# Question #a7f87

Jun 14, 2017

The coefficients in front of the compounds are 1, 3, 2, and 3, respectively so that $F {e}_{2} {O}_{3} \left(s\right) + 3 C O \left(g\right) \to 2 F e \left(s\right) + 3 C {O}_{2} \left(g\right)$.

#### Explanation:

When you balance an equation, you want an equal number of atoms on both sides. Let's begin with $F e$, or iron.

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

There are 2 atoms of $F e$ on the left but only 1 on the right. Add a coefficient of 2 to the $F e$ atom (or multiply $F e$ by 2) on the right to get 2 $F e$ atoms on the right side of the equation:

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

Now let's focus on $O$ atoms since they are next in the equation after $F e$. There are 4 atoms of O on the left side of the equation, while only 2 on the right side, so add the coefficient 2 to the $O$ on the right side (Note: this changes the number of C atoms as well):

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

We need to fix the number of $C$ atoms on the left now, so add 2 to the $C O$ to make the number of Carbons equal (This changes the number of O atoms too):

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

There are now 5 $O$ atoms on the left with 4 on the right, and we can't add a fraction as a coefficient. This means we have to change the numbers so that the numbers of $C$ atoms are equal on both sides of the equation. Let's change the coefficients of the compounds with $C$ to 3:

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

There we go! All of the atoms on both sides are now balanced.