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 converts glucose (C6H12O6) to form ethanol (C2H5OH) and propionic acid (C2H3CO2H) by the following reactions: C_6H_l2O_6 rightarrow 2C_2H_5OH + 2CO_2 C_6H_12O_6 rightarrow C_2H_3CO_2 H + 2H_2O 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 CO2 have been produced and 90 kg of uncreated glucose remains in the broth. What are the weight percent of each of the remaining components in the broth at the end of the process. Assume none of the glucose is retained by the microorganism. Answer: 89.2 wt% water, 3.2 wt% ethanol. 2.9 wt% propionic acid, 2.5 wt% C02,2.3 wt% glucose.

Explanation / Answer

Feed contains 12 wt% solution of glucose. Hence glucose in feed= 4000*12/100 =480 kg

Molar mass of glucose = 180. Moles of glucose used = 480/180 =2.67

From the reaction C6H12O6---à 2C2H5OH+2CO2

Moles of CO2 produced= 120/44= 2.73 kg moles . This is same as moles of ethanol= 2.73, mass of ethanol produced= 2.73*46 =125.58 kg

Moles of C6H12 O6 utilized from reaction -1= 2.73/2= 1.365

Moles of C6H1206 unreacted= 90 kg, moles of glucose unused= 90/180 =0.5 moles

Moles of glucose reacted from reaction-2 is (C6H12O6--à2C2H3COOH+2H2O)= 2.67-1.365-0.5= 0.805

Moles of propionic acid formed= 2*0.805=1.61 kmoles

Molar mass of propionic acid = 72. Mass of propionic acod = 1.61*72=116 kg

Product contains : 90 kg glucose ( unreacted), propionic acid = 116 kg, CO2=120 kg, water = 4000*0.88=3520 kg, ethanol= 125.58 kg

Total products : 90+116+120+3520+125.58=3899.31 kg

Composition :

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