The yield is 25.00% (1/4)
The well-written balanced equation for the complete octane combustion is:
Now, let's first calculate the moles of
These moles have been originated from one eight of octane moles, given that the coefficient of octane (2) is one eigth of
Now, we go to calculate the mole amount of octane that could have reacted.
The balanced equation says that for each 2 moles of octane are needed 25 moles of diatomic oxygen, that is the factor
But, as we see that the available moles of dioxygen (4) are the same of octane moles (4), we deduce that only a small part of octane will be combusted, because there is not enough oxygen.
The maximum amount of octane that could have reacted is then the fraction
If the reaction had been completed, these moles should have yielded
We can then calculate the yield of the actual reaction by taking the ratio between the maximum amount of octane which has actually reacted and the amount that could have reacted or by taking the ratio between the
#Yield = 0.07999" mol"/0.32" mol" = 0.6399" mol"/2.56" mol" = .2500, that is 25%.
The rationale of this problem is that you should calculate the yield of
Only 8% of the added octane could have reacted (0.32 mol by 4.000) and only one quarter of that actually reacted (2%) to yield