Question

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

Answer 1

`3 MgO + 2 Fe = Fe_2O_3 + 3 Mg`

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.

Answer 2

`3MgO+ 2Fe rarrFe_2O_3+Mg`

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 `MgO` by `3`

`rarr3MgO+ Fe rarrFe_2O_3+Mg`

Now, `Mg` has `3` atoms in the left and `1` atom on the right.
So, multiply the right side `Mg` by `3`

`rarr3MgO+ Fe rarrFe_2O_3+3Mg`

Now, `Fe` has `1` atom on the left and `2` atoms in the right, So, multiply the `Fe` in the left hand side by `2`

`rarr3MgO+ 2Fe rarrFe_2O_3+3Mg`

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! ☺

Answer 3

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

Explanation:

`"Magnesium oxide"` is REDUCED to `"magnesium metal:"`

`stackrel(II^+)MgO + 2H^+ +2e^(-) rarr stackrel(0)Mg(s) +H_2O(l)` `(i)`

`"Iron metal"` is OXIDIZED to `"ferric oxide...:"`

`stackrel(0)"Fe" + 3H_2O(l)rarr stackrel(III^+)"Fe"_2"O"_3(s)+ 6H^+ +6e^(-)` `(ii)`...

...and so we take `3xx(i)+(ii)` to get:

`3MgO + 6H^+ +6e^(-) +Fe + 3H_2O(l)rarr 3Mg(s) +"Fe"_2"O"_3(s)+ 6H^+ +6e^(-)+3H_2O(l)`

….and upon cancellation...

`"Fe+ 3MgO" rarr "3Mg(s) +" "Fe"_2"O"_3(s)`

Charge and mass are balanced as required....