How do you balance #MgO + Fe -> Fe_2O_3 + Mg#?
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.
#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
So, multiply the right side
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! ☺
We could split this equation into individual redox reactions....
...and so we take
….and upon cancellation...
Charge and mass are balanced as required....