Calculate the theoretical oxygen demand (mg/L). of a solution containing 450mg of glucose (C6H12O6). in 2 L of distilled water.

ThOD = 239.792 mg/L

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