Butane burns in oxygen. How many grams of water vapor is produced by the combustion of 580 grams of butane at standard conditions?

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  • 4 weeks ago
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    Molar mass of C₄H₁₀ = (12.0×4 + 1.0×10) g/mol = 58.0 g/mol

    Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol

    Equation for the reaction:

    2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

    Mole ratio C₄H₁₀ : H₂O = 2 : 10 = 1 : 5

    Moles of C₄H₁₀ = (580 g) / (58.0 g/mol) = 10 mol

    Moles of H₂O = (10 mol) × 5 = 50 mol

    Mass of H₂O vapor formed = (50 mol) × (18.0 g/mol) = 900 g

    ====

    OR:

    (580 g C₄H₁₀) × (1 mol C₄H₁₀ / 58.0 g H₂O) × (10 mol H₂O / 2 mol C₄H₁₀) × (18.0 g H₂O / 1 mol H₂O)

    = 900 g H₂O (vapor)

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