Fermentation yielded 10 L, 50% ethanol solution (S = 1.2 g / cm3). What mass of glucose was used for this if η = 80%?

decision
m = p * V = 1, 2 * 10 liters = 1, 2 * 10,000 = 12,000 g = 0, 012 kg is the mass of the resulting solution in the standard
m (C2 H5 OH) = m (p – pa) * W percent / 100 percent = 6,000 grams = 0, 006 kg – the practical mass of the standard in solution
m (C2 H5 OH) = 6000 grams * 100 percent / 80 percent = 7500 g = 0.0075 kg – theoretical mass of the standard in solution

x 7500 grams
C6 H12 O6 2 * C2 H5 OH + 2 * CO2
1 mol 2 mol
M = 180 grams / mol M = 46 grams / mol
n = 7500 grams / 46 grams / mol = 7500/46 mol = 163 mol
since the standard is 2 times greater than the glucose consumption, then n (C6 H12 O6) = 163 mol / 2 = 81.5 mol
then m (C6 H12 O6) = M * n = 180 grams / mol * 81.5 mol = 14670 grams = 14.67 kg
answer: 14, 67 kg