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Why does $30~\mathrm{mL}$ of $\ce{C3H7OH}$ burn in 1 second (inside a water jug) compared to a sixth of the amount: $5~\mathrm{mL}$ of $\ce{C3H7OH}$ (on a plate) burn in 180 seconds? Why does the smaller amount take longer to burn the propanol?

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  • $\begingroup$ this compound is propanol* the $\ce {-OH}$ is for alcohol, then the terminology is "ol" $\endgroup$ – Babounet Dec 8 '14 at 8:31
  • $\begingroup$ Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the help center. Do you want to know more about this site, please take the tour.|| While editing I took the opportunity to correct the mistake pointed out by @Babounet. $\endgroup$ – Martin - マーチン Dec 8 '14 at 8:43
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For "burning" you need oxygen because actually when a compound is burning it's what is called an oxidation following this reaction

$$\ce{C3H7OH + 5O2 -> 3CO2 + 4H2O}$$

with propanol called the combustible and oxygen is called the comburant. Both are required as well as a power source to start the combustion process.

So in a plate you have the air surrounding the compound then you have practically an endless amount of oxygen available.

So the reaction will stop when all the Propanol will be consumed.

Hence, in a jug, this time the limiting reactive is the oxygen because you have less oxygen supply. Then the combustion will stop when all the oxygen will be consumed in the jug.

So, to sum up, in the jug the oxygen will be consumed faster than the propanol in the plate, that's why the combustion last longer in the plate even if you have more combustible.

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  • $\begingroup$ thank you for the edit Martin, I was currently looking for the right way to display the reaction :) And also for the english correction, in my native language we often use "y" instead of "i" in this kind of words so I'm often confusing :x $\endgroup$ – Babounet Dec 8 '14 at 8:48
  • $\begingroup$ If this is the reaction I'm thinking of, seen here, then there is also the difference between a liquid+gas reaction and a gas+gas reaction. In the former, the reaction is limited to the portion of alcohol that is vaporized, in the latter, virtually all of the material that will combust is already in the proper state for that reaction to occur (the gas phase, that is.) $\endgroup$ – Jason Patterson Dec 8 '14 at 14:14

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