In which one of the following reactions, the maximum quantity of H2 gas is produced by the decomposition of 1 g of compound by H20/02?

To determine the maximum hydrogen yield per gram of compound, we calculate the moles of H2 produced per gram of reactant. In reaction 1 (Steam Methane Reforming), 1 mole of CH4 (16 g) produces 3 moles of H2 [2]. This yields 3/16 = 0.1875 moles of H2 per gram. In reaction 2 (Water-Gas Shift), 1 mole of CO (28 g) produces 1 mole of H2 [1], yielding 1/28 ≈ 0.0357 moles/g. In reaction 3 (Partial Oxidation), 1 mole of CH4 (16 g) produces 2 moles of H2 [1], yielding 2/16 = 0.125 moles/g. In reaction 4, 1 mole of C12H24 (168 g) produces 12 moles of H2, yielding 12/168 ≈ 0.0714 moles/g. Therefore, the decomposition of methane via steam reforming (Reaction 1) provides the highest quantity of hydrogen gas per gram of the starting compound.

Sources

1. [1] https://www.energy.gov/eere/fuelcells/hydrogen-production-natural-gas-reforming
2. [2] https://ww3.arb.ca.gov/fuels/lcfs/workshops/08142017_h2renewables-attachment.pdf