Question

When 4 mol of glucose is fermented, a mass of 55.2 g of ethanol is produced. Show that the percentage yield of ethanol is 15%
(Mr of C2H5OH = 45)

Answer 1

Theoretical yield =0.15 g

% yield = 15%

reaction

C6H12O6 → 2C2H5OH + 2CO2

from above reaction, we get to know that,

1 mole of glucose → 2 mole ethanol

for 4 mole of glucose → 8 mole of ethanol

\frac{55.2}{8(46)} = yield of ethanol

yield of ethanol = 0.15g

% yield = (experimental/theoretical )*100

% yield of ethanol = 0.15 x 100 =15%

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