Question

According to the following reaction, what volume of 0.244 M KCl solution is required to react exactly with 50.0 mL of 0.210 M Pb(NO3)2 solution? 2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq)

Answer 1

Answer:

86 mL

Explanation:

First find the moles of Pb (NO3)2

n=cv

where

c ( concentration)= 0.210 M

v ( volume in L) =0.05

n= 0.210 × 0.05

n= 0.0105

Using the mole ratio, we can find the moles of KCl by multiplying by 2

n (KCl) =0.0105 ×2

=0.021

v (KCl)= n/ c

= 0.021/ 0.244

=0.08606557377

=0.086 L

= 86 mL

Answer 2

Answer:

The volume of 0.244 M KCl solution is required 86.07 mL.

Explanation:

$Concentration=\frac{Moles}{Volume (L)}$

Moles of lead nitrate in 50.0 mL of 0.210 M solution be n.

Molarity of the lead nitrate solution = 0.210 M

Volume of the solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)

$0.210 M=\frac{n}{0.050 L}$

$n=0.210 M\times 0.050 L=0.0105 mol$

According to reaction given, 1 mole of lead nitrate reacts with 2 moles of KCl.

Then 0.0105 moles of lead nitrate will react with:

$\frac{2}{1}\times 0.0105 mol=0.021 mol$ of KCl

Moles of KCl = 0.021 mol

Volume of KCl solution = V

Molarity of the KCl solution = 0.244 M

$0.244 M=\frac{0.021 mol}{V}$

$V=\frac{0.021 mol}{0.244 M}=0.08607 L = 86.07 mL$

The volume of 0.244 M KCl solution is required 86.07 mL.