Question

How much carbon monoxide will result if `222*kg` of carbon are combusted to give `5/8` equiv of carbon monoxide?

Answer 1

Some of your data are superfluous.......we finally get over `3xx10^5*g` of carbon monoxide.......

Explanation:

We need a stoichiometric equation:

`C(s) + 1/2O_2(g) rarr CO(g)`

And thus there is 1:1 equivalence between carbon and its oxidation product.

`"Moles of carbon"` `=` `"Mass of carbon"/"Molar mass of carbon"`

`=` `(222xx10^3*g)/(12.011*g*mol^-1)=18483*mol`

And thus `18483*mol` result.

Because some of this is lost, we get finally,

`18483*molxx5/8=11552*mol` (that's a heady mix; a lot of people have been poisoned due to carbon monoxide poisoning from leaky flues!).

And this represents a mass of:

`11552*molxx28*g*mol^-1=??*g`

Now clearly, if you received this question in an exam, you would be asked to calculate the volume this gas would occupy under standard conditions.