How would you find a balanced equation if acetylene gas ("C"_2"H"_2) undergoes combustion to form carbon dioxide and water vapor?

Jun 20, 2016

One way is to "balance by inspection".

Explanation:

The unbalanced equation is

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

Let's start with the most complicated formula, ${\text{C"_2"H}}_{2}$. We put a 1 in front of it.

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

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

$\textcolor{red}{1} \text{C"_2"H"_2 + "O"_2 → color(orange)(2)"CO"_2 + "H"_2"O}$

We have also fixed 2 $\text{H}$ atoms in the ${\text{C"_2"H}}_{2}$, so we need 2 $\text{H}$ atoms on the right. We put a 1 in front of the $\text{H"_2"O}$.

$\textcolor{red}{1} \text{C"_2"H"_2 + "O"_2 → color(orange)(2)"CO"_2 + color(blue)(1)"H"_2"O}$

Now we have fixed 5 $\text{O}$ atoms on the right, so we need 5 $\text{O}$ atoms on the left.

Oops! We would need to use 2½ molecules of ${\text{O}}_{2}$.

To avoid fractions, we multiply all coefficients by 2 and get

$\textcolor{red}{2} \text{C"_2"H"_2 + "O"_2 → color(red)(4)"CO"_2 + color(blue)(2)"H"_2"O}$

We now have 10 $\text{O}$ atoms on the right, so we need 10 $\text{O}$ atoms on the left. We can now put a 5 in front of the ${\text{O}}_{2}$.

$\textcolor{red}{2} \text{C"_2"H"_2 + color(orange)(5)"O"_2 → color(red)(4)"CO"_2 + color(blue)(2)"H"_2"O}$

The equation should now be balanced. Let's check.

$\text{Atom"color(white)(m) "On the left"color(white)(m) "On the right}$
stackrel(—————————————)(color(white)(m)"C"color(white)(mmmmm)4color(white)(mmmmmm) 4)
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m m} 4 \textcolor{w h i t e}{m m m m m m} 4$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m l} 10 \textcolor{w h i t e}{m m m m m l l} 10$

The equation is balanced!