# CO : CO₂ ratio when given amounts of carbon and oxygen react with each other

By the reaction of carbon and oxygen a mixture of $\ce{CO}$ and $\ce{CO2}$ is obtained. What is the composition of the mixture by mass obtained when 20 grams of $\ce{O2}$ reacts with 12 grams of carbon?

I wrote the equation as

$$\ce{4C + 3O2 -> 2CO + 2CO2}$$

and identified $\ce{O2}$ as the limiting reagent then the masses of $\ce{CO}$ formed will be 11.66 grams and mass of $\ce{CO2}$ formed will be 18.33 grams, which gives the ratio as 7:11, but the answer is given as 21:11.

Can somebody please tell me where am I going wrong?

• What was your motivation for writing the equation that way? That automatically assumes an (incorrect) answer. – Chet Miller Aug 22 '17 at 15:31

First and foremost refer to the tip given in the answer of @ChesterMiller: use separate reactions for each given process, and don't rely on limiting reagent concept when more than one reaction is possible:

\begin{align} \ce{\underset{$x$}{C} + 0.5 \underset{$0.5x$}{O2} &-> \underset{$x$}{CO}} \tag{1} \\ \ce{\underset{$y$}{C} + \underset{$y$}{O2} &-> \underset{$y$}{CO2}} \tag{2} \end{align}

let $x$ mol be the amount of carbon participating in forming carbon monoxide $\ce{CO}$, and $y$ mol -- the amount of carbon participating in forming carbon dioxide $\ce{CO2}$.

We have the following amounts of reactants in the system:

$$n(\ce{C}) = \frac{m(\ce{C})}{M(\ce{C})} = \frac{\pu{12 g}}{\pu{12 g mol^-1}} = \pu{1 mol}$$

$$n(\ce{O2}) = \frac{m(\ce{O2})}{M(\ce{O2})} = \frac{\pu{20 g}}{\pu{32 g mol^-1}} = \pu{0.625 mol}$$

where $m$ -- mass; $M$ -- molecular weight.

Taking into account reactants ratios, the following system of equations is relevant:

\begin{align} \begin{cases} x + y &= 1 \qquad &\text{total amount of carbon}\\ 0.5x + y &= 0.625 \qquad &\text{total amount of oxygen} \end{cases} \end{align}

Solving this system of equations, one gets $x = 0.75$ (mol) and $y = 0.25$ (mol). Hence the masses of the products are:

\begin{align} m(\ce{CO}) &= x \cdot M(\ce{CO}) &= \pu{0.75 mol} \cdot \pu{28 g mol^-1} &= \pu{21 g} \\ m(\ce{CO2}) &= y \cdot M(\ce{CO2}) &= \pu{0.25 mol} \cdot \pu{44 g mol^-1} &= \pu{11 g} \end{align}

• That's homework, you shouldn't make complete answer - Chester's answer is OK, so yours wasn't needed. – Mithoron Aug 22 '17 at 18:11
• @Mithoron It's hard for me to decide when a homework-type question needs a real answer and when a clue is enough as I have no idea what level of understanding we are dealing with. OP showed his attempt to solve the problem and also struggled to proceed with solving the problem after the initial direction has been given, as seen in the comment section to Chester's answer. So I decided to elucidate it a bit more. If you think this was a bad move, feel free to flag the post for deletion. – andselisk Aug 22 '17 at 18:18
• It's probably pointless by now, just remember in the future. – Mithoron Aug 22 '17 at 18:24
• Is this actually a plausible reaction? Could we also get more $CO_2$ but some unburnt $C$? – badjohn Aug 25 '17 at 8:28
• @badjohn The way the problem is stated (conditions were such that only $\ce{CO}$ + $\ce{CO2}$ are presented in the system) suggests only this solution. – andselisk Aug 25 '17 at 10:04

You should have written:

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

and

$$\ce{C + \frac{1}{2}O2 -> CO}$$

Then you should have determined what fraction of the carbon reacts by by the first reaction and what fraction of the carbon reacts by the second reaction.

• How can I find that? – Gem Aug 22 '17 at 15:51
• Let f be fraction of C that forms CO2 and let (1-f) be the fraction of C that forms CO. Solve for f such that 20 grams of O2 are consumed. – Chet Miller Aug 22 '17 at 15:58