Is it possible to obtain the amount of air required for a combustion analysis?
For the given problem there isn't much $\ce{CO}$ created. Given that the question is multiple choice an exact amount isn't needed. The problem only needs to be solve close enough to determine which choice is the answer. So assume complete combustion and then guess that the answer will be about 5% high. If octane, $\ce{C8H18}$, is completely burned with oxygen then the reaction is:

$$\ce{C8H18 + 12.5O2 -&gt; 8CO2 + 9H2O}$$ 

since the problem states that 1.14 kg of octane was burned, the number of moles is:

$$\dfrac{1140}{114} = 10.0$$

Thus for complete oxidation:


125 moles of oxygen would be needed.  
Air is only 21% oxygen so 125/0.21 = 595 "moles" of air are needed.
At 28.8 g/"mole" that would be 17.1 kg of air. 


In reality there is about 5% less oxygen required since there is some $\ce{CO}$ produced, so $17.1*0.95 = 16.3\ \pu{ kg}$ is a good guess.



I'll leave finding the exact solution to the OP.