Answer
(a) Mg$_3$N$_{2}$ + 6H$_2$O ➔ 2NH$_3$ + 3Mg(OH)$_2$
(b) 4Fe + 3O$_2$ ➔ 2Fe$_2$O$_3$
(c) 3Zn + 2H$_3$PO$_4$ ➔ 3H$_2$ + Zn$_3$(PO$_4$)$_2$
Work Step by Step
(a) Mg$_3$N$_{2}$ + H$_2$O ➔ NH$_3$ + Mg(OH)$_2$
We look first at the magnesium atoms. We have 3 on the left and 1 on the right. We add 3 as a coefficient in front of the Mg(OH)$_2$ on the right to balance the magnesium atoms.
Mg$_3$N$_{2}$ + H$_2$O ➔ NH$_3$ + 3Mg(OH)$_2$
Next, we look at the nitrogen atoms. We have 2 on the left and 1 on the right. We add a 2 in front of the NH$_3$ to balance the nitrogen.
Mg$_3$N$_{2}$ + H$_2$O ➔ 2NH$_3$ + 3Mg(OH)$_2$
We now look at the hydrogen atoms. We have 2 on the left and 12 on the right. We place a 6 in front of the H$_2$O on the left side to balance the hydrogen.
Mg$_3$N$_{2}$ + 6H$_2$O ➔ 2NH$_3$ + 3Mg(OH)$_2$
Finally, we look at the oxygen atoms. We have 6 on the left and 6 on the right. The oxygen atoms are balanced.
Now let's check the numbers of atoms in the reactants and the products.
Reactants/Products:
magnesium: 3/3
nitrogen: 2/2
hydrogen: 12/12
oxygen: 6/6
(b) Fe + O$_2$ ➔ Fe$_2$O$_3$
We look first at the iron atoms. We have 1 on the left and 2 on the right. We add 2 as a coefficient in front of the Fe on the left to balance the iron atoms.
2Fe + O$_2$ ➔ Fe$_2$O$_3$
We now look at the oxygen atoms. We have 2 on the left and 3 on the right. If we add a 3 in front of the O$_2$ on the left and and add a 2 in front of Fe$_2$O$_3$ on the right, then we balance the oxygen atoms.
2Fe + 3O$_2$ ➔ 2Fe$_2$O$_3$
However, now we have an imbalance of iron atoms: 2 on the left and 4 on the right. Let's change the coefficient in front of Fe from a 2 to a 4.
4Fe + 3O$_2$ ➔ 2Fe$_2$O$_3$
Now let's check the numbers of atoms in the reactants and the products.
Reactants/Products:
iron: 4/4
oxygen: 6/6
(c) Zn + H$_3$PO$_4$ ➔ H$_2$ + Zn$_3$(PO$_4$)$_2$
We first look at the zinc atoms. We have 1 on the left and 3 on the right. If we add a 3 in front of Zn, we will be balanced with respect to zinc.
3Zn + H$_3$PO$_4$ ➔ H$_2$ + Zn$_3$(PO$_4$)$_2$
We now look at the hydrogen atoms. We have 3 on the left and 2 on the right. We need to add a 2 in front of H$_3$PO$_4$ and a 3 in front of H$_2$ to balance the hydrogen.
3Zn + 2H$_3$PO$_4$ ➔ 3H$_2$ + Zn$_3$(PO$_4$)$_2$
Next, we look at the phosphate ions, which we treat as a single entity. We have 1 phosphate on the left and 1 on the right, so we are balanced with respect to phosphate.
Now let's check the numbers of atoms in the reactants and the products.
Reactants/Products:
zinc: 3/3
hydrogen: 6/6
phosphate: 2/2
