Mass loss from NaHCO₃ decomposition determines its percentage

Steps:

• Mass loss (1.5 g) equals mass of CO₂ + H₂O from 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O.
• 2 mol NaHCO₃ (168 g) yield 62 g gases (CO₂ 44 g + H₂O 18 g).
• Fraction of NaHCO₃ mass lost = 62/168.
• Let NaHCO₃ mass = x g; then (62/168)x = 1.5 → x = 1.5 × (168/62) = 4.07 g.
• Percentage = (4.07/6) × 100% = 68%.

Why C is correct:

• Matches calculation: 68% NaHCO₃ yields 1.5 g loss per stoichiometry.

Why the others are wrong:

• A: Assumes 100% loss from NaHCO₃, ignoring Na₂CO₃ residue.
• B: Miscalculates gas mass as half of NaHCO₃ input.
• D: Treats entire mixture mass as lost, exceeding actual gases.

Final answer: C