In the anaerobic fermentation of grain, the yeast Saccharomyces cerevisiae digests glucose from plants to form the products ethanol and propenoic acid by the following reactions: Reaction 1: C6H12O6  2 C2H5OH + 2 CO2 Reaction 2: C6H12O6  2 C2H3CO2H + 2 H2O In an open flow reactor 4000 kg of a 12% glucose-water solution flows in. During fermentation, 120 kg of carbon dioxide is produced together with 90 kg of unreacted glucose. What are the weight percents of ethyl alcohol and propenoic acid that exit the broth? Assume that none of the glucose is assimilated into the bacteria.

Explanation:

The first step in order to solve this particular question is to make sure that the two reactions given in the question is balanced. Therefore, we have;

Reaction 1: C6H12O6 > 2 C2H5OH + 2 CO2.

Reaction 2: C6H12O6 > 2 C2H3CO2H + 2 H2O.

Next, we determine the number of moles of water and that of glucose. Recall that we are given from the question that the open flow reactor =  4000 kg of a 12% glucose-water solution flows in that is to say the percentage for water is [100% - 12% = 88%]. Also, the molar mass of water = H₂O = 18 kg/kmol and that for glucose =180 kg/kmol.

Number of moles of water = (4000 kg × 88%) ÷ 18 = 195.6 kmol.

Number of moles of glucose= (4000kg × 12%) ÷ 180 = 2.67 kmol.

Next thing to do is to determine the number of moles in the unreacted glucose . Therefore, the 90 kg of unreacted glucose ÷ 180kg/kmol = 0.5  kmol.

So, we have that During fermentation, 120 kg of carbon dioxide is produced.  Thus, the number of kmol = 120kg÷ 44kg/mol = 2.73kmol.

For reaction 1, we have 2 moles of CO₂ that is to say the extent of the reaction = 2.73kmol / 2 moles of CO₂ = 1.365 kmol.

For reaction 2, we have 2 moles of CO₂ that is to say the extent of the reaction = 2.67kmol - 0.5  kmol - 1.365kmol = 0.805 kmol.

For both reaction, the kmol for outflow of glucose = 2.73 kmol.

Also, 2 × 0.805 + 1.365 × 0 = 1.61kmol.

Hence, 195.6 kmol + 1.61 =197.21 kmol.

The mass of ethanol = 46.1 kg/kmol × 2.73 kmol = 125.853 kg.