[ultrashnotebel]

So yeah, I have an exercise that goes as follows: "0,3 moles of CH4 are burnt in 2,4 moles of air. Knowing that the final gaseous mix is only composed of N2, CO2, CO and water vapor, and by admittinng that the composition of air is 80% of N2 and 20% of O2, what if the final molar fraction of CO2?"


So essentially, I'm supposed to figure out how much CH4 is going to react with O2 to make CO2, and how much is going to make CO. The only problem is, I have no idea how to do that. The only thing I know is that CH4 will react with O2 to make CO if there's not enough O2, but when I think about it carefull, that doesn't actually make sense. If CH4 favors reacting with O2 to make CO2, why does it not simply do it until there's no more O2, and therefore no more reaction possible? Why does it decide to stop producing CO2 and start producing CO? Is it capable of foresight, does it know that it won't be able to be completely combusted if it keeps making CO2?

So yeah, anyone mind explaining?

[thetada]

I would treat it like this: work of the number of moles of CO you could produce with that many methane molecules, then divvy up the remaining oxygen between the CO molecules. Eg, if you had one mole of methane and 1.5 moles of molecular oxygen, you'd have half a mole of oxygen remaining after making one mole of CO. That half a mole of oxygen would be enough to convert half of the CO molecules into CO2. In other words, you'd have made half a mole of CO and half a mole of CO2.

Whether or not that's the question writer's intended meaning is another matter.

[XeLa.]

Firstly, I'm not quite sure how it works chemically how the methane "knows" to react perfectly leaving no oxygen gas left. I don't think that happens in nature (perhaps an assumption used by the author?), but my hands are off - someone more experienced than I would have to answer that.

In regards to solving the question, however, we could start by writing the combustion reactions for methane where only CO₂ is produced [reaction 1] and where only CO is produced [reaction 2].

Now, we know that the final mixture contains only CO, CO₂, N₂ and H₂O...there's no methane or oxygen gas present in the final mixture. What does that imply?

XeLa

[Borek]

Think in terms of

CH₄ + 3CO₂ 4CO + 2H₂O