Question

What pattern do you observe in the coefficients of the hydrocarbon (CxHy) and water (H2O)? How do you explain this pattern?

Answer 1

The coefficients of the hydrocarbon and water is 61 and it is explained below.

Explanation:

We have to  balance the chemical equation that describes the combustion of octane C₈H₁₈. The combustion of octane involves burning this hydrocarbon in the presence of excess oxygen,O 2.  Because octane is a hydrocarbon, that is the compound that contains only carbon and hydrogen, the reaction will produce two products carbon dioxide CO 2 and water H 2 O.

The unbalanced chemical equation is

C₈H₁₈ + O₂ → CO₂ + H₂O

To balance this equation, we have 8 present on the reactant side, so multiply the carbon dioxide by  8  to get  8  atoms of carbon on the products side.

C₈H₁₈ + O₂ → 8 CO₂ + H₂O

Now, we have 18  on the reactant side and  2  on the products side, so multiply the water molecule by  9  to get

C₈H₁₈ + O₂ → 8 CO₂ + 9 H₂O

8 * 2 atoms O + 9 *1 atoms O = 25 atoms O

By adding fractional coefficient to O₂ we get

C₈H₁₈ + $\frac{25}{2}$O₂ → 8 CO₂ + 9 H₂O

Multiply all the coefficients by 2 we get,

2C₈H₁₈ + ($\frac{25}{2}$ * 2) O₂ → 8 * 2 CO₂ + 9* 2 H₂O

The balance equation is

2C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

Add the coefficients to get their sum

= 2 +25 + 16 +18 = 61

Thus, The coefficients of the hydrocarbon and water is 61