Chemistry lab report experiment 1 chemical reactions

EquipmentTest tubes Test tube rack Test tube holders Beakers Alcohol lamp Looped wire Distilled water bottle IV/ Procedure, result, and discussion 1/ Reaction of Cu2+ Procedure: Step 1

VIETNAM NATIONAL UNIVERSITY – HO CHI MINH

CITY INTERNATIONAL UNIVERSITY

CHEMISTRY LAB REPORT

EXPERIMENT 1: CHEMICAL REACTIONS

Instructor: Lê Trần Hồng Ngọc

Name: Võ Thị Thảo Vy ID: BTBTIU22187 Group: 16

Academic year: 2023-2024

I/ Introduction

Chemical reactions play an essential role in human activities, particularly in creating a new substance to produce many types of materials that serve human need In one chemical reaction is the participation of substances called reactants which are

transformed into different substances which are considered as products and they appear with various characteristics In experiments 1, we practiced and observed some types of chemical reactions Chemical changes in these reactions can be the the transformation of color, the formation of a solid, the release of gas, and the production of heat and light Generally, chemical reactions are represented in five types of reactions: synthesis, decomposition, single displacement, double displacement, and combustion

II/ Objectives

To perform different types of chemical reactions (acid-base, precipitate, gas forming, complex compound forming, and oxidation-reduction reactions)

To identify some of the products in these reactions and describe the chemical changes

To write and balance the chemical equations for the reactions observed

III/ Materials and equipment

1/ Materials

0.5M CuSO4

2M NaOH

2M NH4OH

0.5M KCl

0.1M AgNO3

0.5M KBr

0.1M KMnO4 2M H2SO4 3% H2O2 0.1M KI MnO2 0.5M Na2SO3 6M NaOH

2M HCl

Distilled water

LiCl

0.5M FeCl3

CaCl2 2M KOH BaCl2 0.5M FeSO 4 0.5M Al2(SO )4 3

2/ Equipment

Test tubes

Test tube rack

Test tube holders

Beakers

Alcohol lamp Looped wire Distilled water bottle IV/ Procedure, result, and discussion

1/ Reaction of Cu2+

Procedure:

Step 1: Put 10 drops 0.5M CuSO into each two test tube.4

Step 2: Put 10 drops 2M NaOH and 10 drops 2M NH OH into the first test tube and 4 the second test tube, respectively Then observe the phenomenon

Step 3: After a few seconds, continue to put 10 drops 2M NaOH and 10 drops 2M

NH4OH into the first test tube and the second test tube, respectively Then observe the phenomenon

Reaction Observation Chemical equation Image

#1 0.5M CuSO 4

+ 2M NaOH

There appears blue precipitation and colloidal phenomena were formed

CuSO4 + 2NaOH

Na SO2 4 + Cu(OH)2

0.5M CuSO 4

+ 2M NaOH

+ 2M NaOH

The color of the solution is lighter than the previous reaction However, the precipitation is not disappeared

CuSO4 + 2NaOH

Na SO2 4 + Cu(OH)2

#2 0.5M CuSO4

+ 2M NH4OH

There appears blue precipitation

CuSO4 + 2NH OH 4 (NH ) SO4 2 4 + Cu(OH)2

0.5M CuSO4

+ 2M NH4OH

+ 2M NH4OH

The color of the solution becomes darker than the previous reaction The precipitation is dissolved

Cu(OH)2 + 4NH OH 4 4H2O + [Cu(NH ) ](OH)3 4 2

Comments:

#1:

CuSO4 is dissociated: CuSO 4 Cu + SO2+ 4

2-NaOH is dissociated: 2-NaOH Na + OH +

-Cation Cu and Anion OH combine to each other then form the blue precipitation 2+ - Cu(OH) 2

When adding more 2M NaOH into the first test tube, the phenomenon proves that the excess of NaOH does not influence the reaction

#2:

CuSO4 is dissociated: CuSO 4 Cu + SO2+ 4

2-NH4OH is dissociated: NH OH 4 NH + OH 4

-Cation Cu and Anion OH combine to each other then form the blue precipitation 2+ - Cu(OH) 2

When adding more 2M NH OH into the second test tube, the phenomenon proves that 4 there is another reaction happens More specifically, after the precipitation Cu(OH) is 2 formed, it continue to reacts with excess NH OH and form the complex [Cu(NH4 3) ]4 (OH) 2

2/

Reactions of Silver halides

Section 1: Reactions of Potassium Chloride (KCl)

Procedure:

Step 1: Put 10 drops 0.5M KCl into each two test tube

Step 2: Continue to put 10 drops 0.1M AgNO into each two test tube.3

Step 3: Put 10 drops 2M NH OH into one of two test tubes Labeling this test tube is 4

