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3 March, 03:49

Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass of glucose is 180.15 g/mol, the molar mass of ethanol is 46.08 g/mol, and the molar mass of carbon dioxide is 44.01 g/mol. What is the theoretical yield of ethanol from the fermentation of 61.5 g of glucose? theoretical yield: g If the reaction produced 23.4 g of ethanol, what is the percent yield? percent yield:

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  1. 3 March, 04:54
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    The % yield is 74.45 %

    Explanation:

    Step 1: The balanced equation

    C6H12O6⟶2CH3CH2OH+2CO2

    Step 2: Data given

    Molar mass glucose = 180.15 g/mol

    Molar mass of ethanol = 46.08 g/mol

    Molar mass of carbon dioxide = 44.01 g/mol

    Mass of glucose = 61.5 grams

    Mass of ethanol = 23.4 grams

    Step 3: Calculate moles of glucose

    Moles glucose = Mass glucose / Molar mass of glucose

    Moles glucose = 61.5 grams / 180.15 g/mol

    Moles glucose = 0.341 moles

    Step 4: Calculate moles of ethanol

    1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

    0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

    Step 5: Calculate mass of ethanol

    Mass ethanol = moles ethanol * Molar mass ethanol

    Mass ethanol = 0.682 moles * 46.08 g/mol

    Mass ethanol = 31.43 grams = theoretical mass

    Step 6: Calculate % yield

    % yield = actual mass / theoretical mass

    % yield = (23.4 grams / 31.43 grams) * 100%

    % yield = 74.45 %

    The % yield is 74.45 %
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