Question

How many moles of oxygen react with 6 moles of octane?

Answer 1

Octane and oxygen react in a combustion reaction, producing carbon dioxide and water in this reaction (after balancing the equation):

`2"C"_8"H"_18+25"O"_2->16"C""O"_2+18"H"_2"O"`

Multiply both sides by `3`:

`6"C"_8"H"_18+50"O"_2->48"C""O"_2+54"H"_2"O"`

Clearly, `6` moles of octane reacts with `50` moles of oxygen.

This assumes that octane is completely combusted. However, if there is incomplete combustion, carbon monoxide and soot can be produced, and a different number of moles of oxygen will react with the octane.

Answer 2

Approx. `75*mol` `"dioxygen gas...."`

Explanation:

When you deal with hydrocarbon combustion, the standard protocol is to (i) balance the carbons as carbon dioxide; (ii) balance the hydrogens as water; and (iii) balance the dioxygen REACTANT....

We assume complete combustion to carbon dioxide and water...and we write the UNBALANCED equation...

`underbrace(C_8H_18(l) + O_2 rarrCO_2(g) +H_2O(l))_"nothing balanced"`

Balance the carbons.....

`underbrace(C_8H_18(l) + O_2 rarr8CO_2(g) +H_2O(l))_"carbons balanced"`

Then balance the hydrogens as WATER.....

`underbrace(C_8H_18(l) + O_2 rarr8CO_2(g) +9H_2O(l))_"carbons and hydrogens balanced"`

Then balance the oxygens.....

`underbrace(C_8H_18(l) + 25/2O_2 rarr8CO_2(g) +9H_2O(l))_"ALL balanced"`

You can double the entire equation if you like..to remove the half-integral coefficient. I think the arithmetic is easier this way...

AND the question proposes the complete combustion of `6*mol` octane. Given the stoichiometric equation, we require `6xx25/2*mol` dioxygen...i.e. `75*mol` `O_2`. And clearly we could further ask for the mass of dioxygen combusted, and for the VOLUME of carbon dioxide evolved under standard conditions. But all we have done here is use some simple arithmetic to give the appropriate stoichiometry.

We could further propose incomplete combustion to give SOME `CO` or some `C`...i.e....

`underbrace(C_8H_18(l) + 11O_2 rarr6CO_2(g) +CO(g) + C(s)+9H_2O(l))_"incomplete combustion"`

How much carbon or carbon monoxide we get (and certainly we get some in the internal combustion or diesel engines) is determined by experiment.