Question #f3b97

1 Answer
Mar 12, 2016

The yield is 25.00% (1/4)


The well-written balanced equation for the complete octane combustion is:

#2C_8H_18 + 25O_2 -> 16CO_2 + 18 H_2O#

Now, let's first calculate the moles of #CO_2#:

#n_"CO2" = "mass"/"molar mass" = "28.16 g"/"44.01 g/mol" = 0.6399 " mol"#

These moles have been originated from one eight of octane moles, given that the coefficient of octane (2) is one eigth of #CO_2# coefficient (16). So we can be certain that:
#"moles of combusted octane" = "0.6399 moles"/8 = 0.07999 " moles"#

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 #25/2 = 12.5# factor must multiply octane moles to obtain the oxygen moles which should suffice.
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 #2/25# or #1/12.5# or 0.08 = 8% of oxygen moles, that is:
#"maximum octane" = 4 " moles"*2/25 = 0.32 " moles"#.

If the reaction had been completed, these moles should have yielded #0.32*8 = 2.56 " moles"# of #CO_2#, which - as the moles of octane that actually reacted - are pretty more than the true number of #CO_2# moles: 0.6399 mol.

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 #CO_2# that have been formed divided by the maximum expectable #CO_2#:
#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 #CO_2# generated, not by considering all the amount of octane added, but just that limited amount which could have reacted due to the limited amount of oxygen ( limiting reactant).

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 #CO_2#. So the 98% of octane remained unchanged or maybe underwent other transformations that didn't lead to #CO_2#.