplified steam reforming of methane. Reaction B is the water-gas shift reaction. Reaction C is partial oxidation of methane. Reaction D involves a larger hydrocarbon. The amount of hydrogen produced depends on the stoichiometry of the reaction and the molar mass of the starting material.