Calculate the volume of ethanol that can be obtained from alcoholic fermentation of 1.8 kg of glucose. The density of ethanol is 0.79 g / cm3.
1. Let’s compose the equation of the chemical reaction:
C6H12O6 = 2C2H5OH + 2CO2.
2. Find the chemical amount of glucose:
1.8 kg = 1800 g.
n (C6H12O6) = m (C6H12O6) / M (C6H12O6) = 1800 g / 180 g / mol = 10 mol.
3. According to the reaction equation, we find the chemical amount, and then the volume of ethyl alcohol:
n (C2H5OH) = n (C6H12O6) * 2 = 10 mol * 2 = 20 mol.
m (C2H5OH) = n (C2H5OH) * M (C2H5OH) = 20 mol * 46 g / mol = 920 g.
V (C2H5OH) = m (C2H5OH) / ρ (C2H5OH) = 920 g / 0.79 g / ml = 1164.6 ml = 1.16 l.
Answer: V (C2H5OH) = 1.16 l.
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