Answer to Question #144006 in General Chemistry for Javier

Question

How many milliliters of 1.90 M Pb(NO3)2 Pb(NO3)2 will react with 40.0 mL of 1.40 M KCl?

Answer 1

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

n(KCl) = c*V=1,4*40*10^-3= 0,056 mole

n(Pb(NO₃)₂)= 0,056/2=0,028 mole

V(Pb(NO₃)₂)=n/c=(0,028/1,90)*10³=14,74 mL

Answer: 14,74 mL Pb(NO₃)₂