In this article, we'll talk about genetic series of metals. Individual chemical substances usually divided into 2 groups: simple substances and complex.
This diagram provides a simplified view of genetic series of metals.
Above are a group of metals and hydrogen, the structure of which differs from the structure of atoms of other elements. There is 1 electron on the outer level, like in alkali metals, but at the same time, 1 electron is missing to fill the outer level.
Based on genetic series metals form basic oxides. Hydrogen forms a specific amphoteric oxide- water H 2 O, which, when interacting with a basic oxide, gives a base (alkali). Such reactions proceed, as a rule, without a change in the degree of oxidation. With change, only those reactions occur in which complex substances are formed from simple substances:
2 Cu + O 2 = 2 CuO,
Basic oxides react with non-metals, acid oxides, acids, acid salts.
Depending on the acid, metal, or non-metal, various salts are formed. For example:
Cu(HE) 2 + H 2 SO 4 \u003d SO 4 .
Instructions for students in the correspondence course "General Chemistry for Grade 12" 1. Category of students: the materials of this presentation are provided to the student for self-study topics "Substances and their properties", from the course general chemistry 12th grade. 2. Course content: includes 5 topic presentations. Each learning topic contains a clear structure educational material on a specific topic, the last slide is a control test - tasks for self-control. 3. Duration of study for this course: from one week to two months (determined individually). 4. Knowledge control: the student provides a progress report test items- a sheet with options for tasks, indicating the topic. 5. Evaluation of the result: "3" - 50% of tasks completed, "4" - 75%, "5"% of tasks. 6. Learning outcome: pass (fail) of the studied topic.
Reaction equations: 1. 2Cu + o 2 2CuO copper (II) oxide 2. CuO + 2 HCl CuCl 2 + H 2 O copper (II) chloride 3. CuCl NaOH Cu (OH) Na Cl copper (II) hydroxide 4. Cu (OH) 2 + H 2 SO 4 CuSO 4 + 2H 2 O copper (II) sulfate
genetic series organic compounds. If the basis of the genetic series is not organic chemistry are substances formed by one chemical element, then the basis of the genetic series in organic chemistry is made up of substances with the same number of carbon atoms in the molecule.
Reaction scheme: Each number above the arrow corresponds to a specific reaction equation: ethanal ethanol ethene ethane chloroethane ethine Acetic (ethanoic) acid
Reaction equations: 1. C 2 H 5 Cl + H 2 O C 2 H 5 OH + HCl 2. C 2 H 5 OH + O CH 3 CH O + H 2 O 3. CH 3 CH O + H 2 C 2 H 5 OH 4. C 2 H 5 OH + HCl C 2 H 5 Cl + H 2 O 5. C 2 H 5 Cl C 2 H 4 + HCl 6. C 2 H 4 C 2 H 2 + H 2 7. C 2 H 2 + H 2 O CH 3 CH O 8. CH 3 CH O + Ag 2 O CH 3 COOH + Ag
9 cells Lesson #47 Topic: " genetic connection Me, NeMe and their compounds".
Goals and objectives of the lesson:
Understand the concept of genetic connection.
Learn how to make genetic series of metals and non-metals.
Based on the knowledge of students about the main classes of inorganic substances, bring them to the concept of "genetic connection" and the genetic series of metal and non-metal;
To consolidate knowledge about the nomenclature and properties of substances belonging to different classes;
Develop skills to highlight the main thing, compare and generalize; identify and establish relationships;
Develop ideas about the cause-and-effect relationships of phenomena.
Restore in memory the concepts of simple and complex substance, about metals and non-metals, about the main classes of inorganic compounds;
To form knowledge about the genetic relationship and the genetic series, learn how to compose the genetic series of metals and non-metals.
Develop the ability to generalize facts, build analogies and draw conclusions;
Continue developing a culture of communication, the ability to express one's views and judgments.
Cultivate a sense of responsibility for the acquired knowledge.
