Practical work No. 1
Carrying out a chain of chemical transformations
Carry out reactions in which the chemical transformations proposed below are carried out (according to options).
Write down equations for the corresponding reactions. Write ion exchange reactions also in ionic form.
Option 1
MgCO 3 → MgCl 2 → Mg(OH) 2 → MgSO 4
Option 2
CuSO 4 → Cu(OH) 2 → CuO → Cu
Option 3
ZnCl 2 → Zn(OH) 2 → ZnCl 2
↓
Na 2
Practical work No. 2
Preparation and properties of metal compounds
Exercise 1
In mathematics there is a rule - “the sum does not change if the places of the terms are changed.” Is this true for chemistry? Check this out with the following experiment.
Prepare aluminum hydroxide by an exchange reaction and prove its amphoteric nature. To do this you can use the following reaction:
А1Сl 2 + 3NaOH = Al(OH) 3 ↓ + 3NaCl.
Carry out this reaction in two variants, using the same volumes of starting substances in each variant: first, add dropwise a solution of another reagent to a solution of one of the starting substances (reagent), then change the sequence of introducing the reagents into the reaction. Observe in which case a precipitate will form and in which case it will not.
Explain the results and write down the equations for the reactions performed in molecular and ionic forms.
Task 2
Carry out reactions to confirm the qualitative composition of calcium chloride. Write down the reaction equations in molecular and ionic forms.
Task 3
Carry out transformations according to the following scheme 1:
Fe → FeCl 2 → FeCl 3.
- 1 To carry out the second transformation, use chlorine water.
Write the equations for the corresponding reactions and consider them from the standpoint of oxidation-reduction. Carry out qualitative reactions to confirm the presence of reaction products. Write down the reaction equations in molecular and ionic forms.
Task 4
Obtain iron (II) sulfate in at least three ways. Write the equations for ion exchange reactions in ionic and molecular forms, and consider substitution reactions from the standpoint of oxidation-reduction.
Carry out reactions to confirm the qualitative composition of iron (II) sulfate. Write down the equations for the corresponding reactions in molecular and ionic forms.
Practical work No. 3
Experimental tasks for recognizing and obtaining metal compounds
The three test tubes given to you (options 1, 2 or 3) contain solid substances, and the other three (option 4) contain solutions of substances.
Option 1
- a) sodium hydroxide;
b) potassium carbonate;
c) barium chloride.
Option 2
- a) calcium carbonate;
b) sodium sulfate;
c) potassium chloride.
Option 3
- a) barium nitrate;
b) sodium sulfate;
c) calcium carbonate.
- a) sodium chloride;
b) aluminum chloride;
c) iron (III) chloride.
Determine experimentally which test tube contains each of the substances given to you. Write the equations for the corresponding reactions in molecular and ionic forms.
After this part of the work, complete one or two experimental tasks from the following list (as directed by the teacher).
Problem 1
Prove experimentally that the iron sulfate, a sample of which you were given, contains an admixture of iron (III) sulfate. Write the equations for the corresponding reactions in molecular and ionic forms.
Problem 2
Obtain iron(III) oxide starting from iron(III) chloride. Write the equations for the corresponding reactions, and the equation for the reaction involving an electrolyte and in ionic form.
Problem 3
Prepare a solution of sodium aluminate starting from aluminum chloride. Write down the equations of the reactions performed in molecular and ionic form.
Problem 4
Obtain iron (II) sulfate starting from iron. Write down the equations of the reactions performed and analyze redox processes.
Repeat and consolidate practical skills in performing chemical experiments, handling reagents, and observing safety regulations;
- learn to select the reagents necessary for work, assume observed phenomena, and draw conclusions;
- consolidate skills in drawing up equations of ion exchange reactions, drawing up dissociation equations, full and abbreviated ionic equations.
- Developmental: continue to develop self-education skills - work with teaching aids and additional literature.
- Educational:
Continue the formation of ideological concepts about the knowability of nature, the cause-and-effect relationship between the composition, structure and properties of substances;
- students must be able to work carefully and consciously follow established rules (for example, safety precautions).
Equipment: graphic projector with code films, solubility table, TV, programmed teaching aid, tables for filling out a work report and reference tables ( Annex 1), racks with test tubes, trays, waste bottles, hourglasses, indicators - phenolphthalein and litmus, solutions of barium chloride, iron (II) sulfate, sodium carbonate, sulfuric acid, silver nitrate, red blood salt, sodium hydroxide, calcium chloride, copper (II) sulfate, calcium hydroxide, hydrochloric acid. To solve problems of recognizing substances, students are given solutions of sulfuric acid, calcium hydroxide and calcium chloride in numbered bottles.
