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.