Question

A waste contains 20 mg/L of ammonia (NH3) and 100 mg/L of ethyl alcohol (C2H5OH). These substances react according to the following chemical reactions. What is the total theoretical oxygen demand of this waste?

NH3 + 2O2 → NO3- + H+ + H2O
C2H5OH + 3O2 → 2CO2 + 3H2O

Answer 1

The theoretical oxygen demand of a compound can be calculated as follows:

NH3 + 2O2 → NO3- + H+ + H2O

17.031 g   + 64g O2

For 1 g of NH3 we need 64/17.031= 3.75 g O2

Here 20 mg/L or 0.020 g /L NH3 will need

= 0.020 g /L NH3 will need *3.75 g O2/ 1.00 g NH3

= 0.075 g /L

75.0 mg/L O2

For 100 mg/L of ethyl alcohol (C2H5OH):

C2H5OH + 3O2 → 2CO2 + 3H2O

46.068 g   + 96g O2

For 1 g of C2H5OH we need 96/46.068= 2.08 g O2

Here 100 mg/L or 0.100 g /L C2H5OH will need

= 0.100 g /L C2H5OH will need *3.08 g O2/ 1.00 g C2H5OH

= 0.308 g /L

708.0 mg/L O2