Lesson structure:
- Organizing time. 1 min.
- Motivation. 1 min.
- Repetition of methods for determining cations and anions in solutions. 2 minutes.
- A message about the procedure for performing experiments and evaluating the work. 2 minutes.
- A reminder about the structure of the programmed teaching aid. 1 min.
- Completing tasks using a programmed teaching aid. 35 min.
- Summarizing. 3 min.
During the classes
Motivation. An entire science, analytical chemistry, is engaged in recognizing substances and proving their composition. It employs more people than chemical production.
Repetition. Let's remember the methods for determining cations and anions in solutions (you can use the provided reference materials):
- flame coloring (the only way to detect sodium). The teacher shows a fragment of a video film;
- precipitation reactions (little and insoluble substances are formed - white or colored precipitates);
- color reactions - usually a change in the color of indicators in acidic and alkaline solutions;
- reactions that release gases, such as carbon dioxide. The teacher conducts demonstration experiments.
Sequence of work execution.
You have to complete 4 experiments on your own. Each of the first three takes 7 minutes. If the time required is greater, the third experiment may not be performed. Use an hourglass to control time. At the end of the lesson, you give the teacher the answer to the substance recognition task (experiment 4) in the form of two completed tables. At the end of the lesson, you receive two grades: for completing the test experiment and for completing all the work.
Sequence of work with programmed guide(Table 1). You read the first task, printed on the left page of the textbook spread at the top, and write down the missing word, formulated answer, reaction equation on this page. On the left side of the right page of the spread, separated by a vertical line, the necessary explanations and drawings are provided to help you arrive at the correct answer. Having completed the task, turn the page and on the right side of the next spread, find the answer and match what you wrote down with the correct one, printed under the same number.
Once you have received confirmation that your answer is correct, you can move on to the next task, which is printed at the top of the left page of the next spread and is numbered one more than the previous one.
Before conducting experiments, read the safety regulations.
Safety regulations:- Substances should not be touched by hand or tested for taste and smell.
- Do not mix substances you do not know unless instructed by your teacher.
- When performing experiments, use small doses of substances.
- Handle acids and alkalis with care.
- If solutions get on your hands or clothing, wash them off immediately with plenty of water.
- After work, wash your hands with soap.
- Use only clean laboratory glassware.
- Do not empty the remaining substances or pour them back into the container with clean substances.
I have read the safety rules (a) ………………… (signature)
Table 1
Programmed aid
| Left page spread of the manual | Right page spread of the manual | |
| Exercise | Explanation of the task | Answer |
| Experience 1 Confirm the quality composition of barium chloride 1. In an aqueous solution, barium chloride dissociates into ions BaCl 2 = Ba 2+ + 2Cl - Therefore, it is necessary to prove the presence of cations in the solution...... using qualitative reactions. and anions...... |
||
| 2
. According to table 2 ( Annex 1) select appropriate reagents The reagent for barium cations is ...... - anion, ...... The reagent for chloride - anions are cations...... |
1
.
Cl - (chloride anions) |
|
| 3
. To carry out the reaction, pour two samples of the original solution, each 0.5 ml in volume, into two test tubes 4. Add to the first test tube a colorless transparent solution of sulfuric acid......containing sulfate anions BaCl 2 + H 2 SO 4 = BaSO 4 + 2HCl Ba 2+ + 2Cl - + 2H + + SO 4 2- = BaSO 4 + 2H + + 2Cl - Ba 2+ + SO 4 2- = BaSO 4 Checking equations by sum of coefficients: in the molecular equation...... in the complete ionic equation…… in the reduced ionic equation…… |
|
2
. sulfate -, SO 4 2- silver, Ag+ |
| 5
. Add silver nitrate solution......containing silver cations to the second test tube A…… precipitate is formed as a result of the reaction BaCl 2 + 2AgNO 3 = Ba(NO 3) 2 + 2AgCl Ba 2+ + 2Cl - + 2Ag + + 2NO 3 - = Ba 2+ + 2NO 3 - + 2AgCl Ag + + Cl - = AgCl Sum of odds: in the molecular equation...... in the complete ionic equation…… in the reduced ionic equation…… |
|
4 . |
| Conclusion Using precipitation reactions, we proved that the barium chloride solution contains cations ...... and anions ......, thereby confirming the composition of the given salt |
5
. white curd |
|
| Experience 2 Confirm the quality composition of iron (II) sulfate FeSO 4 = Fe 2+ + SO 4 2- Therefore, it is necessary, using qualitative reactions, to prove the presence of cations...... and anions...... in the solution. |
||
| 2
. According to tables 2 and 3 ( Annex 1) select appropriate reagents The reagent for doubly charged iron cations is an alkali solution containing ...... - anions or a solution of red blood salt ...... The reagent for sulfate anions is barium cations...... |
1
.
