Question 88b25

May 15, 2015

The chemical formula for octane is $\text{C"_8"H"_18}$. A combustion reaction is one in which one reactant reacts with oxygen gas, $\text{O"_2}$. Since octane is a hydrocarbon, the products will be carbon dioxide and hydrogen. The balanced equation for the combustion of octane gas is:

"2C"_8"H"_18("g")+"25O"_2("g")$\rightarrow$$\text{16CO"_2("g")+"18H"_2"O(g)}$

Explanation:

First place an 8 in front of ${\text{CO}}_{2}$ because of the 8 carbon atoms in a molecule of octane, and a 9 in front of $\text{H"_2"O}$ because of 18 H atoms divided by 2 atoms of O (in ${\text{O}}_{2}$).

"C"_8"H"_18("g")+"O"_2("g")$\rightarrow$$\text{8CO"_2("g")+"9H"_2"O(g)}$

Balance the oxygen atoms on both sides of the equation. There are 25 O atoms on the right side of the equation. So we place a coefficient of $\frac{25}{2}$ in front of the ${\text{O}}_{2}$ molecule on the left.

"C"_8"H"_18("g")$+$$\frac{25}{2} \text{O"_2("g")}$$\rightarrow$$\text{8CO"_2("g")+"9H"_2"O(g)}$

There cannot be a fraction as a coefficient, so we need to multiply the equation times 2.

"2C"_8"H"_18("g")+"25O"_2("g")#$\rightarrow$$\text{16CO"_2("g")+"18H"_2"O(g)}$