[mwm10620]

2NH4 +3O2 >> 2NO2 +4H +2H2O
- 3.43g O2/g to convert

2NO2 + O2 >> 2NO3
- 1.14g O2/g NO2 oxidized

Total oxidation reaction:
NH4 + 2O2 >> NO3 +2H -H2O
- 4.57g O2/g N oxidized


Can someone please explain how the gram of O2 per gram of ammonia/nitrite is calculated? Thanks!

[chenbeier]

Ammonia is NH₃ and not NH₄, or do you mean NH₄⁺ ammonium.

4 NH₃ + 5 O₂ => 4 NO + 6 H₂O

2 NO + O₂ => 2 NO₂

4 NO₂ + 2 H₂O + O₂ => 4 HNO₃

The mass can calculated by using the mole. For 4 mol ammonia you need 5 Mole oxygen or in other words for 1 mol ammonia you need 1,25 mole oxygen. 17 g  NH₃ to 40 g O₂ or 1g to 2,35 g

[mwm10620]

Yes ammonium. I took those formulas from Metcalf and Eddy. I am wondering how the grams of oxygen to convert was calculated.

[AWK]

Convert mole to mass and divide accordingly (for correctly balanced reaction).

[mwm10620]

Convert mole to mass and divide accordingly (for correctly balanced reaction).

I thought that is what I did, but did not reach the same answer.

2NH4 = 36.0766
3O2 = 95.9964

95.9964/36.0766 = 2.66 (not 3.43) - any advice on my mistake?

[AWK]

3.43=96/28  (O₂/N₂)

[mwm10620]

3.43=96/28  (O₂/N₂)

So I guess you just consider N converted, not the entire NH4 molecule. Thanks for your help

[AWK]

The reaction is not properly balanced (mass balance and charge balance).
The number 3.43 matches your incorrect balance, but this is not a good solution. Something has been distorted here.

[chenbeier]

Yrs it looks like , also for the other two  equations. 32/28 = 1.14 and 64/14 = 4.57

[Borek]

First of all - can you quote the problem in its entirety? For me it is not even clear what the question really is. I can guess, but I can't be sure.

[AWK]

The problem concerns the microbial nitrogen removal (google books helped!). The reactions given therein are as follows:
NH₄⁺ + 3/2O₂ = NO₂^- + H₂O + 2H⁺ (Nitrosomonas)
NO2^- + 1/2O₂ = NO₃^- (Nitrobacter)
So the oxygen to nitrogen mass ratio, despite the incorrect balance of the reaction, is good.