Burning coal for nitrogen (and sulfuric acid)

Suppose I caught a mole of dry soot. The formula which Bergin's crew sent us was C₁₀₀H_75–79O_11–17N_3–4S_1–3. Molar weight of its components: 1201+76+240+56+64 = 1.637 kg. If I burn that in oxygen (or in ozone, oxozone; whatever - pretend I have an infinitude to burn): I get 1201+3200 carbon-dioxide, 76+608 water, and 64+64 sulfur-dioxide (again, not caring about the oxygen). I'd get the same 56g nitrogen in, I hope, gas form.

I concede Concerns with carbon-monoxide, hydrogen-sulfide, hydrogen-cyanide but these should (mainly) be containable until the oxygen does its work. Too much oxygen and we get nitrous oxides, or water-peroxide - but most of all, I have a Feeling about sulfurous even sulfuric acid. In fairness, I can count on that 76:2 hydrogen:sulphur ratio. Two H₂SO₃ (or H₂SO₄) molecules will sequester all the sulfur and leave 72 hydrogens free to make 36 water molecules (72+576). A 1:18 acid:water solution is, I understand, diluted enough to work with. Some of the carbon-compounds will get dissolved into this too. At 295 K I hear 0.07 mol% CO₂ just for water (pdf): so from the heated coal I estimate 0.2 molecules in solution leaving the other 100(-) loose as, mostly, free CO₂ gas. Back to the fluid, don't splash it in your eyes.

So, should my stone space colony want 46.3 tonnes of nitrogen they'll need to burn 1637 × 46.3/56 tonnes of space-coal; a 1353 tonne sootball. Which is a sight better than a million tonnes. Hopefully they can find something like this within the rubble.

And the sulfuric acid looks like a nice little bonus. The pure water alone would make 648 × 46.3/56 = 535.75 tonnes which should offer a bit more in cubic meters. 50×5×2 m³ looks like a swimming pool.