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14 March, 17:56

Calculate the mass of water produced when 42 g of propane, c3h8, is burned with 115 g of oxygen

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  1. 14 March, 18:24
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    The balanced combustion reaction of propane, C₃H₈, is

    C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

    Molar mass of propane: 44 g/mol

    Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane

    Molar mass of oxygen: 32 g/mol

    Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen

    Moles of oxygen needed to completely react propane:

    0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen

    Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.

    Molar mass of water: 18 g/mol

    Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)

    Mass of water produced = 258.768 grams
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