Custom problem: Limiting and excess reactant, fractional conversion, extent of r

Question

Custom problem: Limiting and excess reactant, fractional conversion, extent of reaction Consider the microbial conversion of glucose to L-glutamic acid, as shown below. Assume that initially the bioreactor contains 200 kg of glucose, 15 kg of ammonia and 54 kg of O2 glucose mmonia glutamic acid a) Which reactant is limiting? b) What is the percentage excess of the other reactants? c) If the reaction proceeds to completion, how much of the excess reactants will be left; how much L-glutamic acid will be formed; and what is the extent of reaction ? If the reaction proceeds to a point where the conversion of the limiting reactant 's 33%, how much of each reactant and product is present at the end, and what is the extent of reaction if the reaction proceeds to a point where 35 kg of Q, is left, what is the fractional conversion of glucose? What is the fractional conversion of 02? what is the extent of reaction ? d) ? e)

Explanation / Answer

a)

limiting --> either glucose or ammonia or O2

mol of glucose = mass/MW = 200/180.1559 = 1.11 mol

mol of ammonia = mass/MW = 15/17 = 0.8823

mol of o O2 = mass/MW = 54/32 = 1.6875

raito is:

1:1:1.5

therefore, 0.8823 mol of ammonia requires:

0.8823 mol of glucose ( we hav e)

0.8823*1.5 = 1.32345 mol of ammoniar (we have)

b)

% excess of other reactants:

% glucose = (1.11 -0.8823)/0.8823*100 )= 25 % ecess

% O2 = (1.32345 -1.6875)/1.32345 *100 )= 27 % ecess

c)

excess

mol of glucose = (1.11 -0.8823) = 0.36405

mass of glucose = mol*MW = 0.36405*180 = 65.529 g

mol of O2 = (1.6875-1.32345 ) = 0.2277

mass of O2 = mol*MW = 0.2277*32= 7.2864 g

formation of glutamic acid:

1:1 rtio so

0.8823 mol of acid forms = mass*MW = 0.8823*147.13 = 129.812 g

extent is 0.8823 mol

d)

if limit is 33%, then 0.33 under limiting ratio

mol of glucose = (1.11 -0.8823*0.33)  =0.818841

mass of glucose = mol*MW = 0.818841*180 = 147.4 g

mol of O2 = (1.6875-1.32345*0.33 ) = 1.25

mass of O2 = mol*MW = 1.25*32= 40 g

formation of glutamic acid:

1:1 rtio so

0.8823 mol of acid forms = mass*MW = 0.8823*0.33*147.13 = 42.83 g will be formed

extent is 0.8823 *0.33 = 0.2911 mol

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