# Question #85625

Jun 22, 2016

$2 {\text{C"_ 2"H"_ (2(g)) + 5"O"_ (2(g)) -> 4"CO"_ (2(g)) + 2"H"_ 2"O}}_{\left(l\right)}$

#### Explanation:

The idea here is that because acetylene is a hydrocarbon, which is a compound that contains only carbon and hydrogen, you can expect its complete combustion to produce two products

• carbon dioxide, ${\text{CO}}_{2}$
• water, $\text{H"_2"O}$

In other words, when a hydrocarbon undergoes complete combustion, all the carbon that it contained will be part of the carbon dioxide and all the hydrogen it contained will be part of the water.

You will thus have

${\text{C"_ 2"H"_ (2(g)) + "O"_ (2(g)) -> "CO"_ (2(g)) + "H"_ 2"O}}_{\left(l\right)}$

This is the unbalanced chemical equation that describes the complete combustion of acetylene.

To balance it, start by looking at how many atoms of carbon you have on both sides of the equation.

Since the reactants' side contains $2$ atoms of carbon and the products' side contains $1$ atom of carbon, multiply the carbon dioxide molecule by $\textcolor{red}{2}$ to get

${\text{C"_ 2"H"_ (2(g)) + "O"_ (2(g)) -> color(red)(2)"CO"_ (2(g)) + "H"_ 2"O}}_{\left(l\right)}$

Now focus on hydrogen. The reactants' side contains $2$ atoms of hydrogen, as does the products' side $\to$ hydrogen is balanced.

Finally, look at oxygen. The reactants' side contains $2$ atoms of oxygen and the products' side contains

$\textcolor{red}{2} \times 2 + 1 = \text{5 atoms of O}$

Since molecular oxygen, ${\text{O}}_{2}$, is alone on the reactants' side, you can use a fractional coefficient as a tool to balance the oxygen atoms.

More specifically, multiply the oxygen molecule by $\frac{5}{2}$ to get $5$ atoms of oxygen on the reactants' side

${\text{C"_ 2"H"_ (2(g)) + 5/2"O"_ (2(g)) -> color(red)(2)"CO"_ (2(g)) + "H"_ 2"O}}_{\left(l\right)}$

To get rid of the fractional coefficient, multiply all species by $2$

$2 {\text{C"_ 2"H"_ (2(g)) + (2 xx 5/2)"O"_ (2(g)) -> (2 xx color(red)(2))"CO"_ (2(g)) + 2"H"_ 2"O}}_{\left(l\right)}$

The balanced chemical equation will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{2 {\text{C"_ 2"H"_ (2(g)) + 5"O"_ (2(g)) -> 4"CO"_ (2(g)) + 2"H"_ 2"O}}_{\left(l\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$