# How do you balance MgO + Fe -> Fe_2O_3 + Mg?

May 24, 2018

$3 M g O + 2 F e = F {e}_{2} {O}_{3} + 3 M g$

#### Explanation:

There must be 2 Fe's in the reactants to supply the 2 Fe's needed to form the Iron II Oxide.
There must be 3 MgO 's in the reactants to supply the 3 O's needed to form the Iron II Oxide.
If there are 3 MgO's then there are three Mg's in the products.

May 24, 2018

$3 M g O + 2 F e \rightarrow F {e}_{2} {O}_{3} + M g$

#### Explanation:

color(blue)(MgO+ Fe rarrFe_2O_3+Mg

Balancing the equation means, making the number of molecules of all the different elements in the equation equal on both sides.

First, lets check $O$ (Oxygen). It has $1$ atom in the left and $3$ atoms in the right. So, multiply the left hand $M g O$ by $3$

$\rightarrow 3 M g O + F e \rightarrow F {e}_{2} {O}_{3} + M g$

Now, $M g$ has $3$ atoms in the left and $1$ atom on the right.
So, multiply the right side $M g$ by $3$

$\rightarrow 3 M g O + F e \rightarrow F {e}_{2} {O}_{3} + 3 M g$

Now, $F e$ has $1$ atom on the left and $2$ atoms in the right, So, multiply the $F e$ in the left hand side by $2$

$\rightarrow 3 M g O + 2 F e \rightarrow F {e}_{2} {O}_{3} + 3 M g$

Now, all the elements in the equation have equal atoms in both sides. So it is now a balanced equation

color(purple)(3MgO+ 2Fe rarrFe_2O_3+Mg

Hope that helps! ☺

May 26, 2018

We could split this equation into individual redox reactions....

#### Explanation:

$\text{Magnesium oxide}$ is REDUCED to $\text{magnesium metal:}$

$\stackrel{I {I}^{+}}{M} g O + 2 {H}^{+} + 2 {e}^{-} \rightarrow \stackrel{0}{M} g \left(s\right) + {H}_{2} O \left(l\right)$ $\left(i\right)$

$\text{Iron metal}$ is OXIDIZED to $\text{ferric oxide...:}$

${\stackrel{0}{\text{Fe" + 3H_2O(l)rarr stackrel(III^+)"Fe"_2"O}}}_{3} \left(s\right) + 6 {H}^{+} + 6 {e}^{-}$ $\left(i i\right)$...

...and so we take $3 \times \left(i\right) + \left(i i\right)$ to get:

$3 M g O + 6 {H}^{+} + 6 {e}^{-} + F e + 3 {H}_{2} O \left(l\right) \rightarrow 3 M g \left(s\right) + {\text{Fe"_2"O}}_{3} \left(s\right) + 6 {H}^{+} + 6 {e}^{-} + 3 {H}_{2} O \left(l\right)$

….and upon cancellation...

${\text{Fe+ 3MgO" rarr "3Mg(s) +" "Fe"_2"O}}_{3} \left(s\right)$

Charge and mass are balanced as required....