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

Question

$\pu{12 g}$ 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?

Answer 1

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