Planned results:
Know definitions and classification of inorganic substances.
Be able to classify inorganic substances composition and properties; make up the genetic series of metal and non-metal;
illustrate with equations chemical reactions genetic relationship between the main classes of inorganic compounds.
Competencies:
cognitive skills : systematize and classify information from written and oral sources.
Activity skills : to carry out reflection of one's activity, to act according to the algorithm, to be able to compose the algorithm of a new activity, amenable to algorithmization; understand the language of diagrams.
Communication skills : build communication with other people - conduct a dialogue in pairs, take into account the similarities and differences in positions, interact with partners to obtain a common product and result.
Lesson type:
for the didactic purpose: a lesson in updating knowledge;
according to the method of organization: generalizing with the assimilation of new knowledge (combined lesson).
During the classes
I. Organizational moment.
II. Updating the basic knowledge and methods of action of students.
Lesson motto:"The only way,
leading to knowledge is activity” (B. Shaw). slide 1
At the first stage of the lesson, I update the basic knowledge that is necessary to solve the problem. This prepares students for the perception of the problem. I conduct the work in an entertaining way. I conduct a “brainstorming” on the topic: “Main classes of inorganic compounds” Work on cards
Task 1. “Third extra” slide 2
The students were given cards with three formulas written on them, and one of them was superfluous.
Students identify the extra formula and explain why it is superfluous.
Answers: MgO, Na 2 SO 4, H 2 S slide 3
Task 2. “Name and choose us” (“Name us”) slide 4
| nonmetals | hydroxides | Anoxic acids |
Give the name of the selected substance (“4-5” write down the answers with formulas, “3” with words).
(Students work in pairs, wishing at the blackboard. (“4-5” write down answers in formulas, “3” in words).
Answers: slide 5
1. copper, magnesium;
4. phosphoric;
5. magnesium carbonate, sodium sulfate
7. salt
III. Learning new material.
1. Determining the topic of the lesson together with students.
As a result of chemical transformations, substances of one class are transformed into substances of another: from a simple substance an oxide is formed, an acid is formed from an oxide, a salt is formed from an acid. In other words, the classes of compounds you have studied are interconnected. Let's distribute the substances into classes, according to the complexity of the composition, starting from a simple substance, according to our scheme.
Students express their versions, thanks to which we draw up simple schemes of 2 series: metals and non-metals. Scheme of genetic series.
I draw students' attention to the fact that each chain has something in common - these are the chemical elements metal and non-metal, which pass from one substance to another (as if by inheritance).
(for strong students) CaO, P 2 O 5, MgO, P, H 3 PO 4, Ca, Na 3 PO 4, Ca (OH) 2, NaOH, CaCO 3, H 2 SO 4
(For weak students) CaO, CO 2 , C, H 2 CO 3 , Ca, Ca(OH) 2 , CaCO 3 slide 6
Answers: slide 7
P P2O5 H3PO4 Na3 PO4
Ca CaO Ca(OH)2 CaCO3
What is the name of the carrier of hereditary information in biology? (Gene).
What element do you think will be the “gene” for each chain? (metal and non-metal).
Therefore, such chains or series are called genetic. The topic of our lesson is "Genetic connection of Me and NeMe" slide 8. Open your notebook and write down the date and topic of the lesson. What do you think is the purpose of our lesson? To get acquainted with the concept of "genetic connection". Learn to compose the genetic series of metals and non-metals.
2. Let's define the genetic link.
genetic connection - called the connection between substances of different classes, based on their mutual transformations and reflecting the unity of their origin. slide 9,10
Features that characterize the genetic series: slide 11
1. Substances of different classes;
2. Different substances formed by one chemical element, i.e. represent different forms of the existence of one element;
3.Different substances of the same chemical element connected by transformations.
