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11 April, 19:02

The balanced equation for the combustion of ethanol is 2C2H5OH (g) + 7O2 (g) → 4CO2 (g) + 6H2O (g) How many grams of oxygen gas are required to burn 5.54 g of C2H5OH? (write your answer with 3 sig figs and no units)

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  1. 11 April, 19:28
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    Answer: 13.5g of O2

    Explanation:

    2C2H5OH + 7O2 → 4CO2 + 6H2O

    Molar Mass of C2H5OH = (12x2) + 5+16+1 = 46g/mol

    Mass conc. Of C2H5OH = 2x46 = 92g

    Molar Mass of O2 = 32g/mol

    Mass conc. Of O2 = 7 x 32 = 224g

    From the equation,

    92g of C2H5OH required 224g O O2.

    Therefore, 5.54g of C2H5OH will require = (5.54x224) / 92 = 13.5g of O2
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