Respuesta :

Answer:

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

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