# How do you write a balanced equation for the complete combustion of C_4H_8?

Sep 9, 2016

$2 \text{C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O}$

#### Explanation:

In a combustion reaction, $\text{C}$ and $\text{H}$ are converted to ${\text{CO}}_{2}$ and $\text{H"_2"O}$.

Start by writing the unbalanced equation.

$\text{C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O}$

Put a $1$ in front of the most complicated formula, ${\text{C"_4"H}}_{10}$.

$\textcolor{red}{1} \text{C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O}$

Step 1. Balance $\text{C}$.

Put a $4$ in front of ${\text{CO}}_{2}$.

$\textcolor{red}{1} \text{C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + "H"_2"O}$

Step 2. Balance $\text{H}$.

Put a $5$ in front of $\text{H"_2"O}$.

$\textcolor{red}{1} \text{C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + color(green)(5)"H"_2"O}$

Step 3. Balance $\text{O}$.

We have 13 $\text{O}$ atoms on the right and 2 $\text{O}$ atoms on the left.

We can't balance $\text{O}$ without using fractions.

We start over, multiplying each coefficient by $2$.

$\textcolor{red}{2} \text{C"_4"H"_10 + "O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O}$

Balance $\text{O}$ atoms by putting $13$ in front of ${\text{O}}_{2}$.

$\textcolor{red}{2} \text{C"_4"H"_10 + color(blue)(13)"O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O}$

Every formula now has a coefficient.

Step 4. Check that atoms are balanced.

$\text{Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side}$
$\textcolor{w h i t e}{m} \text{C} \textcolor{w h i t e}{m m m m m l l} 8 \textcolor{w h i t e}{m m m m m m m l} 8$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m m} 20 \textcolor{w h i t e}{m m m m m m m} 20$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m m} 26 \textcolor{w h i t e}{m m m m m m m} 26$

All atoms are balanced.

The balanced equation is

$2 \text{C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O}$