Question

7) Aqueous Aluminum chloride is combined with aqueous Lead (II) nitrate, and a precipitate forms according to the balanced chemical equation shown below: 2 AICI() + 3 Pb(NO3)2() + 2 AI(NO), + 3 PbCl2 a) Determine the volume (in L) of 0.255 M Aluminum chloride solution needed to react completely with 8.95 L of a 0.0679 M Lead (II) nitrate solution: (8)

Answer 1

of ph CNO3)2 Given - 2 AlCl3 cour) + 3 Pb(NO) 12 coop ZAINO3)3 + 3PbCl2 com) 50 2 equv of Alla reacts with 3 equiv Alla reacts with / cepulv Pb (NO3)₂ I equiv af of Now, moles of Phenos, molasity solution Volume of solution in liter 0.0679 amules x 8.95 Litor Liter 0.60770 moles of pb CINO3)2 a 0.608 moles since, mole of AlCl reacts with 3/2 moles of Pb(NO₃)2 0.608 x * 243 8.608 moles af Pb (NO3), will react with moles g Aley 1.216 3 moles of Alch 0.60% moles of Ph cives), chill react with 0.405 moles of Ala,

0.255 M Ald, soln Now, the molali ality of Alla soll is = 0.255 M means 1 Liter 0.255 m soln AlCl3 will have 0.255 moles AlCl3 q 0.405 moles y 1 liter volume needed for reaction with of Pb(NO3)₂ 0.405 moles 0-255 moles 1.589 Liter wolume 0-255 ME 1.59 Liter Alla soin therefore, to react with 8.gr Liter ay og 0-0679m solution of Pb (NO3)2 We need 1.59 Liter ay