# Question #f3b97

##### 1 Answer

#### Answer:

The yield is 25.00% (1/4)

#### Explanation:

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 **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