3. Consider examples of the genetic relationship of Me.
2. A genetic series, where an insoluble base acts as a base, then the series can be represented by a chain of transformations: slide 12
metal→basic oxide→salt→insoluble base→basic oxide→metal
For example, Cu→CuO→CuCl2→Cu(OH)2→CuO
1. 2 Cu + O 2 → 2 CuO 2. CuO + 2HCI → CuCI 2 3. CuCI 2 + 2NaOH → Cu (OH) 2 + 2NaCI
4. Cu (OH) 2 CuO + H 2 O
4. Consider examples of the genetic connection of NeMe.
Among non-metals, two types of series can also be distinguished: slide 13
2. The genetic series of non-metals, where soluble acid. The chain of transformations can be represented as follows: non-metal → acid oxide → soluble acid → salt For example, P → P 2 O 5 → H 3 PO 4 → Ca 3 (PO 4) 2
1. 4P + 5O 2 → 2P 2 O 5 2. P 2 O 5 + H 2 O → 2H 3 PO 4 3. 2H 3 PO 4 +3 Ca (OH) 2 → Ca 3 (PO 4) 2 +6 H 2 O
5. Compilation of the genetic series. Slide 14
1. A genetic series in which alkali acts as a base. This series can be represented using the following transformations: metal → basic oxide → alkali → salt
O 2, + H 2 O, + HCI
4K + O 2 \u003d 2K 2 O K 2 O + H 2 O \u003d 2KOH KOH + HCI \u003d KCl slide 15
2. The genetic series of non-metals, where an insoluble acid acts as a link in the series:
nonmetal→acid oxide→salt→acid→acid oxide→nonmetal
For example, Si→SiO 2 →Na 2 SiO 3 →H 2 SiO 3 →SiO 2 →Si (make equations yourself, who works "4-5"). Self-test. All equations are correct "5", one error "4", two errors "3".
5. Performing differential exercises (self-examination). slide 15
Si + O 2 \u003d SiO 2 SiO 2 + 2NaOH \u003d Na 2 SiO 3 + H 2 O Na 2 SiO 3 + 2НCI \u003d H 2 SiO 3 + 2NaCI H 2 SiO 3 \u003d SiO 2 + H 2 O
SiO 2 +2Mg \u003d Si + 2MgO
1. Carry out transformations according to the scheme. (task "4-5")
Task 1. In the figure, connect the formulas of substances with lines in accordance with their location in the genetic series of aluminum. Write reaction equations. slide 16
Self-test.
4AI + 3O 2 \u003d 2AI 2 O 3 AI 2 O 3 + 6HCI \u003d 2AICI 3 + 3H 2 O AICI 3 + 3NaOH \u003d AI (OH) 3 + 3NaCI
AI(OH) 3 \u003d AI 2 O 3 + H 2 O slide 17
Task 2. "Hit the target." Select the formulas of the substances that make up the genetic series of calcium. Write the reaction equations for these transformations. Slide 18
Self-test.
2Ca + O 2 \u003d 2CaO CaO + H 2 O \u003d Ca (OH) 2 Ca (OH) 2 +2 HCI \u003d CaCI 2 + 2 H 2 O CaCI 2 + 2AgNO 3 \u003d Ca (NO 3) 2 + 2AgCI slide 19
2. Carry out the task according to the scheme. Write the reaction equations for these transformations.
O 2 + H 2 O + NaOH
S SO 2 H 2 SO 3 Na 2 SO 3 or light version
S + O 2 \u003d SO 2 + H 2 O \u003d H 2 SO 3 + NaOH \u003d
SO 2 + H 2 O \u003d H 2 SO 3
H 2 SO 3 + 2NaOH \u003d Na 2 SO 3 + 2H 2 O
IV. AnchoringZUN
Option 1.
Part A.