#2, the other is #1 Observe the phenomenon

Reaction Observation Chemical equation Image

#1 0.5M KCl +

0.1M AgNO 3

There appears white precipitation

KCl + AgNO3 KNO3 + AgCl

#2 0.5M KCl +

0.1M AgNO 3

+ 2M NH4OH

There appears white precipitation after the reaction between KCl and AgNO 3

When put 2M NH OH 4 solution into above solution, the white precipitation disappeared

KCl + AgNO3 KNO3 + AgCl AgCl + 2NH OH →4 [Ag(NH3)2]Cl + H O2

Comments:

#1:

KCl is dissociated: KCl K + Cl+

-AgNO3 is dissociated: AgNO 3 Ag + NO + 3

-Cation Ag and Anion Cl combine to each other then form the blue precipitation AgCl.+ -

#2:

KCl is dissociated: KCl K + Cl+

-AgNO3 is dissociated: AgNO 3 Ag + NO + 3

-Cation Ag and Anion Cl combine to each other then form the blue precipitation AgCl.+ - However, when adding more NH OH, it continues to react with AgCl to form another 4 substances which are [Ag(NH ]Cl and H O, the white precipitation disappeared.3)2 2

Section 2: Reactions of Potassium Bromide (KBr)

Procedure:

Step 1: Add 10 drops of 0.5M KBr into two test tubes ( test tube #1 and test tube #2) Step 2: Add 10 drops of 0.1M AgNO3 into both test tubes

Step 3: Add 10 drops of 2M NH4OH into test tube #2

Step 4: Mix tubes gently and wait at least 2 minutes, then observe the phenomenon Reaction Observation Chemical equation Image

#1

0.5M KBr +

0.1M AgNO3

The light-yellow precipitate was formed AgNO3 + KBr → AgBr + KNO3

#2 0.5M KCl +

0.1M AgNO 3

+ 2M NH4OH

The light yellow precipitate was formed

Then after adding NH4OH, the light yellow precipitate

AgNO3+KBr → AgBr + KNO3 AgBr + 2NH OH →4 4[Ag(NH )3 2]Br +

started to dissolve H O 2 Comments:

#1:

KBr dissociates to form ion K and Br +

-In the solution, ion Br combines with ion Ag to form AgBr precipitate which is- + yellow because it has low solubility

#2:

KBr dissociates to form ion K and Br +

-In the solution, ion Br combines with ion Ag to form AgBr precipitate which is- + yellow

After adding NH OH, The precipitate AgBr reacts with NH OH to create the4 4 [Ag(NH )3 2]Br, which dissolves completely in the solution

3/ Reactions of H2O2

Procedure:

Step 1: Add 1 drop of 0.1M KMnO4; 5 drops of 0.1M KI; 5 drops of 3% H2O2 into test tube #1, #2, and #3, respectively

Step 2: Add 5 drops of 2M H2SO4 into both test tube #1 and #2 Then add a pinch of MnO2 into test tube #3

Step 3: Add 5 drops of 3% of H2O2 into both test tube #1 and #2

Step 4: Mix tubes gently and wait at least 2 minutes, then observe the phenomenon Reaction Observation Chemical

equation

Image

#1

0.1M KMnO4 +

2M H2SO4 +

3% H2O2

The solution transforms lightly purple then the air bubbles appear

2KMnO4 + 3H2SO4 + 5H2O2→ 8H O +2 2MnSO4 + 5O + 2

K SO2 4

#2

0.1M KI + 2M

H2SO4 + 3%

H O2 2

After adding H2O2 the color changed into light-orange color and purple precipitate appeared

2KI + H2SO4+

H O2 2 → K2SO4 + 2H2O + I2

#3 3% H2O2 + A

pinch of

MnO2

After adding MnO , the 2 release of gas and heat appeared and the black-gray solid didn’t dissolve

H O2 2→ O + H O2 2

#1:

In this reaction, H2O2 reacts with H2SO4 to form O : 2

2H O2 2 + 2H2SO4→ 2H O +2SO +O2 2 2

Oxygen on the previous reaction continues to react with KMnO This chemical is 4 dissociated to form Mn2+ and Mn3+:

2H O2 2 + 6H + 2MnO4 + -→ 5H O + 2Mn + 8H O + 5O2 2+ 2 2

#2:

In this reaction, firstly is the reaction between KI and H2SO :4

2KI + H2SO4→ K2SO4 + 2HI

Next, HI reacts with H2O2 to form I gas and H2 2O:

2HI + H2O2→ I + 2H O2 2

So that the precipitate in this reaction is I 2

#3:

