## Chemistry: Principles and Practice (3rd Edition)

(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$_4$)$_2$
(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