a) Balance the reaction.Find oxidant and reducer: Na_(2)SO_(3)+Cl_(2)+H_(2)Oarrow HCl+Nl_(2)SO_(4)

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 />$Na_{2}SO_{3} + Cl_{2} + H_{2}O \rightarrow HCl + Na_{2}SO_{4}$<br /><br />Now, let's balance the equation step by step:<br /><br />1. Balance the sodium (Na) atoms: There are 2 sodium atoms on the left side, so we need 2 sodium atoms on the right side. We already have 2 sodium atoms in $Na_{2}SO_{4}$, so we don't need to change anything.<br /><br />2. Balance the chlorine (Cl) atoms: There is 1 chlorine atom on the left side, but there are 2 chlorine atoms on the right side. To balance the chlorine atoms, we need to put a coefficient of 2 in front of HCl:<br /><br />$Na_{2}SO_{3} + Cl_{2} + H_{2}O \rightarrow 2HCl + Na_{2}SO_{4}$<br /><br />3. Balance the hydrogen (H) atoms: There are 2 hydrogen atoms on the left side, and there are 2 hydrogen atoms on the right side (in 2HCl). So, the hydrogen atoms are already balanced.<br /><br />4. Balance the oxygen (O) atoms: There are 3 oxygen atoms on the left side, and there are 4 oxygen atoms on the right side. To balance the oxygen atoms, we need to put a coefficient of 2 in front of $H_{2}O$:<br /><br />$Na_{2}SO_{3} + Cl_{2} + 2H_{2}O \rightarrow 2HCl + Na_{2}SO_{4}$<br /><br />Now, the equation is balanced:<br /><br />$Na_{2}SO_{3} + Cl_{2} + 2H_{2}O \rightarrow 2HCl + Na_{2}SO_{4}$<br /><br />Next, let's identify the oxidizing and reducing agents:<br /><br />Oxidizing agent: The oxidizing agent is the substance that gains electrons during the reaction. In this case, $Cl_{2}$ is reduced to $HCl$, so $Cl_{2}$ is the oxidizing agent.<br /><br />Reducing agent: The reducing agent is the substance that loses electrons during the reaction. In this case, $Na_{2}SO_{3}$ is oxidized to $Na_{2}SO_{4}$, so $Na_{2}SO_{3}$ is the reducing agent.