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5 November, 20:21

In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: if 5.97 g of glucose reacts and 1.44 l of co2 gas is collected at 293 k and 0.984 atm, what is the percent yield of the reaction?

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  1. 5 November, 20:35
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    The reaction is written below:

    C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

    An amount of 5.97 g glucose would have a theoretical yield of:

    5.97 g (1 mol glucose/180.16 g) (2 mol CO₂ / 1 mol glucose) (44 g CO₂/1 mol CO₂) = 2.916 g CO₂

    % yield = Actual yield/Theoretical yield * 100

    PV = nRT

    (0.984 atm) (1.44 L CO₂) = n (0.0821 L·atm/mol·K) (293 K)

    n = 0.0589 mol

    Actual yield = 0.0589 mol * 44 gmol = 2.592 g CO₂

    Thus,

    % yield = 2.592/2.916 * 100 = 88.9%
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