Question

Question #cf926

Answer 1

Balance the chemical equation!

Explanation:

All you really have to do here is balance the chemical equation that describes the combustion of octane, `"C"_8"H"_18`.

As you know, the complete combustion of octane involves burning this hydrocarbon in the presence of excess oxygen, `"O"_2`. Since octane is a 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 `8` present on the reactants' side, so multiply the carbon dioxide by `8` to get `8` atoms of carbon on the products' side

`"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 `18` on the reactants' side and `2` on the products' side, so multiply the water molecule by `9` to get

`"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 `2` on the reactants' side and

`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 `"O"_2` to get

`"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)`

`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)))`

Answer 2

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 `C_xH_(2x+2)` and the blanced equation of its combustion reaction is as follows

`C_xH_(2x+2)+1/2(3x+1)O_2->xCO_2+(x+1)H_2O`

So the sum of the coefficients becomes

`S=1+1/2(3x+1)+x+(x+1)`

`=>S=1+2x+1/2(3x+1+2)`

`=>S=1+2x+3/2(x+1)`

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

`=2xxS=2(1+2x+3/2(x+1))`

`=2(1+2xx8+3/2(8+1))=2+32+27=61`