Question

Consider the following reaction:

2KI (aq) + Pb(NO3)2(aq) -> 2 KNO3(aq) + Pbl2(s)

What minimum volume of 0.200 M Kl(aq) is required to completely react with 155.0 mL of a 0.112 M Pb(NO3)2 solution?

A. 174 mL
B. 43.4 mL
C. 86.8 mL
D. 348 mL

Answer 1

B. 43.40 mL

Step-by-step explanation:

We are required to determine the volume of 0.200 M Kl(aq).

We have; 155.0 mL of a 0.112 M Pb(NO3)2 solution.

We know that molarity of a solution is contained in 1 L or 1000 mL or 1000 cm³.

Hence,

• 0.112 M Pb(NO3)2 is contained in 1000 mL
• x mol Pb(NO3)2 is contained in 155.0 mL

`x \: mol \: = (155 * 0.112)/(1000) \\ = 0.01736 \: moles`

= 0.01736 moles of Pb(NO3)2

We have a mole ratio of 2: 1 (KI : Pb(NO3)2).

To obtain the number of moles of KI; we divide by half the moles of Pb(NO3)2 (aq);

(0.01736)/2

= 0.00868 moles KI

0.20 moles KI is contained in 1000 mL

0.00868 moles KI is contained in x mL

`x \: ml \: = (0.00868 * 1000)/(0.20)`

= 43.40 mL

Therefore the volume of KI required to react completely with 0.112 M Pb(NO3)2 is 43.40 mL