How do you balance #Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2#?
There are three Ca atoms on the left-hand side (LHS), and one on the right-hand side (RHS). Place a coefficient of
Now there are twelve H atoms on the RHS
Now lets count the number of atoms of each element on each side of the equation.
Since there are the same number of atoms of each element on both sides of the equation, it is balanced.