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1 February, 19:09

11) If 352 moles of butane (CH) is mixed with 8.47 moles of oxygen gas, how many grams of water will be produced?

2C, H, 0 + 130, 800, + 10 H, 0 (balanced)

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  1. 1 February, 20:00
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    117 g

    Explanation:

    We are given the amounts of two reactants, so this is a limiting reactant problem.

    1. We know we will need an equation with masses and molar masses, so let's gather all the information in one place.

    M_r: 63.55 32.00 18.02

    2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

    n/mol: 352 8.47

    2. Calculate the amount of water that each reactant can produce

    (a) From C₄H₁₀

    Moles of H₂O = 352 mol C₄H₁₀ * (10 mol H₂O/2 mol C₄H₁₀) = 704 mol H₂O

    (b) From O₂:

    Moles of H₂O = 8.47 mol O₂ * (10 mol H₂O/13 mol O₂) = 6.515 mol H₂O

    3. Identify the limiting reactant

    The limiting reactant is O₂, because it gives fewer moles of H₂O.

    4. Calculate the mass of water

    Mass of H₂ = 5.515 mol H₂O * (18.02 g H₂) / (1 mol H₂O) = 117 g H₂O

    The reaction produces 117 g H₂O.
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