# Question #cf926

##### 2 Answers

#### Answer:

Balance the chemical equation!

#### Explanation:

All you really have to do here is balance the chemical equation that describes the combustion of octane,

As you know, the **complete combustion** of octane involves burning this hydrocarbon in the presence of excess oxygen, **hydrocarbon**, i.e. a compound that contains only carbon and hydrogen, the reaction will produce **two products**

carbon dioxide,#"CO"_2# water,#"H"_2"O"#

The *unbalanced* chemical equation that describes the combustion of octane will thus be

#"C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> "CO"_ (2(g)) + "H"_ 2"O"_ ((l))#

Now just balance this equation like you would fo any chemical equation.

Start with the atoms of carbon. You have

#"C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + "H"_ 2"O"_ ((l))#

Now look at the atoms of hydrogen. You have

#"C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O"_ ((l))#

Finally, focus on the atoms of oxygen. You have

#overbrace(8 xx "2 atoms O")^(color(blue)("from 8CO"_2)) + overbrace(9 xx "1 atom O")^(color(purple)("from 9H"_2"O")) = "25 atoms O"#

At this point, you can use a little trick to make the balancing easier. Notice that you can add a *Fractional coefficient* to

#"C"_ 8"H"_ (18(l)) + 25/2"O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O"_ ((l))#

The reactants' side now has

#25/2 xx "2 atoms O" = "25 atoms O"#

the same number of atoms of oxygen as the products' side. To get rid of the fractional coefficient, simply multiply **all the coefficients** by

#color(red)(2)"C"_ 8"H"_ (18(l)) + (25/2 xx color(red)(2))"O"_ (2(g)) -> (8 xx color(red)(2))"CO"_ (2(g)) + (9 xx color(red)(2))"H"_ 2"O"_ ((l))#

The balanced chemical equation will thus be

#2"C"_ 8"H"_ (18(l)) + 25"O"_ (2(g)) -> 16"CO"_ (2(g)) + 18"H"_ 2"O"_ ((l))#

Now simply add the coefficients to get their sum

#sum"coefficients" = 2 + 25 + 16 + 18 = color(darkgreen)(ul(color(black)(61)))#

I shall first try to find out a general formula to know the sum of the coefficients of reactants and products for the combustion raction of any alkane.

The general molecular formula of an alkane is

So the sum of the coefficients becomes

It is obvious from above relation that the **Sum (S) will always be a whole number if x is odd**

Otherwise to get the sum as whole number **for even value of x we are to multiply the value of S by 2**

In case of octane x = 8.

So the required sum is