How many moles of precipitate will be formed when 80.0 mL of 2.00 M NaI is reacted with 100.0 mL of 0.900 M Pb(NO₃)₂ in the following chemical reaction? 2NaI(aq) + Pb(NO₃)₂(aq) → PbI₂(s) + 2NaNO₃(aq)

Answer 1

First we need to determine the limiting reactant (LR) by calculating the number of moles of each reactant we have:

For the lead (II) nitrate: (0.900 n/1L)(0.100L) = 0.0900 moles

For the sodium iodide: (2.00 n/1L)(0.0800L) = 0.160 moles

The molar ratio of lead (II) nitrate to sodium iodide is 1:2, so we need two times the number of moles of NaI than Pb(NO₃)₂, and we do not have that ((0.160/0.0900) = 1.78. So, the NaI is the LR.

Finally, since the molar ratio of the NaI to the PbI₂ is 2:1, we should get 1/2 the number of moles of NaI available, or 0.0800 moles of lead (II) iodide.