# What is the chemical equation to show the combustion reaction of ethanol?

Nov 2, 2015

$\text{C"_2"H"_6"O(l)" + color(orange)3"O"_2"(g)}$$\rightarrow$color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")

#### Explanation:

$\text{C"_2"H"_6"O(l)" + "O"_2"(g)}$$\rightarrow$"CO"_2("g") + "H"_2"O(l")

Balance the Equation:

Balance the $\text{C}$ and $\text{H}$ first. There are $2$ carbon atoms on the left side, and $1$ carbon atom on the right side. There are $6$ hydrogen atoms on the left side and $2$ hydrogen atoms on the right side.

Add a coefficient of $\textcolor{red}{2}$ in front of the carbon dioxide, and a coefficient of $\textcolor{p u r p \le}{3}$ in front of the water.

$\text{C"_2"H"_6"O(l)" + "O"_2"(g)}$$\rightarrow$$\textcolor{red}{2} \text{CO"_2("g") + color(purple)3"H"_2"O(l")}$

We now have equal numbers of $\text{C}$ and $\text{H}$ atoms on both sides. However, the $\text{O}$ is not balanced.

We have $7$ $\text{O}$ atoms on the right side, and $3$ on the left side. Place a coefficient of $\textcolor{\mathmr{and} a n \ge}{3}$ in front of the ${\text{O}}_{2}$ on the left side.

$\text{C"_2"H"_6"O(l)" + color(orange)3"O"_2"(g)}$$\rightarrow$color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")

We now have a balanced equation, with $2$ carbon atoms on both sides, $6$ hydrogen atoms on both sides, and $7$ oxygen atoms on both sides.