I'm stuck with a problem regarding to find the amount of air required to burn a certain amount of octane. The reason of my confusion is why am I given the average molecular weight.

The problem is as follows:

[itex]1.14\,kg[/itex] of octane [itex]C_{8}H_{18}[/itex] is burned with a certain amount of air. The resulting products from the combustion are known to have the following percentages by volume: [itex]CO_{2} = 41.18\%[/itex] ; [itex]CO = 5.8%[/itex]$ ; [itex]H_2O_{(vapor)} = 52.94\%[/itex]. Find the weight of required air in kilograms. Assume the composition of the air in this combustion is [itex]21\%[/itex] of [itex]O_{2}[/itex] and [itex]79\%[/itex] of [itex]N_{2}[/itex] by volume and its average molecular weight is[itex]28.8\,\frac{g}{mol}[/itex]$.

The alternatives given are as follows:

[itex]\begin{array}{ll}

1.&9.0\,kg\\

2.&16.4\,kg\\

3.&12.5\,kg\\

4.&45.1\,kg\\

5.&120\,kg\\

\end{array}[/itex]$

What I think to approach this problem was to use the given percentages to get the number of moles using the initial moles of octane.

Using these moles I can obtain the grams of oxygen produced. Since it is given the percentage which is of oxygen in the air I could use this to find the required mass of the air to get that amount of air.

This is shown as follows: (for the sake of brevity I'm ommiting units but they are consistent.)

[itex]\textrm{FW of octane} = 114[/itex]

[itex]n_octane=\frac{1140}{114}=10[/itex]

Then moles of each product:

For [itex]CO_{2} = 41.18\%[/itex]

[itex]\frac{41.18}{100}\times 10 = 4.118[/itex]$

Then in this amount of [itex]CO_{2}[/itex] there must be these grams of oxygen:

[itex]4.118 \times \frac{32\,g\O}{1\,mol\,CO_{2}}= 131.776 \,g[/itex]$

For [itex]CO = 5.8\%[/itex]

[itex]\frac{5.8}{100}\times 10 = 0.058[/itex]

[itex]0.058 \times \frac{16\,g\O}{1\,mol\,CO}= 0.928 \,g[/itex]

For [itex]H_2O_{(vapor)} = 52.94\%[/itex].

[itex]\frac{52.94}{100}\times 10 = 5.294[/itex]

[itex]5.294 \times \frac{16\,g\O}{1\,mol\,H_2O}= 84.704 \,g[/itex]

Adding these together:

[itex]131.776+0.928+84.704=217.408\,g\,O[/itex]

Then using the known percentage of oxygen which is [itex]21\%[/itex]

[itex]\frac{21}{100}x=217.408[/itex]

[itex]x=1035.28 \,g[/itex]

Which is approximately [itex]1.035 kg[/itex]. but this answer doesn't get any close to the specified alternatives. Did I overlooked something. Or could it be that this approach is not applicable?. In this given situation **is it possible to find the required mass of air?.**