The balanced chemical equation for the reaction of Magnesium Nitride ##Mg_3N_2## with water ##H_2O## to produce Magnesium Hydroxide ##Mg(OH)_2## and Ammonia gas ##NH_3## is

##Mg_3N_2 + 6H_2O => 3Mg(OH)_2 + 2NH_3##

If we start with this equation: ##Mg_3N_2 + 2H_2O => Mg(OH)_2 + NH_3##

Initially we can tell that the amount of Magnesium (Mg), Nitrogen (N), Oxygen (O), and Hydrogen (H) is not even on both sides.

It is easiest to start by balancing an element present in only one reactant and one product.

Since the amount of Magnesium must be equal and there is only one reactant with Magnesium (Magnesium Nitride ##Mg_3N_2##) and one product with Magnesium (Magnesium Hydroxide ##Mg(OH)_2##), we know there will always be 3 ##Mg(OH)_2## for every ##Mg_3N_2##.

##Mg_3N_2 + 2H_2O => 3Mg(OH)_2 + NH_3##

At this point the ##Mg## is balanced but Oxygen is still unbalanced.

There are 2 atoms of ##O## on the reactant side and 6 on the product side.

Again, since only one reactant has ##O## and one product has ##O##, we know that for every ##Mg(OH)_2## there must be 2 ##H_2O##.

Since we have ##3Mg(OH)_2## there must be ##6H_2O##.

##Mg_3N_2 + 6H_2O => 3Mg(OH)_2 + NH_3##

##Mg## and ##O## are now balanced.

Nitrogen is still unbalanced and is also only in one term on each side.

There are 2 atoms of ##N## on the reactant side and 1 atom of ##N## on the product side.

By doubling the amount of Ammonia (##NH_3##) on the product side this can be brought into balance.

##Mg_3N_2 + 6H_2O => 3Mg(OH)_2 + 2NH_3##

And this also balances the Hydrogen on both sides, so we are done.