Answer to Question #80427 in General Chemistry for Divine Hope

Question

Butane undergoes combustion as follows.

2 C4H10 + 13 O2 ——> 8 CO2 + 10 H2O

In a particular trial 29 g of C4H10 is combusted in excess oxygen. This results in a recovery of 72 g of CO2. What is the percent yield of the CO2?
Thanks!

Answer 1

At first we must calculate the mole for C4H10.
n(C4H10)=29/58=0.5 moles of C4H10.
From equation we can see that 2 moles of C4H10 are giving 8 moles of CO2. So 0.5 moles of C4H10 will give 2 moles of CO2. Then we must calculate the mass of CO2.
m(CO2)=2*44=88 grams.
Then the yield =(72/88)*100%=81.81%.