Question

Please help!!!! I have less than 20 minutes to figure this out and I have no clue what to do.

Answer 1

See explanation.

Step-by-step 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:

`Pb(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)`

Whereas the one and only unionizable species is the lead (II) sulfate product; thus, the complete ionic equation turns out:

`Pb^(2+)(aq)+2(NO_3)^-(aq)+2Na^+(aq)+SO_4^(2-)(aq)\rightarrow PbSO_4(s)+2Na^+(aq)+2(NO_3)^-(aq)`

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:

`Pb^(2+)(aq)+SO_4^(2-)(aq)\rightarrow PbSO_4(s)`

Best regards!