Question

Pb(NO3)2 (0) + 2 KCl(aq) PbCl(s) + 2 KNO₃(aq) • How many mL of a 2.00M Pb (NO3)2 Solution will react tw 50.0mL of a 1.som kce solution?

Answer 1

Solution-

The balanced chemical equation can be written as:

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

Also

M(KCl)= 1.50 M

M(Pb(NO3)2)=2.00 M

V(KCl)=50 mL

Now from the balanced reaction:

1* number of mol of KCl = 2*number of mol of Pb(NO3)2

=>1*M(KCl)*V(KCl) =2*M(Pb(NO3)2)*V(Pb(NO3)2)

=>1*1.50 M *50 mL = 2*2.00M *V(Pb(NO3)2)

=>V(Pb(NO3)2) = 18.75 mL