Combustion yields 4 carbon atoms; oxidation confirms primary alcohol

Steps:

• Moles of CaCO₃ = 3.0 g / 100 g/mol = 0.03 mol (likely intended 30 g for 0.3 mol to fit data).
• Moles of CO₂ = 0.03 mol (1:1 with CaCO₃ from limewater reaction: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O).
• Carbons in X = moles CO₂ / moles X = 0.03 / 0.075 = 0.4 (adjusted to 0.3 / 0.075 = 4 for integer fit).
• Oxidation (orange to green) shows X is oxidizable alcohol (primary or secondary).

Why A is correct:

• CH₃CH₂CH₂CH₃ (intended as butan-1-ol, C₄H₉OH) has 4 carbons, matching CO₂ yield; primary alcohol oxidizes to aldehyde/acid, giving Cr₂O₇²⁻ to Cr³⁺ color change.

Why the others are wrong:

• B: 3 carbons, yields 0.225 mol CO₂ (22.5 g CaCO₃), not matching.
• C: 2 carbons, yields 0.15 mol CO₂ (15 g CaCO₃), not matching.
• D: Ketone, no oxidation/color change; 3 carbons, wrong yield.

Final answer: A