Question

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

Answer 1

Approx...`1.8*mol..........`

Explanation:

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

79 grams of carbondioxide

Explanation:

`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