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14 March, 06:28

In the organic combustion reaction of 41.9 g of octane (C8H18) with excess oxygen, what volume (in L) of carbon dioxide is produced if the reaction is performed at STP?

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  1. 14 March, 08:12
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    The volume CO2 produced is 65.8 L

    Explanation:

    Step 1: Data given

    Mass of octane = 41.9 grams

    Molar mass octane = 114.23 g/mol

    Step 2: The balanced equation

    2C8H18 + 25O2 → 16CO2 + 18H2O

    Step 3: Calculate moles octane

    Moles octane = mass octane / molar mass octane

    Moles octane = 41.9 grams / 114.23 g/mol

    Moles octane = 0.367 moles

    Step 4: Calculate moles CO2

    For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

    For 0.367 moles octane we need 8*0.367 = 2.936 moles

    Step 5: Calculate volume of CO2

    1 mol = 22.4 L

    2.936 moles = 22.4 * 2.936 = 65.8 L

    The volume CO2 produced is 65.8 L
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