SO 4 2-, sulfate anions |
|
| 3
. To carry out the reaction, pour three samples of the original solution, each 0.5 ml in volume, into three test tubes 4. Add sodium hydroxide solution to the first test tube A precipitate……color is formed as a result of the reaction FeSO 4 + 2NaOH = Na 2 SO 4 + Fe(OH) 2 Fe 2+ + SO 4 2- + 2Na + + 2OH - = 2Na + + SO 4 2- + …… Fe 2+ + 2OH - = …… |
|
2
. OH - , hydroxide – |
| 5
. Add a solution of red blood salt K 3 to the second test tube A precipitate……color is formed as a result of the reaction 3FeSO 4 + 2K 3 = 3K 2 SO 4 + Fe 3 2 3Fe 2+ + 3SO 4 2- + 6K + + 2 2- = 6K + + 3SO 4 2- + Fe 3 2 3Fe 2+ + 2 2- = Fe 3 2 The sums of the coefficients in the above equations are respectively equal to ……, ……, …… (When performing control work, only one qualitative reaction is carried out for the ion being determined) |
|
4
. greenish |
| 6
. Add barium chloride solution to the third test tube...... A precipitate……color is formed as a result of the reaction FeSO 4 + BaCl 2 = BaSO 4 + FeCl 2 Fe 2+ + SO 4 2- + Ba 2+ + 2Cl - = BaSO 4 + Fe 2+ + 2Cl - …… + …… = …… The sums of the coefficients in the given equations are respectively ……, ……, …… |
|
5 . |
| Conclusion Using precipitation reactions, we proved that iron (II) sulfate contains a cation ...... and an anion ...... |
6
.
Ba 2+ + SO 4 2- = BaSO 4 v |
|
| Experience 3 Confirm the quality composition of sodium carbonate 1. In an aqueous solution, this salt dissociates into ions Na 2 CO 3 = …… + …… Therefore, it is necessary, using qualitative reactions, to prove the presence of cations ...... and CO 3 2- (...... - anions) in the solution |
||
| 2
. According to tables 1 and 2 ( Annex 1) select the appropriate qualitative reactions Sodium is determined by the color of the colorless flame of a gas burner (no experiment is carried out during the work). The reagent for carbonate anions is cations...... and acid solutions containing cations...... |
1
.
Na + and (carbonate anions) |
|
| 3
. To carry out qualitative reactions to carbonate ions, pour into two test tubes samples of the initial solution with a volume of 0.5 ml each 4. Add to the first test tube a solution of calcium chloride...... (or calcium hydroxide......) containing cations...... A white precipitate is formed, which dissolves when hydrochloric acid is added...... (at the same time, bubbles of transparent colorless gas appear in the test tube) When a precipitate forms, a reaction occurs Na 2 CO 3 + CaCl 2 = 2NaCl + CaCO 3 2Na + + CO 3 2- + Ca 2+ + 2Cl - = 2Na + + 2Cl - + CaCO 3 …… + …… = …… The sum of the coefficients in the equations is respectively ……, ……, ……. |
|
2 . |
| 5
. Add hydrochloric acid solution to the second test tube...... An odorless gas is released, causing limewater to become cloudy (evidence of CO2 evolution: moisten glass with calcium hydroxide solution and hold over test tube until cloudy) Na 2 CO 3 + 2HCl = 2NaCl + CO 2 + H 2 O 2Na + + CO 3 2- + 2H + + 2Cl - = 2Na + + 2Cl - +CO 2 + H 2 O 2H + + CO 3 2- = CO 2 + H 2 O Sums of coefficients ……, ……, …… |
|
4
. CaCl 2 or Ca(OH) 2 Ca 2+ (calcium) Ca 2+ + CO 3 2- = CaCO 3 v |
| Conclusion Using precipitation reactions and gas evolution reactions, we have proven that sodium carbonate solution contains …… – anions CO 3 2- |
5. | |
| Experience 4.(Substance recognition task) Using characteristic reactions, recognize the solutions of sulfuric acid, calcium hydroxide and calcium chloride contained in three numbered bottles (To recognize means to determine experimentally what substance is in each bottle) 1. Substances found in the solutions provided belong respectively to the classes ......, ....... and ......, and are (strong / weak) ...... electrolytes In an aqueous solution, these substances dissociate into ions H 2 SO 4 = 2H + + SO 4 2- Ca(OH) 2 = Ca 2+ + 2OH - CaCl 2 = Ca 2+ + 2Cl - Therefore, it is necessary, using qualitative reactions, to prove the presence of the following cations in the solution: H +, Ca 2+, and anions: SO 4 2-, OH -, Cl - |
||
| 2
. According to tables 2 and 3 ( Annex 1) select the appropriate reagents Determined ion: Reagent: hydrogen cation H+…… calcium cation Ca 2+…… hydroxide - anion OH - …… sulfate - anion SO 4 2- …… chloride - anion Cl - …… |
1
.
