Question

An industrial wastewater flowing at a rate of 37,000 m3/day is contaminated by 187 mg/L of propanol – CH3CH2CH2OH. As an environmental consultant, you propose to treat the water in an activated sludge process in which propanol is oxidized by oxygen into carbon dioxide – CO2. The reaction is highly favorable and assumed to be complete.
a) Equilibrate the half-reactions and overall reaction that are expected to occur during the treatment process.

b)Determine the theoretical quantity of oxygen (in mol/day) required for the complete oxidation of propanol.

c) Determine the volume of pressurized air (2 atm, 15°C) to be provided (in m3/day) for the complete oxidation of propanol. The partial pressure of O2 in air is 0.2095 atm and the efficiency of oxygen transfer is estimated to be 15%.

Answer 1

(a). 2 CH₃CH₂CH₂OH + 9O₂ 8 H₂O + 6 CO₂

(b).   1 L industrial wastewater cointains =187 mg propanol

37,000 L industrial wastewater cointains = (187 x 37000) mg propanol = 6919 gm of propanol.

2 moles of propanol is oxidized by 9 moles of O₂ requires to oxidize.

120.18 gm of propanol is  oxidized by 32 gm of O₂ requires to oxidize.

6919 gm of propanol is  oxidized by (32 x 6919)/120.18 gm of O₂ requires to oxidize.

6919 gm of propanol is  oxidized by 1842.3 gm of O₂ requires to oxidize.

6919 gm of propanol is  oxidized by (1842.3 / 32.0 = 57.57) moles of O₂

Hence, 57.57 moles per day of oxygen is required.