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3 March, 15:09

The following equation shows the conversion of ethanol c2h5oh in a wine sample into acetic acid c2h4o2 through the action of oxygen from air c2h5oh+o2->c2h4o2+h2o this reaction drastically changes the taste of the wine. If 100.0mL of wine initially contained 12.0g of ethanol, and after a period of time 4.00g of acetic acid were detected, what percent of the ethanol had been converted?

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  1. 3 March, 15:55
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    Given:

    Volume of wine = 100 ml

    Initial Mass of ethanol = 12.0 g

    Mass of acetic acid formed = 4.00 g

    To determine:

    % ethanol converted to acetic acid

    Explanation:

    Molar mass of C2H5OH (ethanol) = 46 g/mol

    Molar mass of C2H4O2 (acetic acid) = 60 g/mol

    The reaction of ethanol in air is:

    C2H5OH + O2 → C2H4O2 + H2O

    Based on the stoichiometry: 1 mole of ethanol produces 1 mole of acetic acid

    Moles of ethanol initially present = mass of ethanol/molar mass

    = 12/46 = 0.2609 moles

    Moles of acetic acid produced = moles of ethanol converted

    = 4/60 = 0.0667 moles

    Mass of ethanol converted = Moles of ethanol * molar mass

    = 0.0667 * 46 = 3.07 g

    % ethanol converted = (mass of ethanol converted/initial mass) * 100

    % converted = (3.07/12.0) * 100 = 25.6%

    Ans: 25.6% of ethanol gets converted to acetic acid
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