base - (alkali) strong |
|
| 3
. To carry out the reactions, pour 0.5 ml of each of the three samples into three clean test tubes Using the solubility table, select the sequence of adding reagents so that in one experiment you can form a precipitate in only one test tube: 5…… (may have no experience) |
|
2
.
CO 3 2-, Na 2 CO 3 litmus or phenolphthalein |
| 4
. Add reagent #1 to three sample tubes. Record your observations in worksheet 2 5. Add reagent #2 to three new sample tubes. Write down your observations in table 2. If using reagents 1 and 2 you have established the qualitative composition of one of the samples, you can write it down in the corresponding line at the bottom of the table. No further experiments are carried out with this sample. 6. Add Reagent #3 to the remaining samples. Record your observations By analogy, continue working with reagents No. 4 and No. 5 |
|
3
. 1 or 2 - BaCl 2 2 or 1 - litmus test 3, 4, 5 - your options |
| 7 . Fill out tables 2 and 3 and submit for verification | ||
Homework. In addition to the abbreviated ionic equations of experiment 4 worksheet, write molecular and full ionic equations in your notebook.
table 2
Results of solving the recognition problem
Table 3
Report on the performance of the recognition task (experiment 4)
Lesson topic: Practical work No. 1 Preparation and properties of metal compounds.
Purpose of the lesson: to review the basic questions of metal chemistry. In practice, consolidate knowledge about the basic properties of metals, qualitative reactions on metals.Equipment: sets of chemical reagents and equipment for practical work.During the classes
1. Organizational part.2. Repetition of safety rules when working with caustic substances.3. Carrying out work according to the instructions in the textbook, pp. 84 – 85, grade 9, Gabrielyan O.S.:Exercise 1 In chemistry, this rule is not true. The result of a reaction is often determined by the order in which the reactants are combined and their ratio. Let's prove it.1) Add an alkali solution drop by drop to a test tube with a solution of aluminum chloride:A1S1 3 + 3NaOH(deficiency) = 3NaCl + Al(OH) 3 ↓ Al 3+ + 3Cl - + 3Na + + 3OH - = A1(OH) 3 ↓ + 3Na + + 3Cl - A1 3+ + 3OH - = Al(OH)3↓We observe the formation of a white precipitate of aluminum hydroxide.2) Add the solution to another test tube with an alkali solutionaluminum chloride. In this case, the alkali is present in excess, so A1(OH) 3 at the beginning it is not formed, sodium aluminate is formed:A1S1 3 + 4NaOH(excess) = NaA1O 2 + 3NaCl + 2H 2 ABOUTA1 3+ + 3Cl - + 4Na + + 40N - = Na + + A1O 2 - + 3Na + + 3Cl - + 2H 2 ABOUT A1 3+ + 4OH - = A1O 2 - + 2H 2 ABOUTOnly after adding excess A1C13 will a precipitate of A1(OH)3 precipitate.3) Let us prove the amphoteric nature of A1(OH) 3 . For this purpose, the resulting precipitate A1(OH) 3 divide into 2 test tubes. Add a solution of any strong acid to one of the test tubes, and an alkali solution (excess) to the other. In both cases we observe the dissolution of the aluminum hydroxide precipitate:A1(OH) 3 + 3НС1 = А1С1 3 + 3H 2 ABOUTA1(OH) 3 + 3H + + 3Cl - = A1 3+ + 3Cl - + 3H 2 ABOUTA1(OH) 3 + 3H + = A1 3+ + 3H 2 ABOUTA1(OH) 3 +NaOH = NaA1О 2 + 2H 2 ABOUTA1(OH) 3 +Na + + HE - = Na + +A10 2 - + 2H 2 ABOUTA1(OH) 3 + HE - = A1O 2 - + 2H 2 ABOUTThus, aluminum hydroxide dissolves in both acids and alkalis, so it is amphoteric.Task 2 Task 2 To prove the qualitative composition of CaC1 2 Let's carry out reactions characteristic of calcium cation and chloride anion. For this purpose, a solution of CaCl 2 pour into 2 test tubes.Add sodium carbonate