Answer:
See explanation.
Explanation:
Hello!
In this case, when writing the complete molecular, ionic and net ionic equations, we must make sure we have the reactants and the most likely products; thus, when lead (II) nitrate react with sodium sulfate, it is noticed that the complete molecular equation is:
[tex]Pb(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)[/tex]
Whereas the one and only unionizable species is the lead (II) sulfate product; thus, the complete ionic equation turns out:
[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+2Na^+(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^+(aq)+2(NO_3)^-(aq)[/tex]
Whereas it is noticed that both sodium and nitrate ions are the spectator ions as they are present at both reactants and products; therefore, the net ionic equation turns out:
[tex]Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)[/tex]
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