Question

3. Calculate the number of moles of precipitate that would be produced when two solutions were mixed. a 1.0 mL 0.10 M Pb(NO3)2 and 5.0 ml, 0.10 M KI b. 2.0 ml 0.10 M Pb(NO3)2 and 4.0 mL, 0.10 M KI c. 3.0 mL 0.10 M Pb(NO3)2 and 3.0 mL, 0.10 M KI d. 4.0 mL 0.10 M Pb(NO3)2 and 2.0 mL, 0.10 M KI e. 5.0 mL 0.10 M Pb(NO3)2 and 1.0 mL, 0.10 M KI

PLEASE DO ALL

Answer 1

The stoicheometric reaction of lead Nitrate with Potassium iodide is as follows.

Pb(NO3)2 + 2KI- - Pb 12 + 2 KNO3

As per the above reaction, one mole of lead nitrate reacts with 2 moles of potassium iodide to precipitate completely and gives one mole of lead iodide..

Calculating number of moles present in given molar solution is as follows

M=(N/V) * 1000

Where M= Molarity, N = number of moles,  V= Volume of solution in mL.

Rearranging the above equation as N=(M * V)/1000, we can calculate number of moles.

A) 1.0mL of 0.10M lead nitrate reacts with 5mL of 0.10M of KI.

Using the above equations,

1) number of moles present in 1.0mL of 0.10M of lead nitrate is 0.1 millimoles.

2) number of moles present in 5mL of 0.10M KI is 0.5 millimoles.

3) 0.1millimoles of lead nitrate reacts with 0.2millimoles of KI to give 0.1millimoles of PbI.Remaining 0.3 millimoles of KI remains unconsumed.

B) 2.0 mL of 0.10 M lead nitrate reacts with 4 mL of 0.10 M of KI.

Using the above equations,

1) number of moles present in 2.0 mL of 0.10M of lead nitrate is 0.2 millimoles.

2) number of moles present in 4mL of 0.10M KI is 0.4 millimoles.

3) 0.2 millimoles of lead nitrate reacts with 0.4 millimoles of KI to give 0.2 millimoles of PbI.  KI is consumed.

C) 3.0 mL of 0.10 M lead nitrate reacts with 3 mL of 0.10 M of KI.

Using the above equations,

1) number of moles present in 3.0 mL of 0.10M of lead nitrate is 0.3 millimoles.

2) number of moles present in 3 mL of 0.10M KI is 0.3 millimoles.

3) As per the stoicheometry of reaction 0.15 millimoles of lead nitrate reacts with 0.3 millimoles of KI to give 0.15 millimoles of PbI. 0.15 millimoles of leadnitrate remains unreacted.

D) 4.0 mL of 0.10 M lead nitrate reacts with 2 mL of 0.10 M of KI.

Using the above equations,

1) number of moles present in 4.0 mL of 0.10M of lead nitrate is 0.4 millimoles.

2) number of moles present in 2 mL of 0.10M KI is 0.2 millimoles.

3) 0.1 millimoles of lead nitrate reacts with 0.2 millimoles of KI to give 0.1 millimles of PbI.  0.3 millimoles of lead nitrate remains unreacted.

E) 5.0 mL of 0.10 M lead nitrate reacts with 1 mL of 0.10 M of KI.

Using the above equations,

1) number of moles present in 5.0 mL of 0.10M of lead nitrate is 0.5 millimoles.

2) number of moles present in 1 mL of 0.10M KI is 0.1 millimoles.

3) 0.05 millimoles of lead nitrate reacts with 0.1 millimoles of KI to give 0.05 millimoles of PbI.  0.45 millimoles of lead nitrate remains unreacted.