What volume of a 0.390 M NH4I solution is required to react with 227 mL of a 0.320 M Pb(NO3)2 solution? How many moles of PbI2 are formed from this reaction?

Answer 1

First, write the correctly balanced equation.

2NH₄I + Pb(NO3)2 ==> 2NH4NO3 + PbI₂(s) … balanced equation

Mols Pb(NO3)2 = 227 ml x 1L/1000 ml x 0.320 mol/L = 0.07264 mols

mols NH4I needed = 0.07264 mol Pb(NO3)2 x 2 mol NH4I/mol Pb(NO3)2 = 0.1453 mol NH4I

vol NH4I needed = 0.1453 mols x 1L/0.390 mol = 0.373 L = 373 mls NH4I needed

mols PbI2 formed = 0.07264 mol Pb(NO3)2 x 1 mol PbI2/mol Pb(NO3)2 = 0.0726 mols PbI2