Question

How would you prepare 9.70 g 0f PbCl2(s) from a 0.100 M solution of Pb(NO3)2 and a 0.200 M solution of CaCl2?

Answer 1

number of mole of PbCl2 = (mass formed)/(molar mass of PbCl2)
molar mass of PbCl2 = 278.1 g/mol
number of mole of PbCl2 = 9.70/278.1
= 0.0349 mole

reaction taking place
Pb(NO3)2 + CaCl2 --> PbCl2 + Ca(NO3)2
according to reaction
number of mole of Pb(NO3)2 = number of mole of PbCl2 =
(molarity of Pb(NO3)2)*(volume in L) = 0.0349
0.100*V1 = 0.0349
V1 = 0.349 L

also,
number of mole of CaCl2 = number of mole of PbCl2 =
(molarity of CaCl2)*(volume in L) = 0.0349
0.200*V1 = 0.0349
V1 = 0.174 L

we can prepare 9.70 g of PbCl2 by mixing 0.349 L of 0.100 M of Pb(NO3)2 and 0.174 L of 0.200 M of CaCl2