Question

Answer the following for the reaction:

Pb(NO3)₂(aq) + 2KCl (aq) ⟶ PbCl₂(s) + 2KNO₃(aq)

a) How many grams of PbCl2 will be formed from 50.0 mL of a 1.50 M KCl solution?

b) How many mL of a 2.00 M Pb(NO3)2 solution will react with 50.0 mL of a 1.50 M KCl solution?

Answer 1

a) No. of mole = molarity volume of solution in liter

no. mole of KCl = 1.50M 0.050 L = 0.075 mole

According to reaction 2 mole of KCl produce 1 mole of PbCl₂ then 0.075 mole of KCl produce 0.075/2 = 0.0375 mole of PbCl₂

molar mass of PbCl₂ = 278.1 gm/mole that mean 1 mole of PbCl₂ = 278.1 gm

then 0.0375 mole of PbCl₂ = 278.1 0.0375 = 10.42875 gm

10.42875 gm of PbCl₂ will be formed from 50.0 mL of a 1.50 M KCl solution.

b)

No. of mole = molarity volume of solution in liter

no. mole of KCl = 1.50M 0.050 L = 0.075 mole

According to reaction 2 mole of KCl react with 1 mole of Pb(NO₃)₂ then 0.075 mole of KCl require 0.075/2 = 0.0375 mole of Pb(NO₃)₂

volume of solution in liter = no. of mole/molarity = 0.0375/2 = 0.01875 L = 18.75 ml

18.75 ml of a 2.00 M Pb(NO₃)₂ solution will react with 50.0 mL of a 1.50 M KCl solution