Comparing oxygen moles in balanced combustion reactions Steps: - Write balanced equation for complete combustion: CxHyOz + O2 → xCO2 + (y/2)H2O, solving for O2 coefficient. - For CH3OH (C1H4O1): CH3OH + 1.5O2 → CO2 + 2H2O (1.5 mol O2). - For CH3COOH (C2H4O2): CH3COOH + 2O2 → 2CO2 + 2H2O (2 mol O2); for C2H6 (C2H6): C2H6 + 3.5O2 → 2CO2 + 3H2O (3.5 mol O2); for C2H2 (C2H2): C2H2 + 2.5O2 → 2CO2 + H2O (2.5 mol O2). - Identify lowest O2 value as 1.5 mol for CH3OH. Why A is correct: - CH3OH requires only 1.5 mol O2 per mole due to its pre-existing oxygen atom, minimizing external O2 per carbon and hydrogen oxidized (per combustion stoichiometry). Why the others are wrong: - B requires 2 mol O2, higher due to two carbons needing more oxidation despite one oxygen. - C requires 3.5 mol O2, highest as hydrocarbon lacks oxygen, demanding full O2 for all C-H bonds. - D requires 2.5 mol O2, more than A from unsaturated bonds still needing substantial O2 for two carbons. Final …