How many moles of PbI2 are formed from this reaction?

Question

What volume of a 0.510 M NH4I solution is required to react with 981 mL of a 0.680 M Pb(NO3)2 solution? How many moles of PbI₂ are formed from this reaction?

Answer 1

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

moles Pb(NO3)2 present = 981 ml x 1 L / 1000 ml x 0.680 mol/L = 0.6671 moles

moles NH₄I required = 0.6671 mols Pb(NO₃)₂ x 2 mol NH₄I / mol Pb(NO₃)₂ = 1.334 mols NH₄I needed

Volume of NH₄I needed = 1.334 mols x 1 L / 0.510 mol = 2.62 L NH₄I needed

Moles of PbI₂ formed = 0.6671 mols Pb(NO₃)₂ x 1 mol PbI₂ / 1 mol Pb(NO₃)₂ = 0.667 mols PbI₂ formed