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

Jan 21, 2017

Some of your data are superfluous.......we finally get over $3 \times {10}^{5} \cdot g$ of carbon monoxide.......

#### Explanation:

We need a stoichiometric equation:

$C \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow C O \left(g\right)$

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

$\text{Moles of carbon}$ $=$ $\text{Mass of carbon"/"Molar mass of carbon}$

$=$ $\frac{222 \times {10}^{3} \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 18483 \cdot m o l$

And thus $18483 \cdot m o l$ result.

Because some of this is lost, we get finally,

$18483 \cdot m o l \times \frac{5}{8} = 11552 \cdot m o l$ (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.