Solutions of calcium iodide and lead nitrate are mixed together. A reaction occurs where a solid forms at the bottom of the beaker. This is a precipitate of lead iodide. Calcium nitrate is also product but it stays in solution as ions because it is soluble. Write the equation for this reaction making sure that it is balanced electronically and atomically. Remember, the result is calcium nitrate and lead (II) iodide.
The answer is one of the following. They all seem balanced, so I do not know what to do. Please explain the process in your response.
1. Ca(NO3) + PbI => CaI + Pb(NO3)
2. CaI2 + Pb(NO3)2 => Ca(NO3)2 + PbI2
3. Ca(NO3)2 + PbI2 => CaI2 + Pb(NO3)2
4. Ca2(NO3) + Pb2I => Ca2I + Pb2(NO3)
5. Ca(NO3) + PbI2 => CaI + Pb(NO3)2
The first thing you need to do is balance the charges. You need to know the charges of all the ions:
Calcium is +2
Iodide is -1
Nitrate is -1
Lead is +2
This solution explains how to balance the reaction between calcium nitrate and lead iodide.