If a #25.6*g# mass of #"octane"# is completely combusted, what mass of carbon dioxide will result?

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Answer 1 · 2017-06-29T07:32:11

Approx...#1.8*mol..........#

We need (i) a stoichiometrically balanced equation.....

#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#

Is this balanced with respect to mass and charge? It must be if we purport to represent chemical reality.

And (ii) we need equivalent quantities of octane.....

#"Moles of octane"=(25.6*g)/(114.23*g*mol^-1)=0.224*mol#

And given the equation, CLEARLY we KNOW that 8 equiv of carbon dioxide result per equiv of octane.....

And thus a molar quantity of #8xx0.224*mol# #CO_2(g)# #=# #??*mol# #=# #??*g#.

Answer 2 · 2017-06-29T07:35:00

79 grams of carbondioxide

#C_8H_18 + 12.5 O_2 -> 8CO_2 + 9H_2O#

This is the chemical reaction.

1 mole of octane mass is 114 grams.

It means if you have 114 grams of octane, you will have 352 grams of #CO_2# after the reaction.

However you have only 25.6 grams of octane. You can get how much carbondioxide after the reaction.

#=(25.6times352)/114#

#=79# grams