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26 October, 00:31

Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 26.7g of carbon dioxide is produced from the reaction of 10.3g of octane and 71.6g of oxygen gas, calculate the percent yield of carbon dioxide.

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  1. 26 October, 00:46
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    96.13% Yield Carbon

    Explanation:

    molecular weights:

    mwOctane = 114 g/mol

    mwO2 = 32 g/mol

    mwCO2 = 44 g/mol

    mwH2O = 18 g/mol

    Balancing the equation:

    2 CH3 (CH2) 6CH3 + 25 O2 = 16 CO2 + 18 H2O

    Present moles:

    gOCtane = 10.3g

    gO2 = 26.7g

    molOctane = gOCtane/mwOctane = > 0.0904 mol

    molO2 = gO2/mwO2 = > 0.8344 mol

    check ratios based on chemical equation:

    Octaneratio = molOctane/2mol = > 0.0452

    O2ratio = molO2/25mol = > 0.0334

    since Oxygen ratio is smaller, the this is the limit reactant:

    25mol of O2 produces 18 moles of Water, then 1.2188 moles of oxygen produces:

    (25mol/16mol) * molO2 = > 1.3037 mol

    of waters produced.

    1.6927 mol*mwCO2 = > 74.4788 g

    should be produced, CO2 Produced,

    Real Produced = 71.6 g

    Ideally Produced = 74.48 g

    Yield = Real Produced/Ideally Produced = > 0.9613

    96.13% Yield Carbon
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