8. When the following Double replacement reaction is balanced, what is the coefficient that would be in front of Potassium bromide? AlBr_(3)+K_(2)SO_(4)-KBr+Al_(2)(SO_(4))_(3) 4 3 2

To balance the double replacement reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation.<br /><br />The unbalanced equation is:<br />$AlBr_{3}+K_{2}SO_{4}-KBr+Al_{2}(SO_{4})_{3}$<br /><br />To balance the equation, we need to find the coefficient for Potassium bromide (KBr).<br /><br />Let's start by balancing the aluminum (Al) atoms. There is 1 Al atom on the left side and 2 Al atoms on the right side. To balance the Al atoms, we need to put a coefficient of 2 in front of $AlBr_{3}$.<br /><br />$2AlBr_{3}+K_{2}SO_{4}-KBr+Al_{2}(SO_{4})_{3}$<br /><br />Next, let's balance the potassium (K) atoms. There are 2 K atoms on the left side and 1 K atom on the right side. To balance the K atoms, we need to put a coefficient of 2 in front of KBr.<br /><br />$2AlBr_{3}+K_{2}SO_{4}-2KBr+Al_{2}(SO_{4})_{3}$<br /><br />Finally, let's balance the bromine (Br) atoms. There are 6 Br atoms on the left side and 2 Br atoms on the right side. To balance the Br atoms, we need to put a coefficient of 6 in front of KBr.<br /><br />$2AlBr_{3}+K_{2}SO_{4}-6KBr+Al_{2}(SO_{4})_{3}$<br /><br />Now, let's check the balance of the oxygen (O) atoms. There are 4 O atoms on the left side and 12 O atoms on the right side. The oxygen atoms are already balanced.<br /><br />The balanced equation is:<br />$2AlBr_{3}+K_{2}SO_{4}-6KBr+Al_{2}(SO_{4})_{3}$<br /><br />Therefore, the coefficient that would be in front of Potassium bromide (KBr) is 6.