G.C.1-5 Lab Report for CHM2045 In this experiment, insoluble cobalt carbonate CoCO3 is prepared by mixing solutions of cobalt chloride and sodium carbonate according to Equation 1. PDF

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Summary

Using the information in Table 1, and videos are posted in your canvas page under module 5 and Instructional Materials page, please complete tables for Part I and Part II....

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G.C. 1-5 LAB REPORT

Name: Z Number:

SUMMARY OF THE VIDEO: Part 1:

Use a bulb and pipet to transfer 2 mL of CoCl2 and 13 mL of Na2CO3 in test tube #1. Tilt test tube by placing thumb on top and slowly tilting the solution back and forth. Place 3 mL of CoCl2 and 12 mL of Na2CO3 in test tube #2 and tilt. Place 4 mL of CoCl2 and 11 mL of Na2CO3 in test tube #3 and tilt. Place 6 mL of CoCl2 and 9 mL of Na2CO3 in test tube #4 and tilt. Place 7.5 mL of CoCl2 and 7.5 of Na2CO3 in test tube #5 and tilt. Place 10 mL of CoCl2 and 5 mL of Na2CO3 in test tube #6 and tilt. Leave test tubes for a couple of minutes to let the precipitate settle down. Then use a ruler to measure how much precipitate there is and how much liquid is on top of precipitate. Measure height of precipitate without disturbing it and record in cm.

Part 2: Put a few drops of the supernatant off the top of each test tube in two wells per tube. Test for excess Co^2+ by adding a few drops of 0.10 M sodium phosphate in the first row of wells. Test for excess CO3^2- by adding a few drops of 0.10 M barium chloride inthe second row of wells.

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In this experiment, insoluble cobalt carbonate CoCO3 is prepared by mixing solutions of cobalt chloride and sodium carbonate according to Equation 1. CoCl2(aq) + Na2CO3 (aq)  CoCO3 (s) + 2NaCl(aq) Solutions of cobalt chloride and sodium carbonate is combined in six different ratios, and the amount of cobalt carbonate formed is compared by measuring the heights of the solids precipitated. The limiting reagent in each case is first predicted and then verified, by making observations and testing some of the supernatant liquid (above the precipitate) with two different chemical experiments.

Verification of the limiting reagent is performed by considering the following: 1. Cobalt ions Co2+ readily react with phosphate ions PO43- in sodium phosphate to form insoluble cobalt phosphate Co 3(PO4)2. If a precipitate is formed with some of the supernatant liquid and a solution containing phosphate ions, we can verify that not all of the Co 2+ ions have been used up in the reaction. In this case, the cobalt ions are in excess and the carbonate is the limiting reagent. 2. Carbonate ions CO32- readily react with barium ions Ba 2+ in barium chloride to form an insoluble barium carbonate BaCO3 precipitate. If a precipitate is formed with some of the supernatant liquid and a solution containing barium ions, we can verify that not all of the CO 32- ions have been used up in the reaction. In this case, the carbonate ions are in excess and the cobalt ion is the limiting reagent. Using the information in Table 1, and videos are posted in your canvas page under module 5 and Instructional Materials page, please complete tables for Part I and Part II. YOU MUST WATCH THE VIDEOs to complete tables and answer all question. YOU MUST SHOW ALL YOUR CALCULATIONS WITH CORRECT SIG. FIG. AND UNITS TO RECEIVE CREDIT FOR YOUR ANSWERS.

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Table1. ASSIGNMENT # Test-tube # 1 Test-tube # 2 Test-tube # 3 Test-tube # 4 Test-tube # 5 Test-tube #6

COCI2 2.00 mL 3.00 mL 4.00 mL 6.00 mL 7.50 mL 10.00 mL

Na2CO3 13.00 mL 12.00 mL 11.00 mL 9.00 mL 7.50 mL 5.00 mL

PART I: FOR COCl2

TEST-

TEST-

TEST-

TEST-

TEST-

TEST-

TUBE #4

TUBE #5

TUBE #6

TUBE #1

TUBE #2

TUBE #3

Concentration Volume

2.00 mL

3.00 mL

4.00 mL

6.00 mL

7.50 mL

10.00 mL

delivered (mL) Moles of CoCl2

0.0002

0.0003

0.0004

0.0006 mol

0.00075

0.001 mol

mol Testtube #5

Test-tube #6

7.50 mL

5.00 mL

0.0009 mol

0.00075

0.0005

Test-tube #4

mol Testtube #5

mol Test-tube #6

Concentration Volume

13.00 mL

12.00 mL

mol Test-tube Test-tube #3 #4 0.10 M 11.00 mL 9.00 mL

delivered (mL) Moles of

0.0013

0.0012

0.0011

mol Test-tube #1

mol Test-tube #2

mol Test-tube #3

For Na2CO3

Na2CO3 Volume

mol mol Test-tube Test-tube #1 #2

Measurements MUST WATCH VIDEO TO FILL THIS PART OF THE TABLE 5.5 cm 6.5 cm 7.0 cm 9.0 cm 9.5 cm Height of

