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
Let’s find the amount of the substance С6Н12О6.
n = m: M.
M (C6H12O6) = 180 kg / kmol.
n = 1.8 kg: 180 kg / kmol = 0.01 kmol.
Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.
С6Н12О6 → 2С2Н5ОН + 2СО2 ↑.
According to the reaction equation, per 1 kmol of С6Н12О6 there is a kmol of С2Н5ОН. The substances are in quantitative ratios of 1: 2.
The amount of C2H5OH substance is 2 times more than C6H12O6.
n (C2H5OH) = 2n (C6H12O6) = 0.01 × 2 = 0.02 kmol.
Let’s find the mass of С2Н5ОН.
M (C2H5OH) = 46 kg / kmol.
m = n × M.
m = 46 kg / kmol × 0.02 kmol = 0.92 kg = 920 g.
Let’s find the volume of alcohol.
V = m: p.
V = 920 g: 0.76 g / cm3 = 1210 cm3.
Answer: 1210 cm3.