1. The growth of an organism on hexadecane (MW=226.4) can be described by the fo

ID: 987012 • Letter: 1

Question

1. The growth of an organism on hexadecane (MW=226.4) can be described by the following stoichiometric equation:

C16H34 + 12.4 02 + 2.09 NH3 -------> 2.42 ( C4.4 H7.3 N0.86O1.2 ) biomass + yH2O + 5.33 CO2

Calculate :

a) The coefficient y,

b) The respiratory quotient in gmol CO2/gmol O2 and g CO2/ g O2

c) Actual yield of mass ( Yxs) and maximamum (Yxs) in gmol of biomasss/ gmol of substrate and g of biomass/ g of substrate.

Note: MW of amonia = 17.03, MW of biomass=91.34, MW of CO2= 44, and MW of O2=32

The given balanced chemical reaction is

C16H34 + 12.4 02 + 2.09 NH3 -------> 2.42 (C4.4 H7.3 N0.86O1.2) biomass + yH2O + 5.33 CO2

Since this is a balanced chemical reaction, moles of H - atom on reactant side is equal to the moles of H - atoms on product side. Hence

34x1 H + 2.09x3 H = 2.42x7.3 H + 2xy H

=> 40.27 = 2y + 17.666

=> y = 11.30 (answer)

(b): Respiratory quotient(RQ) is defined as the ratio of CO2 produced to O2 consumed during the metabolism of food.

Here moles of CO2 produced = 5.33 mol

moles of O2 consumed = 12.4 mol

Hence RQ in terms of gmol = 5.33 / 12.4 = 0.430 (answer)

mass of CO2 produced = 5.33 mol x 44 g/mol = 234.52 g

mass of O2 consumed = 12.4 x 32 g/mol = 396.8 g

Hence respiratory quotient in terms of g = 234.52 g / 396.8 g = 0.591 (answer)

Stoichiometric moles of product biomass(MW = 91.34g/mol) formed = 2.42

Hence gmoles of boimass / gmoles of substrate = 2.42 / 1 = 2.42

Stoichiometric mass of substrate hexadecane (MW=226.4) used = 1 mol x 226.4 g/mol = 226.4 g

Stoichiometric mass of product biomass(MW = 91.34g/mol) formed = 2.42 x91.34g/mol = 221.04 g

Hence g of boimass / g of substrate = 226.4 g / 221.04 = 1.024

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