8.5 cm

precipitate formed (cm) Color of

Clear blue

supernatant

Clear

Cloudy

Cloudy

purple

purple

purple/pink

light pink

Clear pink

liquid observed. Color of

Cloudy

Cloudy

cloudy

Cloudy

Cloudy

Cloudy

precipitate

blue

purple/

purple

purple/

light pink

dark pink

observed. 3

Clear

blue

pink

SHOW YOUR WORK BELOW:

CoCl2: Test tube 1: 2.00 mL x 1 L/1000 mL x 0.1 mol/1 L= 0.0002 mol Test tube 2: 3.00 mL x 1 L/1000 mL x 0.1 mol/1 L= 0.0003 mol Test tube 3: 4.00 mL x 1 L/1000 mL x 0.1 mol/1 L= 0.0004 mol Test tube 4: 6.00 mL x 1 L/1000 mL x 0.1 mol/1 L= 0.0006 mol Test tube 5: 7.5 mL x 1 L/1000 mL x 0.1 mol/1 L= 0.00075 mol Test tube 6: 10.00 mL x 1 L/1000 mL x 0.1 mol/1 L= 0.001 mol NaCO3: Test tube 1: 13.00 mL x 1 L/1000 mL x 0.1 mol/1 L = 0.0013 mol Test tube 2: 12.00 mL x 1 L/1000 mL x 0.1 mol/1 L = 0.0012 mol Test tube 3: 11.00 mL x 1 L/1000 mL x 0.1 mol/1 L = 0.0011 mol Test tube 4: 9.00 mL x 1 L/1000 mL x 0.1 mol/1 L = 0.0009 mol Test tube 5: 7.5 mL x 1 L/1000 mL x 0.1 mol/1 L= 0.00075 mol Test tube 6: 5.00 mL x 1 L/1000 mL x 0.1 mol/1 L = 0.0005 mol

Excess Co2+ or

TESTTUBE #1 CO32-

TESTTUBE #2 CO32-

TESTTUBE #3 CO32-

TESTTUBE #4 CO32-

TESTTUBE #5 N/A

TESTTUBE #6 Co2+

0

0

0

0

0.0005

CO32- present? Moles of Co2+ 0

4

in excess Moles of CO32- 0.0011

0.0009

0.0007

0.0003

0

0

3 x 10-3

4 x 10-4

6 x 10 -4

7.5 x 10-4

5 x 10-4

Testtube #4

Testtube #5

Testtube #6

in excess Theoretical

2 x 10-4

Yield of CoCO3 (moles)

YOU MUST WATCH THE VIDEO PART II: Verification of Limiting Reagent Observations of precipitate formation Reaction of

Testtube #1

Testtube #2

Testtube #3

NR

NR

NR

NR

NR

PPT Purple

Reaction of

WHITE

White PPT

NR

PPT

White PPT

NR

supernatant

White PPT

supernatant liq. with Na3PO4

liquid with BaCl2 SHOW ALL YOUR CALCULATIONS: Excess in Co2+: Test tube 6: 0.001 mol – 0.0005 mol = 0.0005 mol Excess in CO32-: Test tube 1: 0.0013 mol – 0.0002 mol = 0.0011 mol Test tube 2: 0.0012 mol – 0.0003 mol = 0.0009 mol Test tube 3: 0.0011 mol – 0.0004 mol = 0.0007 mol Test tube 4: 0.0009 mol – 0.0006 mol = 0.0003 mol

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.C. 1- 5: DISCUSSION QUESTIONS 1. Were your predictions for which ion was in excess verified by Part II of the experiment? Explain. Yes, when no precipitate forms there will be an excess. 2. Compare the theoretical yield of the CoCO3 precipitate expected with the heights/volume of precipitate. What would explain any anomalies you may find? Did any of your observations about color of the precipitate and liquid give supporting evidence? The height of the precipitate increases as the theoretical yield increases. This is seen in all but the last tube. In this case the precipitate may have been disturbed during the process of measuring it, or the measurement may have been read incorrectly. 3. How would you determine the point at which a reagent from part I becomes the limiting reagent for the reaction? Does the molar ratio have an effect on which reagent is the limiting reagent? Explain.

Whichever compound has the smallest amount of moles would be the limiting reagent. Molar ratio does have an effect because the limiting reagent depends on the number of moles you have.

4. What reagent would you predict to be in excess for reacting 7.50 mL of a 0.10M BaCl2 solution with 7.50 mL of 0.10M KIO 3 solution? Explain. Hint: Balance the equation first. I would predict KIO3 to be in excess. It has two moles and BaCI2 only has one.

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SHOW ALL YOUR CALCULATIONS: .0075L BaCl2 x .10 mol BaCl2/ 1L x 2KCl/1BaCl2 .0075 L KIO3 x .10 mol KIO3/1L x 2KCl/ 2 KIO3

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