solution to one of them: Na 2 C ABOUT 3 + CaC1 2 = CaC ABOUT 3 ↓ + 2 NaCl 2 Na + + CO 3 2- + Ca 2+ + 2C l - = CaCO 3 ↓ + Na + + 2C l - Sa 2+ + CO 3 2- = CaC ABOUT 3 ↓ We observe the release of a white precipitate of calcium carbonate CaCO 3 Pour the silver nitrate solution into another test tube.CaC1 2 + 2 AgN ABOUT 3 = Ca ( N ABOUT 3 ) 2 + 2 AgCl ↓ Sa 2+ + 2C l - + 2 Ag + + 2 N ABOUT 3 - = Ca 2+ + 2 N ABOUT 3 - + 2 AgCl ↓ WITH l - + Ag + = AgCl ↓ We observe the release of a white cheesy sediment.Task 3 It is necessary to carry out the following transformations: Fe → FeCI 2 → FeCl 3 Add a solution of hydrochloric acid to a test tube with iron filings. We observe the dissolution of iron and the release of hydrogen gas: Fe 0 + 2H + C1 = Fe 2+ Cl 2 + N 2 0 Fe 0 - 2e = Fe 2+ 2 1 reducing agent2H + +2e = N 2 0 2 1 oxidizing agent
Let's prove the presence of iron ions(II). To do this, add a solution of red blood salt to the test tube:
Qualitative reaction to iron (II) ion: K 3 + Fe +2 C1 2 = 2 KS 1 + KFe +3 redbloodsaltTurnbulevablue
3 TO+ + Fe 2+ + 2 WITH l - + 3- = KFe ↓ + 2K + + 3 WITH l - K + + Fe 2+ + 3 - = KFe ↓ We observe the formation of a dark blue precipitate of Turnboole blue, therefore, ionsFe 2+ received.To carry out the second transformation, we use chlorine water, which is a solution of chlorine in water, i.e. it is a C1 reagent 2 . 2Fe 2+ Cl 2 + C1 2 0 = 2Fe 3+ Cl 3
Fe 2+ -le= Fe 3+
2 Fe 2+ + C.I. 2 ° = 2 Fe 3+ + 2C l - The color of the solution changes.Let us prove the presence of iron (III) ions. To do this, you can carry out one of the suggested reactions:Qualitative reactions to iron ion ( III ): a) Add a solution of yellow blood salt to the test tube: K 4 + Fe +3 C1 3 = 3KCI + KFe +3 lFe +2 (CN) 6 ]↓ yellowbloodsaltBerlinazure 4 TO+ + Fe 3+ + 3 WITH l - + 4- = KFe ↓ + 3K + + 3 WITH l - TO+ + Fe 3+ + 4- = KFe ↓ We observe the formation of a dark blue precipitate of Prussian blue, which means that iron (III) ions are present in the solution.b) Add to the test tube with the solution F eS1 3 ammonium or sodium thiocyanate: Fe +3 C.I. 3 + NaNCS = [ FeNCS ] Cl 2 + NaCI sodium thiocyanate Fe 3+ + NCS - = FeNCS 2+ Task 4Need to getFeSO 4 three different ways:Pour a dilute solution of sulfuric acid into a test tube with iron filings. We observe the dissolution of iron and the release of hydrogen:Fe° + H 2 + " S0 4 -» Fe +2 S0 4 + H 2 °
reducing agent oxidizer
- 2e = Fe 2+
2H + +2e = H 2°
As a result of the reaction, ferrous sulfate is formed.Add iron to a test tube with copper sulfate solution. We observe a change in the color of the solution; from blue the solution becomes light green, which quickly turns yellow and becomes cloudy. As a result of the reaction, red copper is released.Cu 2+ SO 4 + Fe° = Fe +2 SO 4 + Cu 0 ↓ Blue light green redFe° -2e= Fe 2+ reducing agent
2+ +2e = Cu° oxidizer
For gettingFeSABOUT 4 Let's carry out the following transformations:FeCl 2 → Fe(OH) 2 → FeSABOUT 4 Add an alkali solution to the ferric chloride solution:FeCl 2 + 2NaOH = 2NaCl + Fe(OH) 2 ↓ Fe 2+ + 2 WITHl - + 2Na + + 2 ABOUTH - = 2Na + + 2 WITHl - + Fe(OH) 2 ↓ Fe 2+ + 2ОH - = Fe(ABOUTH) 2 ↓ As a result of the reaction, a white precipitate of iron (II) hydroxide is formed.To the sediment obtained in the previous experimentFe(OH) 2 add a solution of sulfuric acid:Fe(OH) 2 + H 2 SABOUT 4 = FeSABOUT 4 + 2H 2 ABOUTFe(OH) 2 + 2H + + SABOUT 4 2 - = Fe 2+ + SABOUT 4 2- + 2 H 2 ABOUTFe(OH) 2 + 2 H + = Fe 2+ + 2 H 2 ABOUTTask 5To prove the qualitative composition of FeSO4, pour a solution of iron sulfate into 2 test tubes. Add a solution of red blood salt to one of them:K3 + FeS04 = K2S04 + KFe ↓red blood salt turnbull's blueWe observe the formation of a dark blue precipitate of Turnboule blue, which means that iron ions - Fe2+ - are present in the solution.In another test tube, add barium chloride solution:FeSO4+You12 = FeCl2 + BaS04↓Fe2+ + SO42- + Ba2+ + 2WITHl- = Fe2+ + 2WITHl- + BaSO4↓Ba2+ + SO42- = BaSO4↓We observe the release of a white precipitate of barium sulfate BaSO4, which means that SO sulfate ions are present in the solution 4 2- .
