Please answer the questions parts and the calculation part All infor attached Pr

Question

Please answer the questions parts and the calculation part

All infor attached

Preparing Solutions and Stoiehiometry tachground: Acarty all experiments in goneral chemisary involve reagent solations that have already been prepared for yoa. This experiment gives you the opportunity to plan how you will make two reagent solutions. Then you wil perform an experiment using the solutions that you prepate. In this manner you can translate the calculations that you have leaned 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 solutions KI and then examine the stoichiometry when this solutión is mixed with a solution of Ph(NO,,. Then, you will prepare a solution that is 0.1 the caleium ion (Ca) and investigate the stoichiometry when that solun is mixed with a solution containing the carbonate ion (COj 00M i You will be given the follow ing reagents and equipment to work with: Solid KI 0.060 M Pb(NO) Solid CaCly or Cacl-2O 0.10M Na CO, Analytical balance 100.0-mL volumetric flask 250.0-mL volum Transfer pipets (1.00-mL or 5,00-mL.) Centrifuge Part l: I. Prepare a written plan for making up 100.0-ml of a 0.100 M Kl 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 your instructor before you procced with the experiment 2. Prepare a written plan for making up 250.0-mL of a solution that is 0.100 the experiment M in Ca ions. Show this plan to your instructor before you proceed with

Explanation / Answer

Calculation:

1)

Preparation of 100 ml of 0.100 M KI

Molar mass KI = 166.0 g/mol

Moles of KI required = 0.100 L x 0.100 M = 0.0100

Mass of KI = 0.0100 mol x 166.0 g/mol = 1.66 g

Add 1.66 g of KI in enough water to reach the volume of 100 mL

Preparation of 250 ml of 0.100 M Ca2+

Calculate moles in 250 mL of 0.100M solution.

Moles of CaCl2 of required = 0.250 L x 0.100 M = 0.0250

Molar mass of CaCl2 = 110.98 g/mol

Mass of CaCl2 = 0.0250 mol x 110.98 g/mol = 2.77 g

Add 2.77 g of CaCl2 in enough water to reach the volume of 250 mL

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

3) Write out the total ionic equation and cancel spectator ions which are common on both sides

Ca+2(aq) + 2Cl-(aq) + 2Na+(aq) + CO3-2(aq) --> CaCO3(s) + 2Na+(aq) + 2Cl-(aq)

The net ionic equation is

Ca+2(aq) + CO3-2(aq) --> CaCO3(s)

Write the balanced equations for the reaction

Na2CO3 (aq) + 2HCl (aq) -> 2NaCl (aq) + H2O (l) + CO2 (g)

2Na+ (aq) + CO3-2 (aq) + 2H+ (aq) + 2Cl- (aq) -> 2Na+ (aq) + 2Cl- (aq) + H2O (l) + CO2 (g)

Cancel out spectator ions which are common on both sides

The net ionic equation is

CO3-2 (aq) + 2H+ (aq) -> H2O (l) + CO2 (g)

Questions

1)

Since the total amount of solute is the same before and after dilution,

Then

C con × V con = Cdil × V dil

0.560 x 100 = 0.0140 x V dil

V dil = 0.560M x 100ml / 0.0140 M= 4000ml = 4.0L

Use a graduated cylinder to measure the100 mL of 0.560 M KI, transfer it to a flask and add sufficient water to give a total volume of 4.0 L

2. Calculate moles of Na2SO4

0.5 L x 1.8 M = 0.9 mol

Molar mass of Na2SO4 = 142.04 g /mol.

mass = mol x molar mass

= 0.9 mol x 142.04 g /mol = 127.84 g

Add 127.84 g Na2SO4 in enough water to reach the volume of 500 mL.

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