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

(a) 2N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 3N$_2$ + 4H$_2$O (b) 2F$_2$ + 2H$_2$O ➔ 4HF + O$_2$ (c) Na$_2$O + H$_2$O ➔ 2NaOH
(a) N$_2$H$_{4}$ + N$_2$O$_4$ ➔ N$_2$ + H$_2$O We first look at the nitrogen atoms. We have 4 on the left and 2 on the right. We add a 2 in front of the N$_2$ to balance the nitrogen. N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 2N$_2$ + H$_2$O We now look at the hydrogen atoms. We have 4 on the left and 2 on the right. We place a 2 in front of the H$_2$O on the left side to balance the hydrogen. N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 2N$_2$ + 2H$_2$O Finally, we look at the oxygen atoms. We have 4 on the left and 2 on the right. If we replace the 2 in front of the H$_2$O with a 4. The oxygen atoms are balanced. N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 2N$_2$ + 4H$_2$O The hydrogen are now out of balance. We have 4 on the left and 8 on the right. Let's place a 2 in front of N$_2$H$_{4}$. 2N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 2N$_2$ + 4H$_2$O The nitrogen are now out of balance. We have 6 on the left and 4 on the right. If we change the 2 to a 3 in front of 2N$_2$, then the nitrogen will be balanced. 2N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 3N$_2$ + 4H$_2$O Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: nitrogen: 6/6 hydrogen: 8/8 oxygen: 4/4 (b) F$_2$ + H$_2$O ➔ HF + O$_2$ We look first at the hydrogen atoms. We have 2 on the left and 1 on the right. We add 2 as a coefficient in front of the HF on the right to balance the hydrogen atoms. F$_2$ + H$_2$O ➔ 2HF + O$_2$ We now look at the oxygen atoms. We have 1 on the left and 2 on the right. If we add a 2 in front of the H$_2$O on the left, then we balance the oxygen atoms. F$_2$ + 2H$_2$O ➔ 2HF + O$_2$ However, now we have an imbalance of hydrogen atoms: 4 on the left and 2 on the right. Let's change the coefficient in front of HF from a 2 to a 4. F$_2$ + 2H$_2$O ➔ 4HF + O$_2$ Finally, we look at the fluorine atoms. We have 2 on the left and 4 on the right. Let's add a 2 in front of F$_2$ to balance the fluorine atoms. 2F$_2$ + 2H$_2$O ➔ 4HF + O$_2$ Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: fluorine: 4/4 hydrogen: 4/4 oxygen: 2/2 (c) Na$_2$O + H$_2$O ➔ NaOH We first look at the sodium atoms. We have 2 on the left and 1 on the right. If we add a 2 in front of NaOH, we will be balanced with respect to sodium. Na$_2$O + H$_2$O ➔ 2NaOH We now look at the hydrogen atoms. We have 2 on the left and 2 on the right. We are balanced in terms of hydrogen. Next, we look at the oxygen atoms. We have 2 on the left and 2 on the right. We are balanced in terms of oxygen. Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: sodium: 2/2 hydrogen: 2/2 oxygen: 2/2