During alcoholic fermentation of glucose, 200 g of ethanol was obtained with a yield of 90%. Find the glucose mass.

1. Let’s write down the equation of alcoholic fermentation of glucose:

C6H12O6 → 2C2H5OH + 2CO2;

2.Calculate the practical chemical amount of ethanol:

n (C2H5OH) = m (C2H5OH): M (C2H5OH);

M (C2H5OH) = 2 * 12 + 5 + 16 + 1 = 46 g / mol;

npr (C2H5OH) = 200: 46 = 4.3478 mol;

3. find the theoretical amount of ethanol:

ntheor (C2H5OH) = npr (C2H5OH): ν = 4.3478: 0.9 = 4.8309 mol;

4.determine the amount of glucose:

n (C6H12O6) = ntheor (C2H5OH): 2 = 4.8309: 2 = 2.41545 mol;

5.Calculate 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.