Question

Balanced complete ionic and net ionic equations

Answer 1

1) The balanced chemical equation will be : 
    2HClO4 (aq) + Ca(OH)2(aq) → Ca(ClO4)2 (aq) + 2H2O (l) 

 Ionic equation: 
 2H+(aq) + 2ClO4-(aq) +Ca2+(aq) + 2OH-(aq) → Ca2+(aq) + 2ClO4-(aq)  +2H2O(l) 

 Net ionic equation: 
 2H+ (aq) + 2OH-(aq) → 2H2O(l) 

 2) Molecular equation: 
 H2SO4(aq) + Li2SO3(aq) → Li2SO4(aq) + SO2(g) + H2O(l) 

 Ionic equation: 
 2H+ (aq) + SO4 2-(aq) + 2Li+ (aq) + SO3 2-(aq) → 2Li+(aq) + SO4 2-(aq) +  SO2(g) + H2O(l) 

 Net ionic equation: 
 2H+(aq) + SO3 2-(aq) → SO2(g) + H2O(l)

Answer 2

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,

$2H^+(aq)+SO_3^{2-}(aq)\rightarrow H_2O(l)+SO_2(g)$