1. The genetic series of the metal is: a) substances that form a series based on one metal
a)CO 2 b) CO c) CaO d) O 2
3. Determine the substance "Y" from the transformation scheme: Na → Y→NaOH a)Na 2 O b) Na 2 O 2 c) H 2 O d) Na
4. In the transformation scheme: CuCl 2 → A → B → Cu, the formulas of intermediate products A and B are: a) CuO and Cu (OH) 2 b) CuSO 4 and Cu (OH) 2 c) CuCO 3 and Cu (OH) 2 G)Cu(Oh) 2 andCuO
5. The end product in the chain of transformations based on carbon compounds CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate c) sodium carbide d) sodium acetate
E → E 2 O 5 → H 3 EO 4 → Na 3 EO 4 a) N b) Mn in)P d)Cl
Part B.
Fe + Cl 2 A) FeCl 2
Fe + HCl B) FeCl 3
FeO + HCl B) FeCl 2 + H 2
Fe 2 O 3 + HCl D) FeCl 3 + H 2
E) FeCl 2 + H 2 O
E) FeCl 3 + H 2 O
1 B, 2 A, 3D, 4E
a) potassium hydroxide (solution) b) iron c) barium nitrate (solution) d) aluminum oxide
e) carbon monoxide (II) f) sodium phosphate (solution)
Part C.
1. Implement the scheme of transformation of substances: Fe → FeO → FeCI 2 → Fe (OH) 2 → FeSO 4
2Fe + O 2 \u003d 2FeO FeO + 2HCI \u003d FeCI 2 + H 2 O FeCI 2 + 2NaOH \u003d Fe (OH) 2 + 2NaCI
Fe(OH) 2 + H 2 SO 4= FeSO 4 +2 H 2 O
Option 2.
Part A. (questions with one correct answer)
b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances connected by transformations
2. Determine the substance "X" from the transformation scheme: P → X → Ca 3 (PO 4) 2 a)P 2 O 5 b) P 2 O 3 c) CaO d) O 2
a) Ca b)CaO c) CO 2 d) H 2 O
4. In the conversion scheme: MgCl 2 → A → B → Mg, the formulas of intermediate products A and B are: a) MgO and Mg (OH) 2 b) MgSO 4 and Mg (OH) 2 c) MgCO 3 and Mg (OH) 2 G)mg(Oh) 2 andMgO
CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate
6. Element "E", participating in the chain of transformations:
Part B. (tasks with 2 or more correct answers)
1. Establish a correspondence between the formulas of the starting substances and the reaction products:
Formulas of starting substances Formulas of products
NaOH + CO 2 A) NaOH + H 2
Na + H 2 O B) NaHCO 3
NaOH + HCl D) NaCl + H 2 O
1B, 2V, 3 A, 4G
a) sodium hydroxide (solution) b) oxygen c) sodium chloride (solution) d) calcium oxide
e) potassium permanganate (crystalline) e) sulfuric acid
Part C. (with extended answer)
S + O 2 \u003d SO 2 2SO 2 + O 2 \u003d 2 SO 3 SO 3 + H 2 O \u003d H 2 SO 4 H 2 SO 4 + Ca (OH) 2 \u003d CaSO 4 +2 H 2 O
CaSO 4 + BaCI 2 \u003d BaSO 4 + CaCI 2
v.Resultslesson. Grading.
VI.D/Z p.215-216 prepare for project No. 3 Option 1 of task No. 2,4, 6, Option 2 of task No. 2,3, 6. slide 20
VII. Reflection.
Students write down on paper what they did well and what they didn't. What were the difficulties. And a wish to the teacher.
The lesson is over. Thank you all and have a good day. slide 21
If there is time.
A task
Once Yuh conducted experiments to measure the electrical conductivity of solutions of various salts. Chemistry beakers with solutions were on his laboratory table. KCl, BaCl
2
, K
2
CO
3
, Na
2
SO
4
and AgNO
3
. Each glass was neatly labeled. There was a parrot in the lab whose cage didn't lock very well. When Juh, absorbed in the experiment, looked back at the suspicious rustle, he was horrified to find that the parrot, grossly violating safety regulations, was trying to drink from a glass of BaCl 2 solution. Knowing that all soluble barium salts are extremely poisonous, Yuh quickly grabbed a glass with a different label from the table and forcibly poured the solution into the parrot's beak. The parrot was saved. What glass of solution was used to save the parrot?
