The net ionic equation for the reaction for the reaction between AgNO3(aq) and Na2SO4(aq) is
2Ag ^+(aq) + SO4^2-(aq) → Ag2SO4 (s)
Explanation
Step 1: write a balanced molecular equation between AgNO3 and Na2SO4
2AgNO3(aq) + Na2SO4 (aq) → Ag2So4(s) +2NaNO3 (aq)
Step 2: Break all the soluble electrolytes into their ions
Ag^+(aq) +NO3^-(aq) + 2Na^+(aq) +SO4^2- (aq)→
Ag2SO4(s) + 2Na^+(aq) +NO3^-(aq)
Step 3: cancel the spectator ions( ions that exist in the same form on both reactant and product side of equation above)
= 2 Na^+ and NO^-
Step 4: write down the net equation
Ag^+(aq) +SO4^2- (aq)→ Ag2SO4 (s)
The net ionic equation is [tex]\rm 2\;Ag^+(aq)\;+\;2\;SO_4^2^-(aq)\;\rightarrow\;Ag_2SO_4\;(s)[/tex]
The ionic species will be:
[tex]\rm AgNO_3\;(aq)\;\rightarrow\;Ag^+(aq)\;+\;NO_3^-(aq)[/tex]
[tex]\rm Na_2SO_4\;(aq)\;\rightarrow\;Na^+\;(aq)\;+\;SO_4^2^-\;(aq)[/tex]
[tex]\rm Ag_2SO_4^2^-\;\rightarrow\;Ag^+\;(aq)+\;SO_4^2^-\;(aq)[/tex]
The net ionic equation will be the reaction of products forming reactants.
The reaction can be :
[tex]\rm Ag^+(aq)\;+\;NO_3^-(aq)\;+\;Na^+(aq)\;+SO_4^2^-(aq)\;\rightarrow\;Ag_2SO_4^2^-(s)\;+\;Na^+(aq)\;+\;NO_3^-(aq)[/tex]
Eliminating the same species on both the side. The net ionic equation will be:
[tex]\rm 2\;Ag^+(aq)\;+\;2\;SO_4^2^-(aq)\;\rightarrow\;Ag_2SO_4\;(s)[/tex].
For more information about the ionic equation, refer to the link:
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