How would you balance the equation for the combustion of octane: C8H18(l)+O2(g)---->CO2(g)+H2O(l)?

2 Answers
anor277
Nov 11, 2015

The complete combustion of any hydrocarbon gives carbon dioxide and water. I will represent the combustion of hexane.

Explanation:

#C_6H_14(g) + 19/2O_2(g) rarr 6CO_2(g) + 7H_2O(g)#

Is this equation balanced? How do you know? How is the complete combustion of octane, #C_8H_18#, to be represented? I balanced the carbons, and then the hydrogens, and then the oxygens. The order I used is unimportant, it is important that I balance the equation.

Nikka C.
Nov 11, 2015

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

or

#2C_8H_18 (l)# + #25 O_2 (g)# #rarr# #16CO_2 (g)# + #18H_2O (l)#

Explanation:

First, you need to tally all the atoms.

#C_8H_18 (l)# + #O_2 (g)# #rarr# #CO_2 (g)# + #H_2O (l)# (unbalanced)

Based on the subscripts, you have

left side:
C = 8
H = 18
O = 2

right side:
C = 1
H = 2
O = 2 + 1 (do not add this up yet)

Second, find the easiest atom to balance. In this case, the #C# atom. Always remember that in balancing, you are NOT SUPPOSED TO CHANGE THE SUBSCRIPTS, only put coefficients before the substance (as changing the subscripts means that you are changing the molecular structure instead).

#C_8H_18 (l)# + #O_2 (g)# #rarr# #color (red) 8CO_2 (g)# + #H_2O (l)#

left side:
C = 8
H = 18
O = 2

right side:
C = (1 x #color (red) 8#) = 8
H = 2
O = (2 x #color (red) 8#) + 1

Since #CO_2# is a substance, you have to apply the coefficient to both #C# and two #O# atoms as they are all bonded to each other.

Third, balance the next easiest atom.

#C_8H_18 (l)# + #O_2 (g)# #rarr# #8CO_2 (g)# + #color (blue) 9H_2O (l)#

left side:
C = 8
H = 18
O = 2

right side:
C = (1 x 8) = 8
H = (2 x #color (blue) 9#) = 18
O = (2 x 8) + (1 x #color (blue) 9#) = 25

Now all that is left is to balance are the #O# atoms. Since the sum of #O# atoms on the right side is an odd number, I can use my knowledge in fractions to balance the left side of the equation.

Thus,

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

left side:
C = 8
H = 18
O = (2 x #color (green) (25/2)#) = 25

right side:
C = (1 x 8) = 8
H = (2 x 9) = 18
O = (2 x 8) + (1 x 9) = 25

But if you don't want fractions as coefficients, you can always multiply the WHOLE equation by 2.

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

=

#2C_8H_18 (l)# + #25 O_2 (g)# #rarr# #16CO_2 (g)# + #18H_2O (l)# (balance)

Both answers are considered correct.

Related questions
See all questions in Balancing Chemical Equations
Impact of this question
104913 views around the world
You can reuse this answer
Creative Commons License