Write the balanced equations for the following reactions :
(vii) `Ba(OH)_(2)+HBrrarrBaBr_(2)+H_(2)O`

Explanation

Step 1 : Rewrite the given equation as it is
`Ba(OH)_(2)+HBrrarrBaBr_(2)+H_(2)O`
Step 2 : Write the number of atoms of each element in the unbalanced equation on both sides of the equation.

Step 3 : To balance the number of oxygen atoms :

To equalise the number of oxygen atoms, we use 2 as the cofficient of `H_(2)O` in theproduct.
Now, the partly balanced equation becomes as follows :
`Ba(OH)_(2)+HBrrarrBaBr_(2)+2H_(2)O`
Step 4 : Now , balance the number of hydrogen atoms : In the partly balanced equation :

To equalise the number of hydrogen atoms , we use 2 as the cofficient of HBr in the reactants . Now, the equation becomes
`Ba(OH)_(2)+2HBrrarrBaBr_(2)+2H_(2)O`
Now , count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal. Hence , the balanced equation is
`Ba(OH)_(2)+2HBrrarrBaBr_(2)+2H_(2)O`
Now indicate the physical states of the reactants and products.
`Ba(OH)_(2_((aq)))+2HBr_((aq))rarrBaBr_(2_((aq)))+2H_(2)O_((1))`