During the fermentation of 500 g of glucose solution, 13.8 g of ethyl alcohol was formed.

During the fermentation of 500 g of glucose solution, 13.8 g of ethyl alcohol was formed. Determine the mass fraction of glucose (in%) in the original solution.

1. Fermentation of a monosaccharide proceeds according to the scheme:

C6H12O6 → 2C2H5OH + 2CO2 ↑.

2. Knowing the mass of the obtained alcohol, we find its chemical amount:

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

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

n (C2H5OH) = 13.8: 46 = 0.3 mol.

3. The amounts of glucose and alcohol according to the reaction scheme are correlated as 1: 2, we determine how many moles the monosaccharide has:

n (C6H12O6) = n (C2H5OH): 2 = 0.3: 2 = 0.15 mol.

4. Set the glucose mass:

m (C6H12O6) = n (C6H12O6) * M (C6H12O6);

M (C6H12O6) = 6 * 12 + 12 + 6 * 16 = 108 g / mol;

m (C6H12O6) = 0.15 * 180 = 27 g.

5. Let’s calculate the mass fraction of glucose in the initial solution:

w (C6H12O6) = m (C6H12O6): m (solution) = 27: 500 = 0.054 or 5.4%.

Answer: 5.4%.