H O2 2 in this reaction is decomposed and MnO plays a role as catalyst So that MnO2 2 promote the speed of the reaction to form O (gas) so we can observe many bubbles on2 the test tube

4/ Reactions of KMnO4

Procedure:

Step 1: Add 10 drops of 0.5M Na2SO3 into each three test tubes #1, #2 and #3 Step 2: Add 5 drops of 2M H2SO4 into test tubes #1; 5 drops of 6M NaOH into test tube #2; 5 drops of distilled water into test tubes #3

Step 3: Add 5 drops of 0.1M KMnO4 into test tube #1, #2, and #3

Step 4: Mix tubes gently and observe the phenomenon

Reaction Observation Chemical equation Image

#1 0.5M Na2SO3

+ 2M H2SO4

+ 0.1M

KMnO4

The solution changed to light purple color and no participate formed

2KMnO4 + 3H2SO4 + 5Na SO2 3→ 2MnSO4 + K2SO4 + 3H O + 2 5Na SO2 4

#2 0.5M Na2SO3

+ 6M NaOH

+ 0.1M

KMnO4

After putting the three chemicals into the test tube, the solution turned dark green After a few seconds, it turned to brown, then the brown

2KMnO4 + 2NaOH +

Na SO2 3→ KOH +

Na MnO2 4 + Na2SO4 + H O + MnO2 2

precipitate appeared.

#3 0.5M Na2SO3

+ distilled

water + 0.1M

KMnO4

The solution changed to dark-brown color The brown precipitate appeared

2KMnO4 + 3Na2SO3 +H2O → 2MnO + 2 3Na SO2 3 + 2KOH

Comments:

#1:

KMnO4 is a chemical that can oxidize many substances in the acid environment When adding KMnO into the solution includes Na4 2SO3 and H2SO4, KMnO reacts with4

H SO2 4: 4KMnO4 + 6H2SO4 → 2K2SO4 + 4MnSO + 6H O + 5O4 2 2

Na SO2 3 is oxidized to form Na2SO :4

Na SO2 3 + H2SO 4→ Na2SO4 + SO + H O2 2

H SO2 4 in this reaction plays a role as a oxidized substance that promotes the oxidized process

#2:

In this reaction, there is a chain of reactions below:

KMnO4 reacts with NaOH to form K2MnO4 and H O, ion Mn is reduced to Mn :2 7+ 2+ 3KMnO4 + 2NaOH → K2MnO4 + MnO + 2H O2 2

KMnO4 continues to react with Na2SO3 to form MnO , Na2 2SO4, and Na2MnO :4 3KMnO4 + 5Na2SO3 + 2H O → 3MnO + 5Na2 2 2SO4 + 2Na2MnO4 + 2KOH

The precipitate here is MnO 2

#3:

In H O environment, Na2 2SO3 is oxidized to form Na2SO4, KMnO is reduced to form4 MnO2 (the brown precipitate)

5/ Reactions of Fe2+

and Fe3

Section 1: Ferric ion (Fe )3+

Procedure:

Step 1: Put 10 drops 0.5M FeCl3 into each two test tubes

Step 2: Put 5 drops 2M KOH and 5 drops 2M NH OH into the first and the second test4 tubes, respectively

Step 3: Observe the phenomenon

Reaction Observation Chemical equation Image

#1 0.5M FeCl +3

2M KOH

The solution becomes red-brown and the precipitation also has red-brown color

FeCl3 + 3KOH Fe(OH)3 + 3KCl

#2 0.5M FeCl + 3

2M NH4OH

There appears red-brown precipitation on the reaction

FeCl3 + 3NH OH 4 3NH4Cl + Fe(OH)3

Comments:

#1:

FeCl3 is dissociated: FeCl → Fe + Cl3 3+

-KOH is dissociated: -KOH → K + OH+ -

The cation Fe and the anion OH combine together to form the precipitation Fe(OH) 3+

-3

#2:

FeCl3 is dissociated: FeCl → Fe + Cl3 3+

-NH4OH is dissociated: NH OH → NH + OH4 4+

-The cation Fe and the anion OH combine together to form the precipitation Fe(OH) 3+

-3

Section 2: Ferrous ion (Fe )2+

Step 1: Put 10 drops 0.5M FeSO into each two test tubes4

Step 2: Put 5 drops 2M KOH and 5 drops 2M NH OH into the first and the second4 test tubes, respectively

Step 3: Observe the phenomenon

equation

Image

#1 0.5M FeSO +4

2M KOH

The dark green precipitate

is formed after keeping it

on sphere for some minutes

FeSO4 + 2KOH

K SO2 4 + Fe(OH)2

#2 0.5M FeCl + 3

2M NH4OH

There appears red-brown precipitation on the reaction

FeSO +2NH OH4 4 -> (NH4)2SO4 + Fe(OH)2

Comments:

