# Question cf926

Nov 14, 2016

Balance the chemical equation!

#### Explanation:

All you really have to do here is balance the chemical equation that describes the combustion of octane, ${\text{C"_8"H}}_{18}$.

As you know, the complete combustion of octane involves burning this hydrocarbon in the presence of excess oxygen, ${\text{O}}_{2}$. Since octane is a hydrocarbon, i.e. a compound that contains only carbon and hydrogen, the reaction will produce two products

• carbon dioxide, ${\text{CO}}_{2}$
• water, $\text{H"_2"O}$

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

${\text{C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> "CO"_ (2(g)) + "H"_ 2"O}}_{\left(l\right)}$

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

${\text{C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + "H"_ 2"O}}_{\left(l\right)}$

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

${\text{C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O}}_{\left(l\right)}$

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 ${\text{O}}_{2}$ to get

${\text{C"_ 8"H"_ (18(l)) + 25/2"O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O}}_{\left(l\right)}$

The reactants' side now has

$\frac{25}{2} \times \text{2 atoms O" = "25 atoms O}$

the same number of atoms of oxygen as the products' side. To get rid of the fractional coefficient, simply multiply all the coefficients by $\textcolor{red}{2}$

$\textcolor{red}{2} {\text{C"_ 8"H"_ (18(l)) + (25/2 xx color(red)(2))"O"_ (2(g)) -> (8 xx color(red)(2))"CO"_ (2(g)) + (9 xx color(red)(2))"H"_ 2"O}}_{\left(l\right)}$

The balanced chemical equation will thus be

$2 {\text{C"_ 8"H"_ (18(l)) + 25"O"_ (2(g)) -> 16"CO"_ (2(g)) + 18"H"_ 2"O}}_{\left(l\right)}$

Now simply add the coefficients to get their sum

$\sum \text{coefficients} = 2 + 25 + 16 + 18 = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{61}}}$

Nov 14, 2016

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 ${C}_{x} {H}_{2 x + 2}$ and the blanced equation of its combustion reaction is as follows

${C}_{x} {H}_{2 x + 2} + \frac{1}{2} \left(3 x + 1\right) {O}_{2} \to x C {O}_{2} + \left(x + 1\right) {H}_{2} O$

So the sum of the coefficients becomes

$S = 1 + \frac{1}{2} \left(3 x + 1\right) + x + \left(x + 1\right)$

$\implies S = 1 + 2 x + \frac{1}{2} \left(3 x + 1 + 2\right)$

$\implies S = 1 + 2 x + \frac{3}{2} \left(x + 1\right)$

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

$= 2 \times S = 2 \left(1 + 2 x + \frac{3}{2} \left(x + 1\right)\right)$

$= 2 \left(1 + 2 \times 8 + \frac{3}{2} \left(8 + 1\right)\right) = 2 + 32 + 27 = 61$