Identify the metal by calculating oxygen volume required for 1.5 g based on oxide formulas Steps:

• Assume STP (22.4 L/mol) for volume calculation, as room conditions approximate standard in such problems.
• Moles of metal = 1.5 / atomic mass; for Na/K (M₂O): moles O₂ = moles metal / 4; for Mg/Ca (MO): moles O₂ = moles metal / 2.
• Volume (mL) = moles O₂ × 22.4 × 1000.
• Compute: Na (A=23): moles Na=0.0652, O₂=0.0163, V=365 mL; Mg (24): moles=0.0625, O₂=0.03125, V=700 mL; Ca (40): moles=0.0375, O₂=0.01875, V=420 mL; K (39): moles=0.0385, O₂=0.0096, V=215 mL.
• Sodium's 365 mL closest to 300 mL (50% oxygen approximate for basic oxide confirmation).

Why D is correct:

• Volume for Na₂O formation matches 300 mL most closely (365 mL), per stoichiometry law for combustion reactions.

Why the others are wrong:

• A. Calcium: 420 mL exceeds 300 mL for CaO.
• B. Magnesium: 700 mL far exceeds 300 mL for MgO.
• C. Potassium: 215 mL under 300 mL for K₂O.

Final answer: D