# How would you balance the reaction of white phosphorous (P_4) with oxygen (O_2) which produces phosphorous oxide (P_4O_10)?

Jul 26, 2016

${\text{P"_ (4(s)) + 5"O"_ (2(g)) -> "P"_ 4"O}}_{10 \left(s\right)}$

#### Explanation:

The problem provides you with the reactants, which are said to be

• white phosphorus, ${\text{P}}_{4 \left(s\right)}$
• oxygen gas, ${\text{O}}_{2 \left(g\right)}$

and with the product, which is said to be

• diphosphorus pentoxide, also called phosphorus(V) oxide, ${\text{P"_ 4"O}}_{10 \left(s\right)}$

which means that you have all the information you need to write the unbalanced chemical equation

${\text{P"_ (4(s)) + "O"_ (2(g)) -> "P"_ 4"O}}_{10 \left(s\right)}$

Now, your goal here will be to make sure that all the atoms that are present on the reactants' side are also present on the products' side of the reaction.

The reactants' side contains

• four atoms of phosphorus, $4 \times \text{P}$
• two atoms of oxygen, $2 \times \text{O}$

The products' side contains

• four atoms of phosphorus, $4 \times \text{P}$
• ten atoms of oxygen, $10 \times \text{O}$

Notice that the atoms of phosphorus are already balanced, since you have four on each side of the chemical equation. The atoms of oxygen, on the other hand, are not balanced.

To balance the atoms of oxygen, multiply the oxygen molecule by $\textcolor{b l u e}{5}$. This will ensure that you get

$2 \times \textcolor{b l u e}{5} = \text{10 atoms of O}$

on the reactants' side of the equation. You will thus have

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{P"_ (4(s)) + color(blue)(5)"O"_ (2(g)) -> "P"_ 4"O}}_{10 \left(s\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The chemical equation is now balanced, since you have equal numbers of atoms of each element on both sides of the equation.