The balanced equation for the reaction: 3Pb(NO3)2 + 2AlI3 -> 6PbI2 + 2Al(NO3)3 .
The reaction between lead(II) nitrate (Pb(NO3)2) and aluminum iodide (AlI3) yields lead(II) iodide (PbI2) and aluminum nitrate (Al(NO3)3).
Here's the balanced equation for the reaction:
3Pb(NO3)2 + 2AlI3 -> 6PbI2 + 2Al(NO3)3
The reaction proceeds via a double displacement reaction, where the cations and anions of the two reactants switch partners to form new compounds.
Here are the individual steps involved in the reaction:
Ionic dissociation: Both Pb(NO3)2 and AlI3 dissociate into their respective ions in aqueous solution.
Formation of new ionic bonds: Pb2+ ions from Pb(NO3)2 are attracted to I- ions from AlI3, leading to the formation of PbI2.
Precipitation: PbI2 is insoluble in water and precipitates out of the solution as a solid.
Formation of Al(NO3)3: The remaining Al3+ ions combine with NO3- ions from Pb(NO3)2 to form Al(NO3)3, which remains dissolved in the solution.
Overall, this is a simple and straightforward double displacement reaction that yields readily identifiable products.