Percentage yield from gas volume collected

Steps:

• Moles of H₂O₂ = 1.0 mol dm⁻³ × 0.010 dm³ = 0.010 mol.
• Theoretical moles of O₂ = 0.010 mol / 2 = 0.005 mol (from 2:1 stoichiometry).
• Theoretical volume of O₂ at RTP = 0.005 mol × 24 dm³ mol⁻¹ = 0.12 dm³ = 120 cm³.
• Actual volume collected = 90 cm³, so % yield = (90 / 120) × 100% = 75%.

Why D is correct:

• Percentage yield = (actual moles O₂ / theoretical moles O₂) × 100%, matching 75% for 90 cm³ collected versus 120 cm³ theoretical.

Why the others are wrong:

• A: 28.1% assumes incorrect RTP molar volume of ~85 dm³ mol⁻¹.
• B: 37.5% from halving theoretical moles without stoichiometry.
• C: 56.3% from erroneous 100 cm³ actual volume.

Final answer: D