Question #7ba58

Apr 30, 2017

Warning! Long Answer. You follow a systematic procedure to balance the equation.

Explanation:

$\text{C"_8"H"_18 + "O"_2 → "CO"_2 + "H"_2"O}$

A method that often works is to balance everything other than $\text{O}$ and $\text{H}$ first, then balance $\text{O}$, and finally balance $\text{H}$.

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like ${\text{C"_8"H}}_{18}$. We put a 1 in front of it to remind ourselves that the number is now fixed.

$\textcolor{red}{1} \text{C"_8"H"_18 + "O"_2 → "CO"_2 + "H"_2"O}$

Balance $\text{C}$:

We have fixed $\text{8C}$ on the left, so we need $\text{8 C}$ on the right. Put an 8 in front of the ${\text{CO}}_{2}$.

$\textcolor{red}{1} \text{C"_8"H"_18 + "O"_2 → color(blue)(8)"CO"_2 + "H"_2"O}$

Balance $\text{O}$:

We can't balance $\text{O}$ because we have two formulas containing $\text{O}$ that have no coefficients.

Balance $\text{H}$:

We have fixed $\text{18 H}$ on the left, so we need $\text{18 H}$ on the right. Put a 9 in front of $\text{H"_2"O}$.

$\textcolor{red}{1} \text{C"_8"H"_18 + "O"_2 → color(blue)(8)"CO"_2 + color(orange)(9)"H"_2"O}$

Balance $\text{O}$:

We have fixed $\text{25 O}$ on the right, so we need 25 $\text{O}$ on the left.

Oops! That means we would need ${\text{12.5 O}}_{2}$ on the left.

My instructor never let me use fractions when balancing equations.

The solution: Multiply every coefficient by 2.

$\textcolor{red}{2} \text{C"_8"H"_18 + "O"_2 → color(blue)(16)"CO"_2 + color(orange)(18)"H"_2"O}$

Now, we have $\text{50 O}$ on the right, so we can use ${\text{25 O}}_{2}$ on the left.

$\textcolor{red}{2} \text{C"_8"H"_18 + color(purple)(25)"O"_2 → color(blue)(16)"CO"_2 + color(orange)(18)"H"_2"O}$

Every formula now has a fixed coefficient. We should have a balanced equation.

Let’s check:

$\boldsymbol{\text{Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side}}$
$\textcolor{w h i t e}{m} \text{C} \textcolor{w h i t e}{m m m m m m} 16 \textcolor{w h i t e}{m m m m m m m m} 16$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m m m} 36 \textcolor{w h i t e}{m m m m m m m m} 36$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m m m} 50 \textcolor{w h i t e}{m m m m m m m m} 50$

All atoms balance! The balanced equation is

$2 \text{C"_8"H"_18 +25"O"_2 → 16"CO"_2 + 18"H"_2"O}$