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.
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