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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Blueberries116 on January 02, 2020, 04:58:20 PM

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?.

10 moles of octane produce 80 moles of (CO_{2}+CO) and 90 moles of water. So you need 135 moles of O_{2} (assuming no CO is produced) and about 500 moles of N_{2} that gives about 18 kg of air (this number is overestimated about 1020 %)

10 moles of octane produce 80 moles of (CO_{2}+CO) and 90 moles of water. So you need 135 moles of O_{2} (assuming no CO is produced) and about 500 moles of N_{2} that gives about 18 kg of air (this number is overestimated about 1020 %)
printing error  should be 125, though not important in estimation