Answer : The net ionic equation will be,
[tex]3Co^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)[/tex]
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.
The given balanced ionic equation will be,
[tex]3Co^{2+}(aq)+6NO_3^-(aq)+6Na^+(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)+6Na^+(aq)+6NO_3^-(aq)[/tex]
In this equation, [tex]NO_3^-(aq)\text{ and }Na^+(aq)[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
Thus, the net ionic equation will be,
[tex]3Co^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)[/tex]
The net ionic equation for this reaction is given as
[tex]3Co^{2+}(aq)+2PO_4^{3-}(aq)→Co_3(PO_4)_2(s)[/tex]
Spectator ions
These are ions that exists on both side of a reaction i.e present on reactant and product side, and the do not contribute in any fashion to the reactions.
Generally the equation for the balanced ionic equation is mathematically given as
3Co2 (aq) 6NO3(aq) 6Na (aq) 2PO43–(aq) → Co3(PO4)2(s) 6Na (aq) 6NO3(aq)
Therefore, the removal of the spectator ions will give us the net ionic equation for this reaction.
3Co^{2+}(aq)+2PO_4^{3-}(aq)→Co_3(PO_4)_2(s)
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