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A rigid stainless steel chamber contains $\pu{0.32 bar}$ of methane, $\ce{CH4}$, and excess oxygen $\ce{O2}$, at $\pu{200.0 °C}$. A spark is ignited inside the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.

I wrote out the following chemical equation:

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

I know I have $\pu{32 kPa}$ of methane, so using up all the moles of methane would mean that pressure decreases all the way to $0$.

The solution writes the following: it multiplies the 1 mol of $\ce{CH4}$ by $1$ to get the number of moles of $\ce{CO2}$ and by $2$ to get the moles of $\ce{H2O}$ produced.

Why can we do that? I would assume that by multiplying in such a way, you're implying that all of the $\ce{CH4}$ is being used up to create $1$ mol of $\ce{CO2}$. This means that you have no more moles to spare for creating $\ce{H2O}$.

I know my conceptual understanding is wrong, but this is what I thought while doing the question and the solution doesn't intuitively make sense to me.

I was wondering if someone could clarify my misunderstanding for me.

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A mole is a way of counting things, in the case of chemistry those things are molecules.

But the number of things is a chemical reaction are not necessarily preserved: some reactions create more molecules (electrolysis of water (H2O), for example, starts with two molecules of water but delivers one molecule of oxygen gas (O2) and two molecules of hydrogen gas (H2)).

So moles are not conserved if the structure of the molecules changes because of the reaction. The point of writing equations is to match conservation of matter (atoms don't disappear) to the structures of the molecules involved in the reaction. Sometimes reactions just rearrange the atoms and give the same number of moles of product, sometimes they create more molecules, sometimes they create fewer molecules.

In an ideal gas there is a neat relationship between the number of moles and the volume or pressure (at constant pressure the volume is proportional to the number of moles of gas; at constant volume the pressure is proportional to the number of moles of gas). So your question is asking you to compare the number of moles resulting from the reaction to the number going into the reaction. since they are the same, the pressure won't change.

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Are you supposed to consider that these are not ideal gasses? I ask this because if you are allowed to treat them as ideal each mole of methane reacts with 2 moles of oxygen to create 1 mole of CO2 and 2 of water. At 200°C and 0.32 bar the water isn't going to condense so 3 molecules react to form 3 molecules and as PV = nRT for an ideal gas there would be no pressure change.

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    $\begingroup$ No, I'm supposed to consider that these indeed are ideal gases. I don't really understand the stoichiometric logic though. You multiply 1 mole of methane by (2/1) to see how much water you produce. In that case, haven't you produced no carbon dioxide? Or is the logic that, if you use 1 mole of methane, you produce both 1 mole of carbon dioxide, and 2 moles of water? $\endgroup$ – DeepLearner Jan 21 '18 at 21:55
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    $\begingroup$ Yes. Where would the carbon go if it didn't get 'burnt' along with the hydrogen? $\endgroup$ – A. J. deLange Jan 21 '18 at 22:52

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