#1:

FeSO4 is dissociated: FeSO4 Fe + SO2+ 4

2-KOH is dissociated: 2-KOH K + OH +

-The cation Fe and the anion OH combine together to form the precipitation2+ -Fe(OH) 2

#2:

FeSO4 is dissociated: FeSO Fe + SO4 2+ 4

2-NH4OH is dissociated: NH OH NH + OH 4 4+

-The cation Fe and the anion OH combine together to form the precipitation2+ -Fe(OH) 2

6/ Reactions of Al3+

Procedure:

Step 1: Put 10 drops Al2(SO )4 3 into each two test tubes

Step 2: Continue to put 5 drops 2M NaOH into each two test tubes Observe the phenomenon

Step 3: Put 20 drops 2M HCl and 20 drops 2M NaOH into the first and the second test tube, respectively

Step 4: Observe the phenomenon

Reaction Observation Chemical equation Image

#1

and

#2

0.5M

Al (SO2 4)3 +

2M NaOH

The white precipitation is formed

Al (SO )2 4 3 + NaOH

Na SO2 4 + Al(OH)3

#1 0.5M

Al (SO2 4)3 2M

NaOH + 2M

HCl

After the white precipitation is formed

by the reaction between

Al (SO2 4)3 + NaOH, when adding HCl solution, the white precipitation disappears, a no-color solution is form

Al(OH)3 + 3HCl AlCl3 + 3H O 2

#2 0.5M

Al (SO2 4)3 +

2M NaOH +

2M NaOH

The white precipitation disappears

Al(OH)3 + NaOH NaAlO2 + 2H O2

Comments:

#1

Al (SO2 4)3 is dissociated: Al2(SO )4 3 Al + SO3+ 4

2-NaOH is dissociated: 2-NaOH Na + OH+

-The cation Al and the anion OH combine together to form the precipitation Al(OH) 3+

-3 When adding more 2M HCl solution, it continues to react with Al(OH) to form new3 substances which are AlCl (due to ion Al combined to Cl ) and H O The white3 3+ - 2 precipitation which is Al(OH) is disappeared.3

#2:

Al (SO2 4)3 is dissociated: Al2(SO4)3 Al + SO3+

4 2-NaOH is dissociated: 2-NaOH Na + OH+

-The cation Al and the anion OH combine together to form the precipitation Al(OH) 3+ -

3 When adding more 2M NaOH solution, it continues to react with Al(OH) to form3 new substances which are NaAlO (due to ion Na combined AlO ) and H O The2 + 2- 2 white precipitation which is Al(OH) is disappeared.3

7/ Flame tests

Procedure:

Step 1: Prepare 5 test tubes containing LiCl, NaCl, KCl, CaCl2, and BaCl2 solutions Step 2: Light the burner

Step 3: We clean the loop with distilled water

Step 4: Dip a looped into one of the solutions supplied

Step 5: Hold it in the flame, after saw the change in the dominant flame color, record it

Solution Dominant flame color Wavelength

(nm)

Frequency (s )−1

Photon energy (J)

LiCl

Red

701 4.28 x 1014 2.83 x 10-19

NaCl

Yellow

587 5.11 x 1014 3.39 x 10-19

CaCl2

Orange

609 4.93 x 1014 3.27 x 10-19

BaCl2 Light Green 577 5.2 x 10 3.45 x 10 Comments:

We observe the color of the fire when heating LiCl, NaCl, KCl, CaCl and BaCl As2 2 can be seen, the color when heating them vary mostly due to the different rates of electrons in different substances

These are the two equations to calculate the table above

1 C = λ x v

C: the speed of light (3 x 10 m/s)8

λ: the wavelength (nm)

v: frequency (s )-1

2 Ephoton = h x v

Ephoton: the energy per photon (J)

h: (Planck’s constant) 6.626 x 10 (J.s)-34

V/ Suggested questions

1/ What are the purposes of today’s lab report?

The target of Experiment 1 is:

Doing all the chemical reactions consists of acid-base, precipitation, gas forming, complex compound forming and oxidation-reducing reactions

Analyzing the results of these reactions (the products ) and discussing the chemical changes

Complete the chemical equation and balance them for the reactions observed

2/ What is a chemical reaction?

Chemical reaction is the process of converting a set of substances (the reactants) into another set of substances (the products) This reaction is related to the changes of the position of the electrons in forming and breaking the chemical bonds between atoms

3/ Please give examples of different types of chemical reactions