Question

1. Provide degradation stoichiometry of aerobic and anaerobic reactions for the bioremediation of Toluene? Provide the reaction stoichiometry for aerobic and anaerobic degradation separately. After each reactions, provide mass of compound and elements for both side of reactions. In the context of stoichiometry under room temperature,
   1. How many grams of O2 is needed to biodegrade 1 gm of Toluene?
   2. How many grams of NO3- is needed to biodegrade 1 gm of Toluene?
   3. How many grams of Fe3+ is needed to biodegrade 1 gm of Toluene?
   4. How many grams of SO42- is needed to biodegrade 1 gm of Toluene?
      Toluene (T) CH MW = 92.14 g/mol Water solubility = 535 ppm at 25°C Density=0.8669 g/cm3 ThOD for 1 ppm toluene in water = 3.13 ppm CyHg +902 7CO2 + 4H2O CHg +7.2NO, +7.2H* - 7CO2 + 7.6H0 +3.6N2 CHg + 18Mn** + 14H,0 7CO2 + 36H+ + 18Mn? CHg + 36Fe3+ + 14H20 – 7002 + 36H+ + 36Fe2+ CHg +4.550 - + 91 - 7002 + 4.5H2S+4H20 CHg + 5H20-2.5CO2 + 4.5CH4||
      

Answer 1

M.W. of C7H8 = 92.14 g/mol

Moles of toluene in 1g = 1/92.14 mol = 0.01085 mol

A). a). C7H8 + 9 O2 ---> 7 CO2 + 4 H2O

As per reaction stoichiometry, 1 mol C7H8 reacts with 9 mol O2 to form 7 moles CO2 and 4 mol H2O.

For 0.01085 mol C7H8, O2 required = 0.01085 mol

As M.W. of O2 = 32g/mol

O2 required for 1gm C7H8 = 0.01085 mol × 32g/mol = 0.3473 g

b). C7H8 + 7.2 NO3^{​​​​-}+ 7.2 H⁺ ---> 7 CO2 + 7.6 H2O + 3.6 N2

Similarly to part a). , NO3^{​​​​-} required = 0.01085 ×7.2 moles = 0.07814 mol

M.W. of NO3^{​​​​-}= 62 g/mol

NO3^{​​​​-} required = 0.07814 mol × 62 g/mol = 4.845g

c). C7H8 + 36 Fe⁺³ + 14 H2O ----> 7H2O + 36 H⁺ + 36 Fe⁺²

Fe3+ required for 1 g C7H8 ( 0.01085 mol Fe3+) = 36 × 0.01085 mol = 0.39071 mol

M.W. of Fe3+ = 55.8 g/mol

Fe3+ required = 0.39071 mol × 55.8 g/mol = 21.8 g

D). C7H8 + 4.5 SO4^{​​​​2-} + 9H⁺ ---> 7 CO2 + 4.5 H2S + 4 H2O

SO4^{​​​​2-} required with 1 g C7H8 = 0.01085 × 4.5 moles = 0.048825 moles

M.W. of SO4^{​​​​2-} is 96 g/mol

SO4^​​​2- required = 0.048825 mol × 96 g/mol = 4.6872 g