Answer:
BaCl 2 + Na 2 SO 4 \u003d BaSO 4 (precipitate) + 2NaCl (barium sulfate is so slightly soluble that it cannot be poisonous, like some other barium salts).
Attachment 1
9 "B" class F.I.______________________________ (for weak students)
Task 1. “The third extra”.
(4 correct - "5", 3-"4", 2-"3", 1-"2")
| nonmetals | hydroxides | Anoxic acids |
||||
Students define the chosen class and select the appropriate substances from the provided handout.
copper, silicon oxide, hydrochloric, barium hydroxide, coal, magnesium, phosphoric, barium hydroxide, magnesium oxide, iron (III) hydroxide, magnesium carbonate, sodium sulfate.
("4-5" write down the answers with formulas, "3" with words).
12 answers "5", 11-10 - "4", 9-8 - "3", 7 or less - "2"
Task 3.
O 2, + H 2 O, + HCI
For example, K → K 2 O → KOH → KCl (make equations yourself, who works "3", one error "3", two errors "2").
Task 4. Perform the task according to the scheme. Write the reaction equations for these transformations.
O 2 + H 2 O + NaOH
S SO 2 H 2 SO 3 Na 2 SO 3
or light version
H 2 SO 3 + NaOH \u003d
Option 1.
Part A. (questions with one correct answer)
1. The genetic series of a metal is: a) substances that form a series based on one metal
b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances connected by transformations
2. Determine the substance "X" from the transformation scheme: C → X → CaCO 3
a) CO 2 b) CO c) CaO d) O 2
3. Determine the substance "Y" from the transformation scheme: Na → Y→NaOH a) Na 2 O b) Na 2 O 2 c) H 2 O d) Na
4. In the transformation scheme: CuCl 2 → A → B → Cu, the formulas of intermediate products A and B are: a) CuO and Cu (OH) 2 b) CuSO 4 and Cu (OH) 2 c) CuCO 3 and Cu (OH) 2 g) Cu(OH) 2 and CuO
5. The end product in the chain of transformations based on carbon compounds CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate c) sodium carbide d) sodium acetate
6. Element "E", participating in the chain of transformations: E → E 2 O 5 → H 3 EO 4 → Na 3 EO 4 a) N b) Mn c) P d) Cl
Part B. (tasks with 2 or more correct answers)
1. Establish a correspondence between the formulas of the starting substances and the reaction products:
Formulas of starting substances Formulas of products
Fe + Cl 2 A) FeCl 2
Fe + HCl B) FeCl 3
FeO + HCl B) FeCl 2 + H 2
Fe 2 O 3 + HCl D) FeCl 3 + H 2
E) FeCl 2 + H 2 O
E) FeCl 3 + H 2 O
2. A solution of copper sulfate (II) interacts:
a) potassium hydroxide (solution) b) iron c) barium nitrate (solution) d) aluminum oxide
e) carbon monoxide (II) f) sodium phosphate (solution)
Part C. (with extended answer)
1. Implement a scheme for the transformation of substances:
Fe → FeO → FeCI 2 → Fe(OH) 2 → FeSO 4
Annex 2
9 "B" class F.I.______________________________ (for a strong student)
Task 1. “The third extra”. Identify the redundant formula and explain why it is redundant.
(4 correct - "5", 3-"4", 2-"3", 1-"2")
Task 2. “Name and choose us” (“Name us”). Give the name of the selected substance, fill in the table.
