Complete and balance the chemical equation using compound formulas. Include the states. If there is no reaction, enter NR.
Pb(NO3)2(aq) + NaI(aq) →
sine Pb is not reactive, the reaction is NR. Am i correct?
No. Pb^+2 forms PbI2 with the I^- from NaI. PbI2 is insoluble and that is what drives the reaction.
Pb(NO3)2(aq) + 2NaI(aq) ==> PbI2(s) + 2NaNO3(aq)
In reactions of this type, there are three reasons for a reaction occurring.
1. A ppt is formed. For this you need to know the solubility table. Here is a simple one on the web.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
2. A gas is formed. You know the simple ones; i.e., H2, O2, F2, Cl2, CO, CO2, N2, etc.
3. A slightly ionized material is formed; i.e., a weak electrolyte.
4. There are different rules for redox equations.
To determine if there is a reaction between lead(II) nitrate (Pb(NO3)2) and sodium iodide (NaI), we can refer to a solubility chart to check if the resulting compounds are soluble or insoluble.
Lead(II) nitrate is soluble in water, indicated by "(aq)" in the chemical formula. Sodium iodide is also soluble in water.
Since both compounds are soluble in water, we can proceed with the reaction. The balanced chemical equation for the reaction between lead(II) nitrate and sodium iodide is:
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
In this balanced equation, the lead(II) nitrate reacts with sodium iodide to produce lead(II) iodide (PbI2) as a solid precipitate, and sodium nitrate (NaNO3) as a soluble compound in the aqueous state.
Therefore, the correct and balanced chemical equation with compound formulas and states is:
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)