If we know that:

$1g$ of carbon + $1.33g$ of oxygen = carbon monoxide

$1g$ of carbon + $2.66g$ of oxygen = carbon dioxide

My question is: why is that in the second case where we doubled the amount of oxygen we don't get double the amount of the first case (2 times carbon monoxide)

Or why don't we get the same amount of carbon monoxide and the remaining mass of the products is simply oxygen that didn't form any bonds.

Basically, what forces the reaction to occur in a certain way just because the proportions are different but the elements are the same.


In reality, there is not a sudden changeover of $\ce{CO}$ to $\ce{CO2}$ production when the oxygen supply is increased. This is the reason that all dwellings should have carbon monoxide detectors.

As an example, along with $\ce{CO2}$, gasoline (petrol) engines commonly produce 3-7% $\ce{CO}$ before additional air is injected.

The amounts you cite are how much oxygen is in pure samples of each gas, but producing either gas often requires filtering out traces of other gases that were also made in side reactions.

Producer gas is a fuel made by the incomplete oxidation of carbon (from coal or other source, sometimes with the addition of water) and contains both flammable $\ce{CO}$ and nonflammable $\ce{CO2}$ (and $\ce{N2}$ from the air supply). During fuel shortages, gasifiers were cobbled onto vehicles to enable them to run on this type of fuel.

Gasifier producing CO (along with CO2) as fuel

  • $\begingroup$ Does that mean that if I have 100 grams of each element (Carbon and Oxygen) reacting and forming bonds with each other that I will have the formation of both carbon monoxide and carbon dioxide ? And if I have one of them I will have to filter them ? By the way I´m not caring about the reactants that didn´t react. $\endgroup$ – Goncalo Fonseca Feb 11 '18 at 15:33
  • $\begingroup$ Yes, the product is always a mixture of CO and CO2, though with a great excess of oxygen, CO2 predominates. And whether you need to filter the product depends on the applicati0n. $\endgroup$ – DrMoishe Pippik Feb 11 '18 at 19:26
  • $\begingroup$ I dont want to push it, I´m already very thankful for your explanation but I got another question... Why does the CO2 predominates when there is a great excess of oxygen ? Is it because it´s a more stable molecule ? Or it´s another thing ? $\endgroup$ – Goncalo Fonseca Feb 11 '18 at 20:13
  • $\begingroup$ More oxygen pushes equilibrium to the product that contains more oxygen. It's simply that there's more O2 available. $\endgroup$ – DrMoishe Pippik Feb 11 '18 at 21:36

To supplement to DrMoishe's excellent answer, I'll tackle this part of your question:

why is that in the second case where we doubled the amount of oxygen we dont get double the amount of the first case (2 times Carbon Monoxide(2.Carbon Monoxide))

Suppose a reaction: $\ce{3A\+2B->C}$. To produce the double the amount of C, you'll need double the amount of each reactant, i.e., to produce $2$ moles of $\ce{C}$, you'll need $6$ moles of $\ce{A}$ and $4$ moles of $\ce{B}$. Simply only doubling amount of $\ce{B}$ won't work.

As an analogy, suppose you're baking a $1kg$ cake from $250g$ flour and $250g$ cream. Then, to produce $2kg$ of the cake, just using $1.5kg$ flour and $250g$ cream will not work! You'll need to use $250\times2=500g$ cream instead.

In your case also, you need to double the amount of both oxygen and carbon.


As DrMoishe Pippik has pointed out that the net reaction is different in reality.
However, I assumed that you were asking in theoretical chemistry. In this case, in low oxygen condition, Carbon reacts with oxygen to form carbon monoxide with the rate of 2:1 (mole of carbon:mole of oxygen). In free-Oxygen condition, firstly the surface area of outer carbon layer doesn’t have enough oxygen to react with, resulting in the presence of carbon monoxide. Then, there are still unused oxygen, both the remaining carbon and the new carbon monoxide react with oxygen to form carbon dioxide.
P.S: This is my first answer on StackExchange. Please point out any problem or misspellings, thank you!


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