The complete burning of any type of hydrocarbon provides carbon dioxide and also water. Ns will stand for the burning of hexane.

You are watching: Balanced equation for combustion of octane


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

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


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#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 must 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 = 8H = 18O = 2

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

Second, uncover the simplest atom come 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 method that girlfriend are transforming the molecular framework 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 to use the coefficient come both #C# and also two #O# atoms together they are all external inspection to each other.

Third, balance the following 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 come balance are the #O# atoms. Because the amount of #O# atoms on the appropriate side is an odd number, I deserve to use my understanding 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 = 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 girlfriend don"t want fractions as coefficients, you can constantly multiply the totality equation by 2.