The complete burning of any kind of hydrocarbon gives carbon dioxide and water. Ns will stand for the combustion of hexane.

You are watching: In the combustion reaction of octane c8h18 what are the products

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

Is this equation balanced? just how do girlfriend know? just how is the complete burning of octane, #C_8H_18#, to be represented? I balanced the carbons, and then the hydrogens, and then the oxygens. The stimulate I provided is unimportant, the is vital that i balance the equation.

#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)#

First, you have to tally every the atoms.

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

Based ~ above the subscripts, friend have

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

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

Second, uncover the simplest atom to balance. In this case, the #C# atom. Constantly remember that in balancing, you room NOT supposed TO adjust THE SUBSCRIPTS, just put coefficients before the substance (as transforming the subscripts way that girlfriend are transforming the molecular structure instead).

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

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

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

Since #CO_2# is a substance, you have actually to use the coefficient to both #C# and two #O# atoms as they room all external inspection to every other.

Third, balance the next easiest atom.

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

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

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

Now all that is left is to balance room the #O# atoms. Because the sum of #O# atom on the ideal side is one odd number, I deserve to use my expertise in fractions to balance the left next 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 = 8H = 18O = (2 x #color (green) (25/2)#) = 25

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

But if you don"t want fractions together coefficients, you can always multiply the entirety equation by 2.