Answer:
ThOD =239.792 mg/L
Step-by-step explanation:
Theorical Oxigen demand (ThOD):
is the theoretical amount of oxygen
required to oxidize the organic fraction of a
waste up to carbon dioxide and water.
⇒ C sln = 450 mg C6H12O6 / 2 L H2O = 225 mg/L sln
∴ mm C6H12O6 = 180.156 g/mol
balanced reaction:
- C6H12O6 + 6O2 → 6CO2 + 6H2O
∴ mol C6H12O6 = 1 mol
⇒ mass C6H12O6 = (180.156 g/mol)( 1 mol) = 180.156 g
∴ the value of ThOD is determined when 180.156 g C6H12O6 consume mass O2 = 6(32) = 192 g Oxygen; then in a solution of 225 mg/L, you have:
⇒ ThOD = (192/180.156)×225 mg/L
⇒ ThOD = 239.792 mg/L