Calculate the volume of ethanol that can be obtained from alcoholic fermentation of 1.8 kg

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.