Question

Oxygen consumed by a living organism in aerobic reactions is used in adding mass to the organism and/or the production of chemicals and carbon dioxide. Since we may not know the molecular compositions of all species in such a reaction, it is common to de line the ratio of moles of CO_2 produced per mole of O_2 consumed as the respiratory quotient. RQ, where RQ = n_CO_2/n_O2 (or n_CO_2/n_O2) Since it generally is impossible to predict values of RQ, they must be determined from operating data. Mammalian cells are used in a bioreactor to convert glucose to glutamic acid by the reaction C_6H_12 O_6 + a NH_3 + bO_2 rightarrow pC_5 H_9NO_4 + qCO_2 + rH_2O The feed to the bioreactor comprises 1.00 times 10^2 mol C_6H_12O_6/day, 1.20 times 10^2 mol NH_3/day, and 1.10 times 10^2 mol O_2/day. Data on the system show that RQ = 0.45 mol CO_2 produced/mol O_2 consumed. Determine the live stoichiometric coefficients and the limiting reactant. Assuming that the limiting reactant is consumed completely, calculate the molar and mass flow rates of all species leaving the reactor and the fractional conversions of the non-limiting reactants.

Answer 1

The balanced reaction is

C₆H₁₂O₆+ NH₃ +1.5O₂---àC₅H₉NO₄ +CO₂ +3H₂O

Moles of C6H12O6= 100 mol/day,

NH3= 120 mol/day

And O2= 110 moles/day

Molar ratio of C6H12o6: NH3 : O2= 1:1:1.5

Actual ratio = 100:120:110 = 1:1.2:1.1

Limiting reactants is Oxygen.

Since all the limiting reactant is consumed, given

Rq= moles of CO2 produced/ mole of O2 consumed = 0.45

Moles of CO2 produced = 0.45*110=49.5 moles

Moles of excess reactant consumed = 49.5 moles

Products contains :CO2= 49.5 moles/day, C6H9NO4= 49.5 moles/day, H2O= 3*49.5= 148.5 moles/day

NH3= 120-49.5= 70.5 moles/day, O2=110-49.5*1.5=35.75moles/day

Fractional conversion of NH3 =100*49.5/120 = 41.25%

Fractional conversion of C6H12O6= 100*49.5/100=49.5%