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, + 3Pb(NOx- 2 AINO) + 3 PCI 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) 8) Hydrobromic acid and Iron (II) hydroxide react in the balanced chemical equation shown below: 3 HBr + Fe(OH), + 3H2O + Febr, It takes 4.5 L of the HBr solution to react completely with 235 grams of Fe(OH), Calculate the Molarity of the HBr solution: (MM Fe(OH), = 106.88 g/mol (8) -5-

Answer 1

According to HOMEWORKLIB RULES we are entitled to solve max one question.

Q7

2AlCl3(aq) + 3Pb(NO3)2(aq) =====> 2Al(NO3)3(aq) + 3PbCl2(aq)

According to above equation for complete titration

3*moles of AlCl3 = 2*moles of Pb(NO3)2

Moles of AlCl3 = Volume(L)*Molarity(NaOH)

= V(L)*0.255 moles

Moles of Pb(NO3) = Volume(L)*Molarity(Pb(NO3))

=8.95*0.0679 moles

3*moles of AlCl3 = 2*moles of Pb(NO3)2

3*(V(L)*0.255) = 2*(8.95*0.0679)

V(AlCl3) = 2*(8.95*0.0679)/(3*0.255) L

= 1.59 Litres