Please explain how to write a balanced net ionic equation for this!?

Question

Pb(ClO[sbscrpt4])[sbscrpt2] + NaI ---> PbI[sbscrpt2] + NaClO[sbscrpt4]

Step by step would be greatly appreciated since then perhaps I'll understand the concept and be able to do the rest on my own. Thank you !

EDIT: What's really throwing me off is, for example, something like H[sbscrpt2].

Is the charge of H[sbscrpt2] just 1+ or 2+ since there are two H atoms?

Answer 1

To get to the net ionic equation, you need to start with the balanced complete ionic equation. For your reaction, you need to recognize that all of the substances except for PbI2 are soluble, and so in aqueous solution, they exist as independent ions.

So, you start with:
Pb2+(aq) + 2 ClO4-(aq) + 2 Na+(aq) + 2 I-(aq) -->
PbI2(s) + 2 Na+(aq) + 2 ClO4-(aq)

To get to the net ionic equation, you cancel anything that is in exactly the same form on both sides of the equation, which will leave you with:

Pb2+(aq) + 2 I-(aq) --> PbI2(s)

And you're done.

Answer 2

Pb(CIO4)2 +NaI so thats 1 Pb, 2 Cl, 8 O 1 Na and 1I
other side
PbI2 + NaClO4 so 2 Pb, 2 I, 1 Na 1 Cl and 4 O

so in order to have the 8O on the products side that we started with need to multiply that compound by 2, that now means that the Cl and balances as well but we now have an extra Na on the products side. So we then put twice as much of the NA compound into the reactants...and that also gives us the extra I.

Pb(CIO4)2 +2NaI ---------->PbI2 + 2NaClO4

but the lead still doesn't balance so

2Pb(CIO4)2 +4NaI ---------->PbI2 + 4NaClO4

Answer 3

H(sbscrt2) has charge 0. Beacause it neutral, it's a molecule, not ion.
But I don't know to solve this equation.