Chemistry: Principles and Practice (3rd Edition)

(a) C$_5$H$_{12}$ + 8O$_2$ ➔ 5CO$_2$ + 6H$_2$O (b) 4NH$_3$ + 3O$_2$ ➔ 2N$_2$ + 6H$_2$O (c) 2KOH + H$_2$SO$_4$ ➔ K$_2$SO$_4$ + 2H$_2$O
(a) C$_5$H$_{12}$ + O$_2$ ➔ CO$_2$ + H$_2$O We look first at the carbon atoms. We have five carbon atoms on the left and one carbon on the right. We add 5 as a coefficient in front of the CO$_2$ on the right to balance the carbon atoms. C$_5$H$_{12}$ + O$_2$ ➔ 5CO$_2$ + H$_2$O We now look at the hydrogen atoms. We have 12 on the left and two on the right. We place a 6 in front of the H$_2$O on the right side, so we have 12 hydrogen atoms on each side of the equation. C$_5$H$_{12}$ + O$_2$ ➔ 5CO$_2$ + 6H$_2$O Finally, we look at the oxygen atoms. We have 2 on the left and 16 on the right. If we add an 8 in front of the O$_2$ on the left, we will balance the oxygen atoms. C$_5$H$_{12}$ + 8O$_2$ ➔ 5CO$_2$ + 6H$_2$O Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: carbon: 5/5 hydrogen: 12/12 oxygen: 16/16 (b) NH$_3$ + O$_2$ ➔ N$_2$ + H$_2$O We look first at the nitrogen atoms. We have 1 nitrogen atom on the left and 2 on the right. We add 2 as a coefficient in front of the NH$_3$ on the left to balance the nitrogen atoms. 2NH$_3$ + O$_2$ ➔ N$_2$ + H$_2$O We now look at the hydrogen atoms. We have 6 on the left and 2 on the right. We place a 3 in front of the H$_2$O on the right side, so we have 6 hydrogen atoms on each side of the equation. 2NH$_3$ + O$_2$ ➔ N$_2$ + 3H$_2$O Finally, we 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 remove the 3 in front of H$_2$O and replacing it with a 6, we balance the oxygen atoms. However, now we have an imbalance of hydrogen atoms. 2NH$_3$ + O$_2$ ➔ N$_2$ + 6H$_2$O We now have 6 hydrogen on the left and 12 on the right. If we change the coefficient of the NH$_3$ to 4, then we have balanced the hydrogen atoms. 4NH$_3$ + O$_2$ ➔ N$_2$ + 6H$_2$O But now, we have 4 nitrogen on the left and 2 on the right. We simply put a 2 in front of the N$_2$ to balance the nitrogen atoms. 4NH$_3$ + O$_2$ ➔ 2N$_2$ + 6H$_2$O Now the oxygen atoms are out of balance. We have 2 on the left and 6 on the right. We add a 3 in front of O$_2$ to balance the oxygen atoms. 4NH$_3$ + 3O$_2$ ➔ 2N$_2$ + 6H$_2$O Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: nitrogen: 4/4 hydrogen: 12/12 oxygen: 6/6 (c) KOH + H$_2$SO$_4$ ➔ K$_2$SO$_4$ + H$_2$O We look first at the potassium atoms. We have 1 potassium atom on the left and 2 on the right. We add 2 as a coefficient in front of the KOH on the left to balance the potassium atoms. 2KOH + H$_2$SO$_4$ ➔ K$_2$SO$_4$ + H$_2$O We now look at the hydrogen atoms. We have 4 on the left and 2 on the right. We need to add a 2 in front of H$_2$O to balance the hydrogen. 2KOH + H$_2$SO$_4$ ➔ K$_2$SO$_4$ + 2H$_2$O Next, we look at the sulfate ions, which we treat as a single entity. We have 1 sulfate on the left and 1 on the right, so we are balanced with respect to sulfate. 2KOH + H$_2$SO$_4$ ➔ K$_2$SO$_4$ + 2H$_2$O Finally, we look at oxygen. We have 2 oxygen on the left and 2 on the right. We are balance with respect to oxygen. Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: potassium: 2/2 hydrogen: 4/4 sulfate: 1/1 oxygen: 2/2