# Question #12721

Feb 19, 2015

$F {e}_{\left(s\right)} + {O}_{2 \left(g\right)} \to F {e}_{2} {O}_{3 \left(s\right)}$

The idea behind balancing chemical equations is that the number of atoms an element has on the reactants' side must be equal to the number of atoms it has on the products' side.

These atoms will become a part of different compounds once the reaction is completed, but they must always be in equal numbers on both sides.

So, look at iron first. One atom reacts, but two are produced - notice the 2 subscript iron has in $F {e}_{2} {O}_{3}$. This means you must double the number of atoms on the reactants' side to reach an equality.

$2 F {e}_{\left(s\right)} + {O}_{2 \left(g\right)} \to F {e}_{2} {O}_{3 \left(s\right)}$

Now look at oxygen. Two atoms react, but three are produced. The trick here is to find a common multiple that will make the number of atoms equal on both sides.

The easiest way to do this is to multiply the atoms that react by 3, which will give you 6 oxygen atoms that react, and the atoms that are produced by 2 - this will get you 6 oxygen atoms produced.

$2 F {e}_{\left(s\right)} + 3 {O}_{2 \left(g\right)} \to 2 F {e}_{2} {O}_{3 \left(s\right)}$

However, notice that the iron atoms are unbalanced again. You have 2 that react, but 4 that are produced $\to$ multiply the atoms that react by 2 again, which will give you

$4 F {e}_{\left(s\right)} + 3 {O}_{2 \left(g\right)} \to 2 F {e}_{2} {O}_{3 \left(s\right)}$

Now the chemical equation is balanced. See more on how to balance chemical equations here:

http://socratic.org/questions/how-do-you-balance-chemical-equations-step-by-step