# Ratio of carbon monoxide and carbon dioxide when carbon and oxygen are made to react [duplicate]

$$\pu{12 g}$$ of $$\ce{C}$$ reacts with $$\pu{64 g}$$ of $$\ce{O2}$$ to give a mixture of $$\ce{CO}$$ and $$\ce{CO2}$$. Find amount of $$\ce{CO}$$ and $$\ce{CO2}$$ at the end of reaction.

What I tried was to let x moles of carbon react to give $$\ce{CO2}$$ and y moles react to give $$\ce{CO}$$.

$$\ce{ \underset{x}{C} + \underset{x}{O2} -> \underset{x}{CO2}}$$ $$\ce{\underset{y}{2C}} + \underset{y/2}{\ce{O2}}\ce{ -> \underset{y}{2CO}}$$

Solving for $$x + y = 1$$ and $$x + 0.5y = 2$$ , we get $$x=3$$ and $$y= -2$$ which is meaningless.

I followed the method used in this answer.

Can anyone tell me where I went wrong?

• Comments are not for extended discussion; this conversation has been moved to chat. Apr 20, 2021 at 8:42
• This is a bad question. The answer is indeterminate since the problem doesn't define the conditions for completion. Some mixture of CO, CO2 and O2 would seem likely. The only way to have a definite "solution" for the problem would be to make an explicit assumption that all the carbon goes to CO2 since O2 is in excess.
– MaxW
Apr 20, 2021 at 9:39

The answer that you got is meaningless because it is wrong.

You missed one reaction here. $$\require{cancel}\ce{CO}$$ can further react with $$\ce{O2}$$ to give $$\ce{CO2}$$

$$\ce{2CO + O2 -> 2CO2}$$

Now we have three reactions to consider,

$$\ce{2C + O2 -> 2CO} \tag{1}$$ $$\ce{2CO + O2 -> 2CO2}\tag{2}$$ $$\ce{C + O2 -> CO2} \tag{3}$$

(1) and (2) can be combined here to give:

$$\ce{2C + O2 + \cancel{2\ce{CO}} + O2 -> \cancel{\ce{2CO}} + 2CO2}$$

Which gives:

$$\ce{C +O2 ->CO2} \tag{3}$$

These reactions progress in the following manner, (1) + (2) = (3).

Now writing the three equations and using the above statement(which means we can find the value for just (1) first), we get that $$\pu{1 mol}$$ of $$\ce{C}$$ reacts with $$\pu{0.5 mol}$$ of $$\ce{O2}$$ to give $$\pu{1 mol}$$ of $$\ce{CO}$$. We still have $$\pu{1.5 mol}$$ of $$\ce{O2}$$ left, so the reaction goes further.

Now onto (2), $$\pu{1 mol}$$ of $$\ce{CO}$$ reacts with $$\pu{0.5 mol}$$ of $$\ce{O2}$$ to give $$\pu{1 mol}$$ of $$\ce{CO2}$$ and we still have $$\pu{1 mol}$$ of $$\ce{O2}$$ left which is in excess.

You can use this same method for this question as well