Homework. Complete the work by completing all reaction equations.§ 14 (to the end), ex. 2, 3, 7
Chemistry lesson plan, grade 9.
Topic: Practical work No. 2 Preparation and properties of metal compounds
Lesson location: 9th grade. Topic I I. Metals
Lesson type : practical work
Goal of the work:Educational :
Preparation of metal compounds experimentally;
Application of theoretical knowledge in solving experimental problems;
Improving skills in conducting ion exchange reactions;
Review the properties and some methods of obtaining the main classes of chemicals;
Developmental – promote the development of logical thinking, the development of observation, the ability to explain, analyze, compare, and conduct a chemical experiment;
Educational - fostering interest in the subject.
Equipment: test tube devices, test tube holder, measuring spoon, glass rod, alcohol lamp.
Materials - aluminum chloride, sodium hydroxide, calcium chloride, sodium carbonate, silver nitrate, iron filings, hydrochloric acid, sulfuric acid, iron (III) chloride, copper sulfate, sodium sulfate.
During the classes
1. Organizational moment.
2.Updating knowledge
Today we have an unusual lesson - practical work. In mathematics there is a rule: Rearranging the places of the terms does not change the sum. Do you think this rule applies in chemistry?
II. Setting goals and objectives for the lesson. Motivation for learning activities.
What is the topic of practical work?
What will we do in class? Formulate the purpose of practical work. (Consider the properties of metal compounds and find out whether a mathematical rule applies in chemistry)
III. Formation of skills based on their application in standard conditions.
Let's open the textbooks and see what experiments we will conduct (studying the instructions for conducting experiments).
Defining a plan of action.
What is the task before us?
What instruments and materials are needed to conduct experiments?
What safety rules should we follow when performing work?
IV. Formation of differentiated generalized skills.
P Under the guidance of teachers, they formulate the topic and purpose of practical work (based on the instructions) and write it down in notebooks.
V. Problem analysis.
Students receive tables to record observations:
When conducting practical work, we must fill out the table
Experiment No. 1 “Preparation of aluminum hydroxide”
Using equal volumes of starting substances: first, a solution of another reagent was added dropwise to a solution of one of the starting substances (reagent), then the sequence of introduction and reaction of the reagents were changed.
Experiment No. 2 “Confirmation of the qualitative composition of calcium chloride”
Conducted reactions confirming the qualitative composition of calcium chloride
A) A few drops of Na 2 CO 3 solution were added to a test tube with a CaCL 2 solution
B) A few drops of AgNO 3 solution were added to a test tube with a CaCL 2 solution
Experience No. 3 “Implementing a chain of transformations”
The transformations were carried out according to the following scheme
Fe-->FeCl2--->Fe(OH)2.
