PLEASE HELP ASAP THESE 2 QUESTIONS?

1) The number of moles of carbon dioxide produced by the combustion of 1.1 mol of butane, C4H10, in presence of excess oxygen is:

2C4H10 + 13O2 → 8CO2 + 10H2O

2) How many moles of KCl are produced from 1 mol of Cl2 and excess K in the following reaction?

2K + Cl2 → 2KCl

1 Answer

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  • 3 weeks ago

    1)

    2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

    Mole ratio C₄H₁₀ : CO₂ = 2 : 8

    Moles of CO₂ produced = (1.1 mol) × (8/2) = 4.4 mol

    OR:

    (1.1 mol C₄H₁₀) × (8 mol CO₂ / 2 mol C₄H₁₀)

    = 4.4 mol CO₂

    ====

    2)

    2K + Cl₂ → 2KCl

    Mole ratio Cl₂ : KCl = 1 : 2

    Moles of KCl produced = (1 mol) × 2 = 2 mol

    OR:

    (1 mol Cl₂) × (2 mol KCl / 1 mol Cl₂)

    = 2 mol KCl

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