# Question #dcfef

Well, the simple explanation is that combustion involves using ${\text{O}}_{2}$, oxygen gas, to burn... something. Could be a hydrocarbon, could be an alcohol, could be something else.
$\text{CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O} \left(g\right)$