MDCAT Chemistry MCQ #107

How many grams of methanol (CH3OH) are needed to react with 4.8 g of O2 in 2CH3OH + 3O2 → 2CO2 + 4H2O? (Molar masses: CH3OH=32 g/mol, O2=32 g/mol)

MDCAT Chemistry MCQ #107

  1. Question 1

    Q1. How many grams of methanol (CH3OH) are needed to react with 4.8 g of O2 in 2CH3OH + 3O2 → 2CO2 + 4H2O? (Molar masses: CH3OH=32 g/mol, O2=32 g/mol)

    • A) 3.2 g
    • B) 4.8 g
    • C) 6.4 g
    • D) 9.6 g

    Answer: 3.2 g

    Explanation: Molar ratio 2CH3OH:3O2. Moles O2 = 4.8 / 32 = 0.15 mol. Moles CH3OH = (2/3)(0.15) = 0.1 mol. Mass = 0.1 × 32 = 3.2 g. Option B swaps reactants.