A chemical engineer is conducting a pilot plant study to examine the economic fe

Question

A chemical engineer is conducting a pilot plant study to examine the economic feasibility of anaerobically fermenting grain to form ethanol and acrylic acid, two high-value products, using the yeast Saccharomyces cerevisiae as the catalyst. The yeast digests glucose from the grain to form ethanol and carbon dioxide in one reaction, and acrylic acid and water in a second reaction. 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. Assuming that none of the glucose is retained by the microorganisms, what are the wt% of ethanol and acrylic acid in the broth at the end of the fermentation process?

Explanation / Answer

First reaction: C6H12O6 (glucose)--yeast--->2 C2H5OH(ethanol)+2 CO2 (carbon dioxide)

Second reaction: C6H12O6 (glucose)--yeast--->2 CH2=CHCOOH (acrylic acid)+2H2O(water)

mass of glucose fed into the tank=12% of 4000 kg=0.12*4000 kg*1000g=480000g

moles of glucose fed =mass/molar mass=480000g/180.16g/mol=2664.30 moles

moles of glucose remaining after the end of process=90kg=90000g/180.16g/mol=499.55 moles

moles of glucose reacted=2664.30 moles-499.55moles=2164.74 moles

mass of carbon dioxide formed=120 kg=120000g

moles of CO2 formed=120000g/44.01 g/mol=2726.65 moles

as in the first reaction 2 moles of CO2 is produced for every mole of glucose reacted,so you can calculate the moles of glucose taking part in first reaction from the moles of CO2 produced

Thus,moles of glucose taking part in first reaction=1/2* moles of CO2 produced=1/2*(2726.65 moles)=1363.33 moles

So ,,moles of glucose taking part in the second reaction=total moles of glucose -,moles of glucose taking part in first reaction=2164.74 moles-1363.33 moles=801.41 moles

therefore moles of ethanol produced (as per the first reaction)=2*(moles of glucose taking part in first reaction)=2*1363.33 moles=2726.65moles

mass of ethanol=2726.65moles*46.07 g/mol=125.616 kg

Also moles of acrylic acid produced=2*(moles of glucose taking part in the second reaction)=2*801.41moles=1602.82 moles

mass of acrylic acid produced=1602.82 moles*72.06g/mol=115.499 kg

wt% of ethanol=(125.616kg/4000kg )*100=3.14wt%

wt% of acrylic acid=(115.499kg/4000kg)*100=2.89wt%

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