# Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Exercises - Page 183: 65

a. $2HClO_4(aq) + Mg(OH)_2(s) -- \gt Mg(ClO_4)_2(aq) +2 H_2O(l)$ $_{2H^+(aq) + 2ClO_4^-(aq) + Mg(OH)_2(s) -- \gt Mg^{2+}(aq) + 2ClO_4^-(aq) +2 H_2O(l)}$ $2H^+(aq) + Mg(OH)_2(s) -- \gt Mg^{2+}(aq) + 2H_2O(l)$ b. $HCN(aq) + NaOH(aq) -- \gt NaCN(aq) + H_2O(l)$ $_{H^+(aq) + CN^-(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + CN^-(aq) + H_2O(l)}$ $H^+(aq) + OH^-(s) -- \gt H_2O(l)$ c. $HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$ $_{H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + Cl^-(aq) + H_2O(l)}$ $H^+(aq) + OH^-(aq) -- \gt H_2O(l)$

#### Work Step by Step

a. 1. Write an double replacement reaction between the ionic compounds: $HClO_4(aq) + Mg(OH)_2(s) -- \gt Mg(ClO_4)_2(aq) + H_2O(l)$ - Balance it: $2HClO_4(aq) + Mg(OH)_2(s) -- \gt Mg(ClO_4)_2(aq) +2 H_2O(l)$ 2. Separate the aqueous compounds into ions: $2H^+(aq) + 2ClO_4^-(aq) + Mg(OH)_2(s) -- \gt Mg^{2+}(aq) + 2ClO_4^-(aq) +2 H_2O(l)$ - This is the complete ionic equation. 3. Remove the spectator ions: (Those that don't change after the reaction) $2H^+(aq) + Mg(OH)_2(s) -- \gt Mg^{2+}(aq) + 2H_2O(l)$ -------------- b. 1. Write an double replacement reaction between the ionic compounds: $HCN(aq) + NaOH(aq) -- \gt NaCN(aq) + H_2O(l)$ - It is already balanced 2. Separate the aqueous compounds into ions: $H^+(aq) + CN^-(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + CN^-(aq) + H_2O(l)$ - This is the complete ionic equation. 3. Remove the spectator ions: (Those that don't change after the reaction) $H^+(aq) + OH^-(s) -- \gt H_2O(l)$ ---------------------------- c. 1. Write an double replacement reaction between the ionic compounds: $HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$ - It is already balanced 2. Separate the aqueous compounds into ions: $H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + Cl^-(aq) + H_2O(l)$ - This is the complete ionic equation. 3. Remove the spectator ions: (Those that don't change after the reaction) $H^+(aq) + OH^-(aq) -- \gt H_2O(l)$

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