Calculate how much ethanol alcohol is obtained during alcoholic fermentation of 90 g of glucose, the product yield is 60% of the theoretically possible.
Let’s find the amount of substance С6Н12О6.
n = m: M.
M (C6H12O6) = 180 g / mol.
n = 90g: 180 g / mol = 0.5 mol.
Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.
С6Н12О6 → 2С2Н5ОН + 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 C2H5OH substance is 2 times more than C6H12O6.
n (C2H5OH) = 2n (C6H12O6) = 0.5 × 2 = 1 mol.
Let’s find the mass of С2Н5ОН.
M (C2H5OH) = 46 g / mol.
m = n × M.
m = 46 g / mol × 1 mol = 46 g.
46g – 100%,
X – 60%,
X = (46 × 60%): 100% = 27.6 g.
Answer: 27.6 g.
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