# How do you balance the chemical equation: Fe+H_2O = Fe_3O_4+H_2?

Apr 28, 2016

$3 F e + 4 {H}_{2} O = F {e}_{3} {O}_{4} + 4 {H}_{2}$

#### Explanation:

The $F {e}_{3}$ in $F {e}_{3} {O}_{4}$ on the right implies that there must be some multiple of $3$ $F e$ on the left side.

Similarly the ${O}_{4}$ on the right side implies that there must be some multiple of $4$ ${H}_{2} O$ on the left side.

$\left(3 m\right) F e + \left(4 n\right) {H}_{2} O = \left(p\right) F {e}_{3} {O}_{4} + \left(q\right) {H}_{2}$

Since
$\textcolor{w h i t e}{\text{XXX}} \left(3 m\right) F e \rightarrow \left(m\right) F {e}_{3} {O}_{4}$ (there's no place else for the $F e$ to go)
and
$\textcolor{w h i t e}{\text{XXX}} \left(4 n\right) {H}_{2} O \rightarrow \left(n\right) F {e}_{3} {O}_{4}$ (there's no place else for the $O$ to go)
$\Rightarrow m = n$

Further
$\textcolor{w h i t e}{\text{XXX}} \left(4 n\right) {H}_{2} O \rightarrow \left(q\right) {H}_{2}$
$\Rightarrow q = 4 n$

Therefore we have
$\textcolor{w h i t e}{\text{XXX}} \left(3 n\right) F e + \left(4 n\right) {H}_{2} O = \left(n\right) F {e}_{3} {O}_{4} + \left(4 n\right) {H}_{2}$

Using the simplest version: $n = 1$ gives
$\textcolor{w h i t e}{\text{XXX}} 3 F e + 4 {H}_{2} O = F {e}_{3} {O}_{4} + 4 {H}_{2}$