(ii) `Ag_(s) +HCI_(I) to AgCI darr + H_(2) uarr `

Explanation

Step 1 : Rewrite the given equation as it is
` Ag_(s) + HCI_(I) to AgCI darr + H_(2)`
Step 2 : Write the number of atoms of each element in the unbalanced equation on both sides of equations.

The number of silver and chlorine atoms on both sides of the equation are same, therefore, equalise the number of hydrogen atoms.
Step 3: To balance the number of hydrogen atoms .

To equalise the number of hydrogen atoms, we use 2 as the factor in the product HCI, now the unbalanced equation become
` Ag + 2HCI to AgCI + H_(2)`
Step 4 : To balance the number of chlorine atoms :

To equalise the number of chlorine atoms, we use 2 as the factor in the product , AgCI , now the unbalanced equation becomes
` Ag + 2HCI to 2AgCI + H_(2)`
Now count the atoms of each element on both sides of the equation, there are less number of silver atoms in the reactant. Now equalise the silver atoms, the balanced equation becomes.
`2Ag + 2HCI to 2AgCI + H_(2)`
Now indicate the physcial states of the reactants and products
` 2Ag_((s)) + 2HCI_(4(aq)) to 2AgCI darr + H_(2) uarr `