![\mathrm{CO}_2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/oh2vro35vlyqph0mtg0y12c9z3vrorddaa.png)
has 2 oxygens for every carbon.
![\mathrm{CO}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/dgcwg8wc7cuhaq10rli1plceefe6z0e3zt.png)
will contribute one of each for every molecule. You then have to have an even multiple of
![\mathrm{CO}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/dgcwg8wc7cuhaq10rli1plceefe6z0e3zt.png)
molecules to not end up with an odd number of oxygens.
So if you take
![2\mathrm{CO}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ug29o5r15obiiu80d8d6p1fmn00qrc1twq.png)
, you should end up with 2 carbons on the RHS.
![2\mathrm{CO}_2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/asprke2ti9xyq7jovhrb7t6bbv5ecf48r2.png)
then has 4 oxygens. 2 of them are already accounted for, so you just need one
![\mathrm O_2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/i6vdehnfek4trhkqss2og9ima3kht3z992.png)
to balance out the reaction.
![1\mathrm O_2+2\mathrm{CO}\Rightarrow2\mathrm{CO}_2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/3au521in3nacxf6mlpttqwrkicp2rlzija.png)