# During fermentation of 200 g of a 36% glucose solution, 5.6 liters of Co2 were released.

**During fermentation of 200 g of a 36% glucose solution, 5.6 liters of Co2 were released. Determine the yield in this reaction and the mass fractions of substances in the resulting mixture.**

1. We write down the equation for the reaction of glucose fermentation:

C6H12O6 = 2C2H5OH + 2CO2 ↑;

2. find the mass of glucose:

m (C6H12O6) = w (C6H12O6) * m (solution);

m (C6H12O6) = 0.36 * 200 = 72 g;

3.Calculate the chemical amount of glucose:

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

M (C6H12O6) = 6 * 12 + 12 + 6 * 16 = 180 g / mol;

n (C6H12O6) = 72: 180 = 0.4 mol;

4.determine the theoretical amount of carbon dioxide:

ntheor (CO2) = n (C6H12O6) * 2 = 0.4 * 2 = 0.8 mol;

5.Calculate the practical chemical amount of carbon dioxide:

npractic (CO2) = V (CO2): Vm = 5.6: 22.4 = 0.25 mol;

6.Calculate the product yield:

ν (CO2) = npract (CO2): ntheor (CO2) = 0.25: 0.8 = 0.3125 or 31.25%;

7.Calculate the mass of released carbon dioxide:

m (CO2) = n (CO2) * M (CO2) = 0.25 * 44 = 11 g;

8.Let’s find the mass of reacted glucose:

nproreag (C6H12O6) = 0.3125: 2 = 0.15625 mol;

mproreag (C6H12O6) = 0.15625 * 180 = 28.125 g;

9.determine the mass of the remaining glucose:

bridge (C6H12O6) = m (C6H12O6) – mproreag (C6H12O6) = 72 – 28.125 = 43.875 g;

10. find the amount of ethanol obtained:

n (C2H5OH) = ntract (CO2) = 0.25 mol;

11.Calculate the mass of ethanol:

m (C2H5OH) = 0.25 * 46 = 11.5 g

12.determine the mass of the resulting mixture:

m (mixture) = bridge (C6H12O6) + m (C2H5OH) = 43.875 + 11.5 = 55.375 g;

13.Calculate the mass fractions of the components:

w (C2H5OH) = m (C2H5OH): m (mixtures) = 11.5: 55.375 = 0.2077 or 20.77%;

w (C6H12O6) = bridge (C6H12O6): m (mixtures) = 43.875: 55.375 = 0.7923 or 79.23%.

Answer: the yield of the reaction product is 31.25%; 20.77% C2H5OH, 79.23% C6H12O6.