Preparing Solutions and Stoichiometry Background: Nearly all experiments in gene

Question

Preparing Solutions and Stoichiometry Background: Nearly all experiments in general chemistry require reagent solutions that have already been prepared for you. This experiment gives you the opportunity to plan how you will make two reagent solutions. Then you will perform an experiment using the solutions that you prepare. In this manner, vou can translate the calculations that you have learned in lecture into preparing solutions that can be used in the laboratory In the first part of the experiment, you will prepare a 0.100 M solution of KI and then examine the stoichiometry when this solution is mixed with a solution of Pb(NO3)2. Then, you will prepare a solution that is 0.100 M in the calcium ion (Ca2+) and investigate the stoichiometry when that solution is mixed with a solution containing the carbonate ion (COj2-) You will be given the following reagents and equipment to work with: Solid KI 0.060 M Pb(NO3)2 Solid CaCl or CaCl2 2H2O 0.10M Na2CO3 Analytical balance 100.0-mL volumetric flask 250.0-mL volumetric flask Transfer pipets (1.00-mL or 5.00-mL) Centrifuge Part 1: 1. Prepare a written plan for making up 100.0-mL of a 0.100 M KI solution. Include in your plan the exact amount of KI that you will need and the actual steps that you will take to prepare the solution. Show this plan to youir instructor before you proceed with the experiment 2. Prepare a written plan for making up 250.0-mL of a solution that is 0.100 M in Ca2+ ions. Show this plan to your instructor before you proceed with the experiment

Explanation / Answer

2)Aqueous solution of lead nitrate is colorless. When solutions of lead nitrate and potassium iodide are mixed together, a yellow precipitate of lead iodide is formed. Equation for reaction is

Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)

First write full ionic equation of reaction

Pb2+(aq)+2NO3(aq)+ 2K+(aq)+2I(aq) -----> PbI2(s)+2K+(aq)+2NO3(aq)

Now cancel spectator ions (spectator ions are those which are the same, with the same state symbol, on both sides of the equation), and get the net ionic equation.

Net ionic equation

Pb2+(aq)+2I(aq) -----> PbI2(s)

Find the limiting reactant

Concentration of Pb(NO3)2 = 0.060M

Volume of Pb(NO3)2 = 1ml = 0.001L

Concentration of KI = 0.100M

Volume of KI = 2ml = 0.002L

Calculate moles of Pb(NO3)2

Calculate moles of Pb(NO3)2 = Concentration of Pb(NO3)2 x volume of Pb(NO3)2

= 0.060M x 0.001 L = 0.00006 moles

Calculate moles of KI

Calculate moles of = Concentration of KI x volume of KI

= 0.100M x 0.002L = 0.0002 moles

2 moles of KI reacts with 1mol of Pb(NO3)2

0.0002 moles of KI will reacts 0.0002/2 = 0.0001 moles of Pb(NO3)2, but there is only 0.00006 moles of Pb(NO3)2, so it is limiting reactant.

1 mol of Pb(NO3)2 produces 1 mol of PbI2

So, 0.00006 mol of Pb(NO3)2 produces 0.00006 mol of PbI2

Mass of yellow precipitate:

Mol of yellow precipitate (PbI2) = 0.00006

Molar mass of (PbI2) = 461.01 g/mol

Now multiply mol of PbI2 by its molar mass to get mass of yellow precipitate

0.00006 mol x 461.01 g/mol = 0.028 g

Mass of yellow precipitate is 0.028 g

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---