Answer :
Answer:
The net ionic equation for the given reaction :
[tex]3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)[/tex]
Explanation:
[tex]3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)[/tex]...[1]
[tex]MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)[/tex]..[2]
[tex]Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)[/tex]...[3]
[tex]NaI(aq)\rightarrow Na^+(aq)+I^-(aq)[/tex]
Replacing [tex]MnI_2(aq)[/tex] , NaI and [tex]Na_3PO_4(aq)[/tex] in [1] by usig [2] [3] and [4]
[tex]3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)[/tex]
Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:
[tex]3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)[/tex]