Question

Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 8.000 moles of O2.
2 C8H18 + 25 O2 ? 16 CO2 + 18 H2O

a)12.50%
b)25.00%
c)20.00%
d)50.00%

Answer 1

moles of CO2 formed = 28.16/44
= 0.64
so moles of O2 consumed = 25/16*.64
= 1 mole
% yeild = 1/8*100
= 12.5%


so the ans is (A), 12.5%

Answer 2

2 C8H18 +25 O2 --->16 CO2 +18 H2O

8.000 moles C8H18 requires 8.000 x (25/2) = 100 mole O2
you have 4.000 mole which is not enough so the O2 is the limiting reactaant

4 mole O2 gives 4 x (16/25) = 2.56 mole CO2 = 2.56 x (12 + 32) = 112.64 gram CO2

You got 28.16 gram so percent yield is (28.16/112.64) x 100% = 25%

Hence, d is an answer

Answer 3

It's 25.00%, but I don't know how, I just happen to have had this question as well and was trying to understand.

Answer 4

Percent Yield.pdf

The answer is 12.50, every one that keeps saying it's 25% is just outright wrong, if you need proof look at the pdf file I uploaded. I literally looked everywhere and researched how to do it and this is the outcome. I tried my best to layout my thoughts to best demonstrate how I reached my outcome and I really hope that I helped.