Students define the chosen class and select the appropriate substances from the provided handout.
copper, silicon oxide, hydrochloric, barium hydroxide, coal, magnesium, phosphoric, barium hydroxide, magnesium oxide, iron (III) hydroxide, magnesium carbonate, sodium sulfate. ("4-5" write down the answers with formulas, "3" with words).
12 answers "5", 11-10 - "4", 9-8 - "3", 7 or less - "2"
Task 3.
Si→SiO 2 →Na 2 SiO 3 →H 2 SiO 3 →SiO 2 →Si (make equations yourself, who works "4-5"). Self-test. All equations are correct "5", one error "4", two errors "3".
Task 4. In the figure, connect the formulas of substances with lines in accordance with their location in the genetic series of aluminum. Write reaction equations. All equations are correct "5", one error "4", two errors "3".
Task 5. "Hit the target." Select the formulas of the substances that make up the genetic series of calcium. Write the reaction equations for these transformations. All equations are correct "5", one error "4", two errors "3".
Option 2.
Part A. (questions with one correct answer)
1. The genetic series of a non-metal is: a) substances that form a series based on one metal
b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances connected by transformations
2. Determine the substance "X" from the transformation scheme: P → X → Ca 3 (PO 4) 2 a) P 2 O 5 b) P 2 O 3 c) CaO d) O 2
3. Determine the substance "Y" from the transformation scheme: Ca → Y→Ca(OH) 2
a) Ca b) CaO c) CO 2 d) H 2 O
4. In the conversion scheme: MgCl 2 → A → B → Mg, the formulas of intermediate products A and B are: a) MgO and Mg (OH) 2 b) MgSO 4 and Mg (OH) 2 c) MgCO 3 and Mg (OH) 2 g) Mg (OH) 2 and MgO
5. The end product in the chain of transformations based on carbon compounds:
CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate
c) sodium carbide d) sodium acetate
6. Element "E", participating in the chain of transformations:
E → EO 2 → EO 3 → H 2 EO 4 → Na 2 EO 4 a) N b) S c) P d) Mg
Part B. (tasks with 2 or more correct answers)
1. Establish a correspondence between the formulas of the starting substances and the reaction products:
Formulas of starting substances Formulas of products
NaOH + CO 2 A) NaOH + H 2
NaOH + CO 2 B) Na 2 CO 3 + H 2 O
Na + H 2 O B) NaHCO 3
NaOH + HCl D) NaCl + H 2 O
2. Hydrochloric acid does not interact:
a) sodium hydroxide (solution) b) oxygen c) sodium chloride (solution) d) calcium oxide
e) potassium permanganate (crystalline) f) sulfuric acid
Part C. (with extended answer)
Implement the scheme of transformation of substances: S → SO 2 → SO 3 → H 2 SO 4 → CaSO 4 → BaSO 4
Appendix 3
Answer sheet "4-5":
Task 1. MgO, Na 2 SO 4, H 2 S
Task 2.
1. copper, magnesium;
3. silicon oxide, magnesium oxide;
4. phosphoric,
5. magnesium carbonate, sulfate;
6. barium hydroxide, iron (III) hydroxide;
7. sodium hydrochloride
Task 3.
SiO 2 + 2NaOH \u003d Na 2 SiO 3 + H 2 O
Na 2 SiO 3 + 2НCI \u003d H 2 SiO 3 + 2NaCI
H 2 SiO 3 \u003d SiO 2 + H 2 O
SiO 2 +2Mg \u003d Si + 2MgO
Task 4.
4AI + 3O 2 \u003d 2AI 2 O 3
AI 2 O 3 + 6HCI \u003d 2AICI 3 + 3H 2 O
AICI 3 + 3NaOH \u003d AI (OH) 3 + 3NaCI
AI (OH) 3 \u003d AI 2 O 3 + H 2 O
Task 5.
