Using the balanced reaction, enter the formula, charge(if any), and state of the species (exclude H2O) in solution if neither reactant is present in stoichiometric excess. Separate your answers by comma.

Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)

To determine the formulas, charges (if any), and states of the species in solution, we need to break down the balanced reaction:

Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)

1. Pb(NO3)2(aq): This is lead(II) nitrate in aqueous solution. The formula is Pb(NO3)2. Since it is an ionic compound, it dissociates into ions when in solution. The lead (Pb) ion has a charge of +2, and the nitrate (NO3) ion has a charge of -1.

2. 2NaI(aq): This is sodium iodide in aqueous solution. The formula is NaI. Similar to Pb(NO3)2, it dissociates into ions in solution. The sodium (Na) ion has a charge of +1, and the iodide (I) ion has a charge of -1.

3. PbI2(s): This is lead(II) iodide, which is a solid (s). The formula is PbI2. When a solid is formed in a chemical reaction, it means that the ions combine in a fixed ratio to form the solid compound.

4. 2NaNO3(aq): This is sodium nitrate in aqueous solution. The formula is NaNO3. Like the previous compounds, it dissociates into ions in solution. The sodium (Na) ion has a charge of +1, and the nitrate (NO3) ion has a charge of -1.

Putting it all together:

Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)

The formulas, charges (if any), and states of the species in the reaction are:

Pb(NO3)2(aq), +2 charge, aqueous (dissolved in water)
2NaI(aq), +1 charge for Na+ ion and -1 charge for I- ion, aqueous
PbI2(s), no charge, solid (formed as a product)
2NaNO3(aq), +1 charge for Na+ ion and -1 charge for NO3- ion, aqueous