A) HCL solution was added to iron filings
B) NaOH solution was added to the FeCL 3 solution
Experiment No. 4 “Obtaining ferrous sulfate”
A) a solution of H 2 SO 4 was added to a solution of Fe (OH) 3
B) a solution of H 2 SO 4 was added to the iron filings
VI. Doing practical work
Exercise 1 In chemistry, this rule is not true. The result of a reaction is often determined by the order in which the reactants are combined and their ratio. Let's prove it.
1) Add an alkali solution drop by drop to a test tube with a solution of aluminum chloride:
А1С1 3 + 3NaOH(deficiency) = 3NaCl + Al(OH) 3 ↓
Al 3+ + 3Cl - + 3Na + + 3OH - = A1(OH) 3 ↓ + 3Na + + 3Сl -
A1 3+ + 3OH - = Al(OH)3↓
We observe the formation of a white precipitate of aluminum hydroxide.
2) Add the solution to another test tube with an alkali solution
aluminum chloride. In this case, the alkali is present in excess, so A1(OH) 3 is not formed at the beginning, sodium aluminate is formed:
A1C1 3 + 4NaOH (excess) = NaA1O 2 + 3NaCl + 2H 2 O
А1 3+ + 3Сl - + 4Na + + 40Н - = Na + + А1О 2 - + 3Na + + 3Сl - + 2Н 2 О
A1 3+ + 4OH - = A1O 2 - + 2H 2 O
Only after adding excess A1C13 will a precipitate of A1(OH)3 precipitate.
3) Let us prove the amphoteric nature of A1(OH) 3. To do this, divide the resulting precipitate A1(OH) 3 into 2 test tubes. Add a solution of any strong acid to one of the test tubes, and an alkali solution (excess) to the other. In both cases we observe the dissolution of the aluminum hydroxide precipitate:
A1(OH) 3 + 3HC1 = A1C1 3 + 3H 2 O
A1(OH) 3 + 3H + + 3Cl - = A1 3+ + 3Cl - + 3H 2 O
A1(OH) 3 + 3H + = A1 3+ + 3H 2 O
A1(OH) 3 + NaOH = NaA1O 2 + 2H 2 O
A1(OH) 3 + Na + + OH - = Na + +A10 2 - + 2H 2 O
A1(OH) 3 + OH - = A1O 2 - + 2H 2 O
Thus, aluminum hydroxide dissolves in both acids and alkalis, so it is amphoteric.
Task 2
To prove the qualitative composition of CaCl 2, we will carry out reactions characteristic of the calcium cation and chloride anion. To do this, pour the CaCl 2 solution into 2 test tubes.
Add sodium carbonate solution to one of them:
Na 2 CO 3 + CaC1 2 = CaCO 3 ↓ + 2NaCl
2Na + + CO 3 2- + Ca 2+ + 2Cl - = CaCO 3 ↓ + Na + + 2Cl -
Ca 2+ + CO 3 2- = CaCO 3 ↓
We observe the release of a white precipitate of calcium carbonate CaCO 3
Pour the silver nitrate solution into another test tube.
CaС1 2 + 2AgNO 3 = Ca(NO 3) 2 + 2AgCl↓
Ca 2+ + 2Cl - + 2Ag + + 2NO 3 - = Ca 2+ + 2NO 3 - + 2AgCl↓
Сl - + Ag + = AgCl↓
We observe the release of a white cheesy sediment.
Task 3
It is necessary to carry out the following transformations:
Fe → FeCI 2 → FeCl 3
Add a solution of hydrochloric acid to a test tube with iron filings. We observe the dissolution of iron and the release of hydrogen gas:
Fe 0 + 2H + C1 = Fe 2+ Cl 2 + H 2 0
Fe 0 - 2е = Fe 2+ 2 1 reducing agent
2Н + +2е = Н 2 0 2 1 oxidizing agent
Let's prove the presence of iron ions (II). To do this, add a solution of red blood salt to the test tube:
Qualitative reaction to iron (II) ion:
K 3 + Fe +2 C1 2 = 2KS1 + KFe +3
3К + + Fe 2+ + 2Сl - + 3- = KFe ↓ + 2K + + 3Сl -
K + + Fe 2+ + 3 - = KFe ↓
We observe the formation of a dark blue precipitate of Turnboole blue, therefore, ions Fe 2+ received.
