Stoichiometry, yield, and the ideal gas law Stoichiometric equations can be used

Question

Stoichiometry, yield, and the ideal gas law Stoichiometric equations can be used to represent the growth of microorganisms provided a 'molecular formula' for the cells is available. The molecular formula for biomass is obtained by measuring the amounts of C, N, H, O, and other elements in cells. For a particular bacterial strain, the molecular formula was determined to be C4.4H7.3O1.2Ng 0.86. These bacterial cells are grown under aerobic conditions with hexadecane (Ci6H34) as substrate. The reaction equation describing growth is: C16H34+ 16.28 02 +1.42 NH3 rightarrow 1.65 C4.4H73O1.2N0.86 + 8.74 C02+ 13.11 H20 You have been put in charge of a small batch fermented for growing the bacteria and aim to produce 8.2 kg of cells for inoculation of a pilot-scale reactor. What minimum amount of hexadecane substrate (in kg) must be contained in your culture medium? Assume 100% conversion of hexadecane to cells. What must be the minimum concentration of hexadecane (in kg/m3) in the medium if the fermenter working volume is 5.8 cubic meters? What minimum volume (in m3) of air at 30 degree C and 1.3 atm pressure must be pumped into the fermenter during growth to produce the required amount of cells?

Explanation / Answer

a) As per the above equn

one mole of C6H34 ---------> 1.65 mole of cell

226g of hexadecane give 1.65 * 91.84 = 151g

? ---------- 8.2 kg of cell (8200g)

= 8200 * 226 /151 = 12229g =12.25kg

b) apply M= Wt /Mol.wt *1000/V

= 12.2/226 * 1000/5.8 m3

= 10.1Kg/m3

c) Weight of O2 is 16.28 of O2 (32) required to produce 1.65 of 91.84g of cell

? < -------------------------- 8.2kg

so, 28.19Kg

Apply Ideal gas equn  Pv=nRT

1.3 X v= 28.19/226 x 8.206 * 303

= 238kg/m3

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