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

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Answer 1 · 2017-01-21T09:51:41

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

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.

How much carbon monoxide will result if 222*kg222*kg of carbon are combusted to give 5/85/8 equiv of carbon monoxide?