a) Balance the reaction. Find oxidant and reducer: K_(2)Cr_(2)O_(7)+H_(2)S+HClleftarrows S+CrCl_(1)+KCl+H_(2)O

To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's start by writing down the unbalanced equation:<br /><br />$K_{2}Cr_{2}O_{7}+H_{2}S+HCl\rightleftarrows S+CrCl_{1}+KCl+H_{2}O$<br /><br />Now, let's balance the equation step by step:<br /><br />1. Balance the potassium (K) atoms:<br />$K_{2}Cr_{2}O_{7}+H_{2}S+HCl\rightleftarrows S+CrCl_{1}+2KCl+H_{2}O$<br /><br />2. Balance the chromium (Cr) atoms:<br />$K_{2}Cr_{2}O_{7}+H_{2}S+HCl\rightleftarrows S+2CrCl_{1}+2KCl+H_{2}O$<br /><br />3. Balance the chlorine (Cl) atoms:<br />$K_{2}Cr_{2}O_{7}+H_{2}S+6HCl\rightleftarrows S+2CrCl_{1}+2KCl+H_{2}O$<br /><br />4. Balance the hydrogen (H) atoms:<br />$K_{2}Cr_{2}O_{7}+3H_{2}S+6HCl\rightleftarrows 3S+2CrCl_{1}+2KCl+H_{2}O$<br /><br />5. Balance the sulfur (S) atoms:<br />$K_{2}Cr_{2}O_{7}+3H_{2}S+6HCl\rightleftarrows 3S+2CrCl_{1}+2KCl+H_{2}O$<br /><br />Now, let's identify the oxidizing agent and reducing agent in the reaction.<br /><br />The oxidizing agent is the substance that gains electrons during the reaction. In this case, the oxidizing agent is $K_{2}Cr_{2}O_{7}$, as it is reduced to $CrCl_{1}$.<br /><br />The reducing agent is the substance that loses electrons during the reaction. In this case, the reducing agent is $H_{2}S$, as it is oxidized to $S$.<br /><br />So, the balanced equation with the oxidizing agent and reducing agent identified is:<br /><br />$K_{2}Cr_{2}O_{7}+3H_{2}S+6HCl\rightleftarrows 3S+2CrCl_{1}+2KCl+H_{2}O$<br /><br />Oxidizing agent: $K_{2}Cr_{2}O_{7}$<br />Reducing agent: $H_{2}S$