## Chemistry (7th Edition)

(a) $Mg + 2HNO_3 --\gt H_2 + Mg(NO_3)_2$ (b) $CaC_2 + 2 H_2O -- \gt Ca(OH)_2 + C_2H_2$ (c) $2S + 3O_2 --\gt 2 SO_3$ (d) $UO_2 + 4HF -- \gt UF_4 + 2H_2O$
1. Identify and try to balance the elements that appear on exactly 1 compound in each side. (a) Mg is already balanced. But nitrogen is not: There are 2 nitrogens on the right side but only one on the left. Therefore, we should put a 2 as the coefficient of the compound with the nitrogen on the left side. $Mg + 2HNO_3 --\gt H_2 + Mg(NO_3)_2$ (b) Calcium is already balanced, but oxygen is not. There are 2 oxygens on the right side but only one on the left. Therefore, we should put a 2 as the coefficient of the compound with the oxygen on the left side. $CaC_2 + 2 H_2O -- \gt Ca(OH)_2 + C_2H_2$ (c) S is already balanced. But oxygen is not: There are 3 nitrogens on the right side, but only 2 on the left. Therefore, we should put a 3 as the coefficient of the compound with the oxygen on the left side and a 2 on the right one. $S + 3O_2 --\gt 2 SO_3$ (d) U is already balanced. But fluorine is not: There are 4 fluorines on the right side, but only 1 on the left. Therefore, we should put a 4 as the coefficient of the compound with the oxygen on the left side. $UO_2 + 4HF -- \gt UF_4 + H_2O$ 2. For the ones that are not balanced yet, try to balance them. (a) Balanced. (b) Balanced. (c) The S is not balanced, but if we put a 2 on the sulfur on the left side, the equation will be balanced. $2S + 3O_2 --\gt 2SO_3$ (d) The oxygen and the hydrogen aren't balanced yet. But if we put a 2 as the coefficient for $H_2O$, the equation will be totally balanced. $UO_2 + 4HF -- \gt UF_4 + 2H_2O$