During fermentation of 300 g of glucose, 100 g of ethyl alcohol was obtained. Determine the alcohol yield.

| January 12, 2021 | education

During fermentation of 300 g of glucose, 100 g of ethyl alcohol was obtained. determine the alcohol yield
Let’s find the amount of the substance С6Н12О6.
n = m: M.
M (C6H12O6) = 180 g / mol.
n = 300 g: 180 g / mol = 1.67 mol.
Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.
С6Н12О6 → 2 С2Н5ОН + 2СО2 ↑.
According to the reaction equation, there is 2 mol of C2H5OH per 1 mol of С6Н12О6. The substances are in quantitative ratios of 1: 2.
The amount of the substance С2Н5ОН is 2 times more than С6Н12О6.
n (C2H5OH) = 2n (C6H12O6) = 1.67 × 2 = 3.34 mol.
Let’s find the mass of С2Н5ОН.
M (C2H5OH) = 46 g / mol.
m = n × M.
m = 46 g / mol × 3.34 mol = 153.64 g (theory).
153.64 – 100%
100 g – x%,
X = (100 g x 100%): 153.64 g = 65.087%.
Answer: 65.087%.



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