Question

500 kg of n-propanol (CH3CH2CH2OH) are accidently discharged into a body of water containing 108 L of H2O. By how much is the BOD (in mg L-1 ) of this water increased? Assume the following reaction: C3H8O + 9/2 O2 --> 3CO2 + 4 H2O

Answer 1

BOD = miligrams of O₂ consumed per litre

moles of propanol = mass prop. / molar mass, molar mass of propanol is 60

moles = 500 / 60 = 8.33 kilomoles of n-propanol or 8 333 moles of propanol

reaction goes as follows:

C3H8O + 9/2 O2 --> 3CO2 + 4 H2O , let´s multiply everything by 2

2 C₃H₈O + 9 O₂ --> 6 CO₂ + 8 H₂O

2 moles of propanol needs 9 moles of oxygen so:

this is equal to 37,498.5 moles of oxygen , lets calculate the mass of oxygen

mass = moles * molar mass, molar mass oxygen = 32 g/gmol

mass = 37498.5 * 32 = 1, 199,952 grams of oxygen

we want mg so you need to multiply by 1000

1 199 952 000 mg of oxygen

just divide this by the liters of water 108 L

BOD = 1 199 952 000 mg / 108 L = 11 110 666.6 mg/L or 11.11 x 10⁶ mg/L

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