molecular weights: glyverol = 92 g/gmol NH3 = 17 g/gmol Biomass (C4H7O2N) = 101g

Question

molecular weights: glyverol = 92 g/gmol

NH3 = 17 g/gmol

Biomass (C4H7O2N) = 101g/gmol

A process for microbial synthesis of 1, 3-propanediol is being developed for the manufacture of 'green' polyester fabric from renewable resources. Under anaerobic conditions, a selected strain of Klebsiella pneumoniae converts glycerol (C_3 H_8 O_3) to 1, 3-propanediol (C_3 H_8 O_2) and acetic acid (C_2 H_4 O_2), with minimal formation of other fermentation products such as butyric acid, ethanol, and H_2 gas. The fermentation and cell growth equation can be written as: 68 C_3 H_8 O_3 + 3 NH_3 rightarrow 3 C_4 H_7 O_2 N + 49 C_3 H_8 O_2 + 15 C_2 H_4 O_2 + 15 CO_2 + 40 H_2 O where C_4 H_7 O_2 N represents the biomass. a. If the medium contains 100 g/L glycerol, what minimum concentration of ammonia (NH_3) is required to achieve complete conversion of the glycerol? Express your answer in units of gmol/L b. If the bioreactor holds 40 liters of medium and there is complete conversion of the glycerol, what mass of C_4 H_7 O_2 N will be generated?

Explanation / Answer

Given

Molecular weight of glycerol = 92 g/mol

Molecular weight of NH3 = 17 g/mol

(a) 100 g/L of glycerol

Let us consider 1 L of solution

mass of glycerol = 100 g

No. of moles of glycerol = 100 g /92 g/mol = 1.08696 moles

No. moles of NH3 = (3/68) * 1.08696 =0.04795 moles (3/68 represents the stoichometric ratio)

mass of NH3 required = no. of moles of NH3 * molar mass of NH3 = 0.04795 * 17 = 0.815 g

so 0.815 g of NH3 is required for 1 L of solution

so 0.04795 mol of NH3 is required for 1 L of solution

concentration of NH3 required = 0.815 g/L

concentration of NH3 required = 0.04795 gmol/L Answer

(b)Given volume of 40 L of medium

so mass of glycerol = 100 g/L * 40 L = 4000 g of glycerol will be present

No. of moles of glycerol = 4000 g /92 g/mol = 43.48 moles

No. of moles of C4H7O2N = (3/68) * 43.48 =1.918 moles   (3/68 represents the stoichometric ratio)

mass of C4H7O2N = 1.918 mol * 101 g/mol =193.73 g

so 193.73 g of C4H7O2N will be produced

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