Answer to Question #144005 in General Chemistry for Javier

Question

How many grams of PbCl2 will be formed from 40.0 mL of 1.40 M KCl and excess Pb (NO3)2?

Answer 1

The precipitation reaction is

Pb(NO3)2 + 2KCL=PbCl2 + 2KNO3

According to reaction ,2mole KCL react with 1mole Pb(NO3)2, 40mL of 1.40(M) KCL=0.056mole react with 0.028 mole Pb(NO3)2 produce 0.028 mole PbCL2=(278*0.028)=7.784 g PbCL2