# Question #f633a

Jun 12, 2016

$F {e}_{2} {O}_{3} + 3 C O \to 2 F e + 3 C {O}_{2}$

#### Explanation:

Forget what you have written above and start again. You made several mistakes in the second line of your equation, making it impossible to solve.

Most importantly: the molecules in the equation should not be taken apart as you did! So keep them clustered as in the equation you start with:

$\ldots F {e}_{2} {O}_{3} + \ldots C O \to \ldots F e + \ldots C {O}_{2}$

From here it is a combination of common sense and trial and error:

Step 1 balance out the iron, leave the $F {e}_{2} {O}_{3}$ as it is, because you have enough room to play around with the $C$ and $O$ on both sides:

$1 F {e}_{2} {O}_{3} + \ldots C O \to 2 F e + \ldots C {O}_{2}$

Step2 then try to get the number of $O$ on the left side to an even number. You have $3 \cdot O$ on the left, need $3$ more to make it even:

$1 F {e}_{2} {O}_{3} + 3 C O \to 2 F e + \ldots C {O}_{2}$

Step3 now make the number of $C$ equal on both side:

$1 F {e}_{2} {O}_{3} + 3 C O \to 2 F e + 3 C {O}_{2}$

Finally, check the equation and you'll see it is balanced now!