How do we represent the complete combustion of #"octane"#, #C_8H_18(l)#?

2 Answers
Mar 4, 2016

Answer:

#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#

Explanation:

The above equation is stoichiometrically balanced: garbage out equals garbage in. If you like you can double the equation to remove the non-integral coefficient. I have never seen the need to do so, inasmuch the arithmetic is easier when you use this form with the 1 equivalent of hydrocarbon reactant.

So 2 questions with regard to this reaction: (i) How does energy transfer in this reaction; (ii) does this represent an oxidation reduction reaction?

Mar 4, 2016

Answer:

We need to have the same number of moles of each substance on each side. The balanced equation is:

#C_8H_18 + 25/2# #O_2 = 8# #CO_2 + 9# #H_2O# or #2# #C_8H_18 + 25# #O_2 = 16# #CO_2 + 18# #H_2O#

Explanation:

We start out with the unbalanced equation:

#a# #C_8H_18 + b# # O_2 = c# # CO_2 + d# # H_2O#

We need to find the values of the coefficients #a, b, c# and #d# to balance the equation.

For the moment, let's leave #a# as 1. There are 8 moles of C on the left, so to balance we need 8 moles of C on the right, so let's make #c = 8#, because each #CO_2# contains 1.

#C_8H_18 + b# # O_2 = 8# # CO_2 + d # #H_2O#

There are 18 moles of H on the left so we need 18 on the right, but each #H_2O# contains 2, so we make #d = 9#.

#C_8H_18 + b # #O_2 = 8# # CO_2 + 9 # #H_2O#

Now we turn our attention to the right side. There are two O in each of 8 #CO_2# for a total of 16 in the carbon dioxide and one in each of 9 #H_2O# for a total of 9 in the water, so we need 25 O all together.

On the left, each #O_2# contains two O, so one way to balance the equation is to take #25/2# of them:

#C_8H_18 + # #25/2 O_2 = # #8# # CO_2 + 9 # #H_2O#

If the fraction makes you uncomfortable, another way is to multiply all the coefficients by two:

#2 # #C_8H_18 + 25# # O_2 = 16# # CO_2 + 18 # #H_2O#