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

1 Answer
Jan 21, 2017

Answer:

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.