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

Question

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

Noah Aldrich · Accepted Answer

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

#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.

Peyton Adams · Answer

#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)# You must count each and every atom first. #C_8H_18 (l)# + #O_2 (g)# #rarr# #CO_2 (g)# + #H_2O (l)# (unbalanced) The subscripts indicate that you have left side: O = 2, H = 18, and C = 8. right side: (do not add this up yet) C = 1 H = 2 O = 2 + 1. 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: O = 2, H = 18, and C = 8. 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, find the next easiest atom to balance. #C_8H_18 (l)# + #O_2 (g)# #rarr# #8CO_2 (g)# + #color (blue) 9H_2O (l)# left side: O = 2, H = 18, and C = 8. 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. Consequently, #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: O = (2 x 8) + (1 x 9) = 25 H = (2 x 9) = 18 C = (1 x 8) = 8. However, you can always multiply the entire equation by two if you don't want fractions to be the coefficients. #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 responses are regarded as accurate.

Natalie Anton · Answer

undefined 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l)

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

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