# Question #d444d

May 4, 2016

$4 \text{K" + "O"_2 -> 2"K"_2"O}$

#### Explanation:

The idea here is that you need to make sure that all the atoms that are present on the reactants' side are also present on the products' side.

${\text{K"_ ((s)) + "O"_ (color(blue)(2)(g)) -> "K"_ color(red)(2)"O}}_{\left(s\right)}$

In this case, you have two reactants, potassium metal, $\text{K}$, and oxygen gas, ${\text{O}}_{2}$. Notice that the reactants' side contains

• one atom of potassium, $1 \times \text{K}$
• two atoms of oxygen, $\textcolor{b l u e}{2} \times \text{O}$

On the other hand, the products' side, which features potassium oxide, $\text{K"_2"O}$, contains

• two atoms of potassium, $\textcolor{red}{2} \times \text{K}$
• one atom of oxygen, $1 \times \text{O}$

Now, notice that potassium is present on its own on the reactants' side, which means that you can balance it last.

So, to balance the oxygen atoms, you need to have twice as many atoms of oxygen present on the products' side. Multiply the potassium oxide by $2$ to get

${\text{K"_ ((s)) + "O"_ (color(blue)(2)(g)) -> 2"K"_ color(red)(2)"O}}_{\left(s\right)}$

The reactants' side is unchanged, but the products' side now contains

• four atoms of potassium, $\left(2 \times \textcolor{red}{2}\right) \times \text{K}$
• two atoms of oxygen, $2 \times \text{O}$

All you have to do now is make sure that you have enough atoms of potassium on the products' side. To do that, multiply the potassium atom by $4$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4 {\text{K"_ ((s)) + "O"_ (color(blue)(2)(g)) -> 2"K"_ color(red)(2)"O}}_{\left(s\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now the reactants' side and the products' side contain the same number of atoms of each element that takes part in the reaction.