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It says in my textbook that:

$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$$

which seems like a displacement reaction to me. Shouldn't the reaction be:

$$\ce{CH4 + O2 -> CO2 + 2 H2}$$

since $\ce{4H}$ are displaced by $\ce{O2}$ (becoming $\ce{2 H2}$), which makes the reaction balanced?

Why is the reaction $\ce{CH4 + O2 -> CO2 + H2O}$, which is further balanced as my textbook says?

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up vote 13 down vote accepted

The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Then not only methane is burnt, but also the arised hydrogen.

So in "first" step:

$\ce{CH_4 + O_2 -> CO_2 + 2H_2}$

but then the hydrogen will be also combusted in reaction :

$\ce{2H_2 + O_2 -> 2H_2O}$

so the overall reaction will be:

$\ce{CH_4 + 2O_2 -> CO_2 + 2H_2O}$

As was pointed out in the others answers the real mechanism of a methane combustion at a low oxygen pressure is more complicated:

49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida

49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida,

another: COMBUSTION AND FLAME 120:160–172 (2000) Intermediate Species Profiles in Low-Pressure Methane/Oxygen Flames Inhibited by 2-H Heptafluoropropane: Comparison of Experimental Data with Kinetic Modeling,

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Jaroslav, you might want to have a look at mhchem for easier typesetting of chemistry in LaTeX. I like that package a lot! – Klaus Warzecha Mar 28 '14 at 9:36
@ Klaus Warzecha - thanks I will @ user2468338 - It is an oxidation reaction: An oxidation/ reduction reaction is one where there is a change in the oxidation states of two or more species – Jaroslav Kotowski Mar 28 '14 at 9:40
for example reaction of natrium hydroxide with hydrochloric acid is displacement reaction and not oxidation/reduction reaction. – Jaroslav Kotowski Mar 28 '14 at 9:57
@JaroslavKotowski Looks strange to me: with little oxygen, you fully oxidize carbon ($\ce{CO2}$). I've never thought that methane combustion could occur via two steps. Could you please provide reference for that ? – mannaia Mar 28 '14 at 17:16
@mannaia The world "first" is quoted because it is not real 1st step. The reaction in low oxygen pressure is a bit more complicated:… – Jaroslav Kotowski Mar 29 '14 at 9:35

Look at the enthalpies of formation of $\ce{CO}$, $\ce{CO2}$, and $\ce{H2O}$, then the net enthalpies of reaction. In a deficiency of oxygen, and trimmed by the temperature and kinetics versus equilibrium conditions, are you more likely to obtain $\ce{CO}$ plus $\ce{H2O}$ than free hydrogen? Given less oxygen, soot and water? Note reaction energetics below,

enthalpies of formation

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There's nothing wrong. It's incomplete combustion. It is one of a number of oxidizing reactions that can and will be occurring if you are burning this hydrocarbon in an oxygen-restricted environment. But the dominant reaction will be the one where the lowest energy state is reached. So you will likely wind up with one CO molecule, one water molecule and one hydrogen molecule instead.

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