# Determining amount (in mols) of oxygen needed to fully combust coal

I'm trying to calculate the amount of oxygen needed to fully combust some coal.

As an example, assume I have $$\pu{1 kg}$$ of coal which is, by mass fraction, 60% carbon ($$\ce{C}$$) 40% hydrogen ($$\ce{H}$$). I know this is unrealistic but it's just for the sake of the example.

There is then $$\pu{0.6 kg}$$ of $$\ce{C}$$ and $$\pu{0.4 kg}$$ of $$\ce{H}$$. I can convert to $$\pu{mol}$$ by using the MW of both species, $$\pu{0.012 kg/mol}$$ for $$\ce{C}$$ and $$\pu{0.001 kg/mol}$$ for $$\ce{H}$$. Which means I have $$\pu{50 mol}$$ of $$\ce{C}$$ and $$\pu{400 mol}$$ of $$\ce{H}$$.

The reaction stoic for the oxidation of both species is as follows:

$$\ce{C + O2 -> CO2}$$

$$\ce{4H + O2 -> 2H2O}$$

Would it then be correct to say that I need $$\pu{100 mol}$$ of $$\ce{O2}$$ to combust the $$\ce{H}$$ and $$\pu{50 mol}$$ of $$\ce{O2}$$ to combust the $$\ce{C}$$?

• It should be indeed "just for the sake of the example," because your example is not realistic ($\ce{CH8}$ empirical formula is not real). Nonetheless, your calculation is acceptable. – Mathew Mahindaratne May 7 at 20:29
• Why don't you use reasonable examples ? – Maurice May 7 at 20:30
• It would be too tedious to type out the 10 different mass fractions of the real sample and list all of the reactions happening (plus I bet no one would read it). I just wanted to check my understanding really quick. I don't do this stuff very often. – user8151560 May 7 at 20:45
• Your calculations look broadly correct. But your assumption for what coal is looks way off. and you calculations also assume complete combustion which is rarely true for coal except under ideal conditions. Coal fires, for example, produce a lot of CO from incomplete combustion. – matt_black May 8 at 14:13