To carry out the second transformation, we use chlorine water, which is a solution of chlorine in water, i.e. it is a C1 2 reagent.
2Fe 2+ Cl 2 + C1 2 0 = 2Fe 3+ Cl 3
Fe 2+ -le = Fe 3+ 2 reducing agent
Cl 2° + 2e = 2Cl - 1 oxidizing agent
2Fe 2+ +CI 2 ° = 2Fe 3+ + 2Cl -
The color of the solution changes.
Let us prove the presence of iron (III) ions. To do this, you can carry out one of the proposed reactions:
Qualitative reactions to iron (III) ion:
a) Add a solution of yellow blood salt to the test tube:
K 4 + Fe +3 C1 3 = 3KCI + KFe +3 lFe +2 (CN) 6 ]↓
yellow blood salt Prussian blue
4K + + Fe 3+ + 3Сl - + 4- = KFe ↓ + 3K + + 3Сl -
K + + Fe 3+ + 4- = KFe ↓
We observe the formation of a dark blue precipitate of Prussian blue, which means that iron (III) ions are present in the solution.
b) Add ammonium or sodium thiocyanate to the test tube with FeCl 3 solution: Fe +3 CI 3 + NaNCS = Cl 2 + NaCI
sodium thiocyanate
Fe 3+ + NCS - = FeNCS 2+
Task 4
It is necessary to obtain FeSO 4 in three different ways:
Pour a dilute solution of sulfuric acid into a test tube with iron filings. We observe the dissolution of iron and the release of hydrogen:
Fe° + H 2 + "S0 4 -" Fe +2 S0 4 + H 2 °
Fe° - 2e = Fe 2+ 1 reducing agent
2H + +2e = H 2 ° 1 oxidizing agent
As a result of the reaction, ferrous sulfate is formed.
Add iron to a test tube with copper sulfate solution. We observe a change in the color of the solution; from blue the solution becomes light green, which quickly turns yellow and becomes cloudy. As a result of the reaction, red copper is released.
Cu 2+ SO 4 + Fe° = Fe +2 SO 4 + Cu 0 ↓
Blue light green red
Fe° -2e= Fe 2+ 1 reducing agent
Cu 2+ +2е = Cu° 1 oxidizing agent
To obtain FeSO 4 we carry out the following transformations: FeCl 2 → Fe(OH) 2 → FeSO 4
Add an alkali solution to the ferric chloride solution:
FeCl 2 + 2NaOH = 2NaCl + Fe(OH) 2 ↓
Fe 2+ + 2Сl - + 2Na + + 2ОH - = 2Na + + 2Сl - + Fe(OH) 2 ↓
Fe 2+ + 2ОH - = Fe(ОH) 2 ↓
As a result of the reaction, a white precipitate of iron (II) hydroxide is formed.
To the Fe(OH) 2 precipitate obtained in the previous experiment, add a solution of sulfuric acid:
Fe(OH) 2 + H 2 SO 4 = FeSO 4 + 2H 2 O
Fe(OH) 2 + 2H + + SO 4 2 - = Fe 2+ + SO 4 2- + 2H 2 O
Fe(OH) 2 + 2H + = Fe 2+ + 2H 2 O
Task 5
To prove the qualitative composition of FeSO4, pour a solution of iron sulfate into 2 test tubes. Add a solution of red blood salt to one of them:
K3 + FeS04 = K2S04 + KFe ↓
red blood salt turnbull's blue
We observe the formation of a dark blue precipitate of Turnboule blue, which means that iron ions - Fe2+ - are present in the solution.
In another test tube, add barium chloride solution:
FeSO4 + BaС12 = FeCl2 + BaS04↓
Fe2+ + SO42- + Ba2+ + 2Сl- = Fe2+ + 2Сl- + BaSO4↓
Ba2+ + SO42- = BaSO4↓
We observe the release of a white precipitate of barium sulfate BaSO4, which means that sulfate ions SO 4 2- are present in the solution.
VII . Self-control of work performance.
Students fill out the table and draw conclusions for each experiment.
VIII . Lesson summary. Reflection.
The mathematical rule does not apply in chemistry by rearranging the positions of the terms. Sometimes the outcome of a reaction depends on the order in which the solutions are combined, as in the case of aluminum chloride and sodium hydroxide.
IX . Homework: complete the practical work in your notebook