In the anaerobic (absence of oxygen) fermentation of grain, the yeast saccharomy

Question

In the anaerobic (absence of oxygen) fermentation of grain, the yeast saccharomyces cerevisiae is used as the catalyst to form ethanol and acrylic acid. In one reaction the yeast digests glucose to form ethanol and carbon dioxide. In another acrylic acid and water is formed. In a process, a tank is initially charged with 4000 kg of a 12 wt% solution of glucose in water. After fermentation, 120 kg of carbon dioxide have been produced and 90 kg of unreacted glucose remain in the broth. If none of the glucose is retained, what are the wt% of ethanol and acrylic acid in the broth at the end of the fermentation process? In the anaerobic (absence of oxygen) fermentation of grain, the yeast saccharomyces cerevisiae is used as the catalyst to form ethanol and acrylic acid. In one reaction the yeast digests glucose to form ethanol and carbon dioxide. In another acrylic acid and water is formed. In a process, a tank is initially charged with 4000 kg of a 12 wt% solution of glucose in water. After fermentation, 120 kg of carbon dioxide have been produced and 90 kg of unreacted glucose remain in the broth. If none of the glucose is retained, what are the wt% of ethanol and acrylic acid in the broth at the end of the fermentation process?

Explanation / Answer

Mass of glucose present = 12% of 4000 kg = 480 kg

Mass of water = 3520 kg

Reaction 1:

C6H12O6 ---yeast-----> 2C2H5OH + 2CO2

Moles of CO2 formed = Mass formed/MW = 120,000/44 = 2727.27 moles

Thus, moles of glucose consumed in reaction 1 = 2727.27/2 = 1363.63 moles (= 1363.63*180/1000 = 245.45 kg)

Moles of ethanol formed = 2*1363.63 = 2727.26 moles

Mass of ethanol formed = 2727.26*46/1000 = 125.45 kg

Reaction 2:

C6H12O6 ---yeast-----> 2C3H4O2 + 2H2O

Mass of glucose left unreacted = 90 kg

Thus, mass of glucose consumed in reaction 2 = 480-(245.45+90) = 144.55 kg

Moles consumed in reaction 2 = 144.55*1000/180 = 803 moles

Moles of acrylic acid formed = 2*803 = 1606 moles

Mass of acrylic acid formed = 1606*72/1000 = 115.63 kg

Moles of water formed = 2*803 = 1606 moles

Mass of water formed = 1606*18/1000 = 28.9 kg

Overall mass after reaction = 90 + 3520 + 120 + 125.45 + 115.63 + 28.9 = 4000 kg

%wt ethanol = 125.45/4000*100 = 3.13%

%wt acrylic acid = 115.63/4000*100 = 2.89%

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