During alcoholic fermentation of glucose, 200 g of ethanol was obtained with a yield of 90%.
During alcoholic fermentation of glucose, 200 g of ethanol was obtained with a yield of 90%. Find the mass of glucose. That is, 200g. ethanol is 90%
1. Alcoholic fermentation of glucose proceeds according to the reaction equation:
C6H12O6 → 2C2H5OH + 2CO2 ↑;
2.Calculate the practical chemical amount of ethanol obtained:
ntrack (C2H5OH) = m (C2H5OH): M (C2H5OH);
M (C2H5OH) = 2 * 12 + 5 + 16 + 1 = 46 g / mol;
ntract (C2H5OH) = 200: 46 = 4.3478 mol;
3. Determine the theoretical amount of ethanol:
ntheor (C2H5OH) = ntract (C2H5OH): ν = 4.3478: 0.9 = 4.8309 mol;
4.According to the reaction equation, the chemical amount of glucose is two times less than the alcohol produced:
n (C6H12O6) = ntheor (C2H5OH): 2 = 4.8309: 2 = 2.41545 mol;
5. find the mass of glucose:
m (C6H12O6) = n (C6H12O6) * M (C6H12O6);
M (C6H12O6) = 6 * 12 + 12 + 6 * 16 = 180 g / mol;
m (C6H12O6) = 2.41545 * 180 = 434.781 g
Answer: 434.781 g.