Answer :
(1) The net ionic equation will be,
\(H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)\)
(2) The net ionic equation will be,
\(2H^+(aq)+SO_3^{2-}(aq)\rightarrow H_2O(l)+SO_2(g)\)
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
(1) The balanced ionic equation will be,
\(HClO_4(aq)+Ca(OH)_2(aq)\rightarrow Ca(ClO_4)_2(aq)+2H_2O(l)\)
The ionic equation in separated aqueous solution will be,
\(H^+(aq)+ClO_4^-(aq)+Ca^{2+}(aq)+2OH^-(aq)\rightarrow Ca^{2+}(aq)+2ClO_4^-(aq)+2H_2O(l)\)
In this equation, \(Ca^{2+}\text{ and }ClO_4^-\) are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
\(H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)\)
(2) The balanced ionic equation will be,
\(H_2SO_4(aq)+Li_2SO_3(aq)\rightarrow Li_2SO_4(aq)+H_2O(l)+SO_2(g)\)
The ionic equation in separated aqueous solution will be,
\(2H^+(aq)+SO_4^{2-}(aq)+2Li^+(aq)+SO_3^{2-}(aq)\rightarrow 2Li^+(aq)+SO_4^{2-}(aq)+H_2O(l)+SO_2(g)\)
In this equation, \(Li^+\text{ and }SO_4^{2-}\) are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
\(2H^+(aq)+SO_3^{2-}(aq)\rightarrow H_2O(l)+SO_2(g)\)
