Question

Lead (II) nitrate and ammonium iodide react to form lead (II) iodide and ammonium nitrate according to the reaction

Pb(NO3)2(aq)+ 2 NH4I(aq)-----> PbI2(s)+2NH4NO3(aq)

What volume of a 0.710 M NH4I solution is required to react with 135 mL of a 0.400 M Pb (NO3)2 solution?

How many moles of PbI2 are formed from this reaction.

Answer 1

we know that

Number of mole = molarity volume of solution in liter

Number of mole of Pb(NO₃)₂ = 0.400M 0.135L = 0.054 mole

According to reaction 1 mole of Pb(NO₃)₂ react with 2 mole of NH₄I then 0.054 mole of Pb(NO₃)₂ require

0.0542 = 0.108 mole of NH₄I

we know that

Volume of solution in liter = no. of mole of solute / molarity

= 0.108/0.710 = 0.15211 L

0.15211L = 152.11 ml

152.11 ml of 0.710M NH₄I solution require to react with 135 ml of 0.400M Pb(NO₃)₂