Question

reaction : pb(no3)2 + 2KCl ----> PbCl2 +2KNO3

1) how many grams of PbCl2 can be formed from 50.0 mL os a 1.5 M KCl solution ?

2) how many mL of a 2.00 M Pb(NO3)2 solution are required to react with 50.0 mL of a 1.5 M KCl solution ?

3) what is the molarity of 20.0 mL of a KCl solution that reacts completely with 30.mL of a 0.400 M Pb(NO3)2 ?

Answer 1

Ans :

1)

Number of moles of KCl = molarity x volume (L)

= 1.5 x 0.050

= 0.075 mol

Number of moles of PbCl2 formed = 0.075 / 2 = 0.0375 mol

Mass of PbCl2 formed = moles x molar mass

= 0.0375 x 278.1

= 10.4 grams

2)

2 moles of KCl require 1 mol of PbCl2

So 0.075 mol of KCl will require : 0.075 / 2

= 0.0375 mol of Pb(NO3)2

mol = molarity x volume

0.0375 = 2.00 x V

V = 0.01875 L

= 18.75 mL

3)

Number of moles of Pb(NO3)2 = 0.400 x 0.030

= 0.012 mol

So number of mol of KCl required = 2 x 0.012 = 0.024 mol

Molarity = mol / volume (L)

M = 0.024 / 0.020

= 1.2 M