What volume of a 0.170 M NH4I solution is required to react with 853 mL of a 0.440 M Pb(NO3)2 solution?

Answer 1

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

moles Pb(NO₃)₂ present = 0.853 L x 0.440 mol/L = 0.3753 moles

moles NH₄I needed to react = 0.3753 mol Pb(NO₃)₂ x 2 mol NH4I/mol Pb(NO₃)₂ = 0.751 moles NH₄I

Volume NH₄I: (x L)(0.170 mol/L) = 0.751 moles

x = 4.42 L of NH₄I needed