Combustion produces 5 moles of CO₂ per mole of pentan-1-ol; only CO₂ is gaseous at STP Steps:

• Molar mass of C₅H₁₂O (pentan-1-ol) is 88 g/mol.
• Moles of pentan-1-ol = 37 g / 88 g/mol ≈ 0.42 mol.
• Balanced equation: C₅H₁₂O + 7.5 O₂ → 5 CO₂ + 6 H₂O(l); 5 mol CO₂ produced per mol alcohol.
• Moles CO₂ = 5 × 0.42 = 2.1 mol; volume at STP (22.4 dm³/mol) = 2.1 × 22.4 ≈ 47 dm³ (48 dm³). Why C is correct:
• Matches volume of 5 mol CO₂ per mol C₅H₁₂O using STP molar volume (22.4 dm³/mol) and Avogadro's law for ideal gases. Why the others are wrong:
• A: Volume for ~1 mol CO₂ (0.2 mol alcohol, ~18 g), half the actual mass.
• B: Approximate 1 mol gas at room temperature/pressure, ignores stoichiometry and STP conditions.
• D: Results from using room temperature molar volume (~24 dm³/mol): 2.1 × 24 ≈ 50 dm³, but question specifies STP.

Final answer: C