CaO + H 2 O \u003d Ca (OH) 2
Ca (OH) 2 +2 HCI \u003d CaCI 2 + 2 H 2 O
CaCI 2 + 2AgNO 3 \u003d Ca (NO 3) 2 + 2AgCI
Self-assessment sheet.
| Full name of the student | job number | |
Each such row consists of a metal, its basic oxide, a base, and any salt of the same metal:
To go from metals to basic oxides in all these series, reactions of combination with oxygen are used, for example:
2Ca + O 2 \u003d 2CaO; 2Mg + O 2 \u003d 2MgO;
The transition from basic oxides to bases in the first two rows is carried out by the hydration reaction known to you, for example:
CaO + H 2 O \u003d Ca (OH) 2.
As for the last two rows, the MgO and FeO oxides contained in them do not react with water. In such cases, to obtain bases, these oxides are first converted into salts, and then they are converted into bases. Therefore, for example, to carry out the transition from MgO oxide to Mg (OH) 2 hydroxide, successive reactions are used:
MgO + H 2 SO 4 \u003d MgSO 4 + H 2 O; MgSO 4 + 2NaOH \u003d Mg (OH) 2 ↓ + Na 2 SO 4.
The transitions from bases to salts are carried out by reactions already known to you. So, soluble bases (alkalis), which are in the first two rows, are converted into salts under the action of acids, acid oxides or salts. Insoluble bases from the last two rows form salts under the action of acids.
Genetic series of non-metals and their compounds.
Each such series consists of a non-metal, an acid oxide, the corresponding acid, and a salt containing the anions of this acid:
To go from non-metals to acidic oxides, in all these series, reactions of combination with oxygen are used, for example:
4P + 5O 2 \u003d 2 P 2 O 5; Si + O 2 \u003d SiO 2;
The transition from acid oxides to acids in the first three rows is carried out by the hydration reaction known to you, for example:
P 2 O 5 + 3H 2 O \u003d 2 H 3 PO 4.
However, you know that the SiO 2 oxide contained in the last row does not react with water. In this case, it is first converted into the corresponding salt, from which the desired acid is then obtained:
SiO 2 + 2KOH = K 2 SiO 3 + H 2 O; K 2 SiO 3 + 2HСl \u003d 2KCl + H 2 SiO 3 ↓.
The transitions from acids to salts can be carried out by reactions known to you with basic oxides, bases, or with salts.
It should be remembered:
Substances of the same genetic series do not react with each other.
Substances of genetic series different types react with each other. The products of such reactions are always salts (Fig. 5):
Rice. 5. Scheme of the relationship of substances of different genetic series.
This scheme displays the relationship between different classes of inorganic compounds and explains the variety of chemical reactions between them.
Topic task:
Write the reaction equations that can be used to carry out the following transformations:
1. Na → Na 2 O → NaOH → Na 2 CO 3 → Na 2 SO 4 → NaOH;
2. P → P 2 O 5 → H 3 PO 4 → K 3 PO 4 → Ca 3 (PO 4) 2 → CaSO 4;
3. Ca → CaO → Ca(OH) 2 → CaCl 2 → CaCO 3 → CaO;
4. S → SO 2 → H 2 SO 3 → K 2 SO 3 → H 2 SO 3 → BaSO 3;
5. Zn → ZnO → ZnCl 2 → Zn(OH) 2 → ZnSO 4 → Zn(OH) 2;
6. C → CO 2 → H 2 CO 3 → K 2 CO 3 → H 2 CO 3 → CaCO 3;
7. Al → Al 2 (SO 4) 3 → Al(OH) 3 → Al 2 O 3 → AlCl 3;
8. Fe → FeCl 2 → FeSO 4 → Fe(OH) 2 → FeO → Fe 3 (PO 4) 2;
9. Si → SiO 2 → H 2 SiO 3 → Na 2 SiO 3 → H 2 SiO 3 → SiO 2;
10. Mg → MgCl 2 → Mg(OH) 2 → MgSO 4 → MgCO 3 → MgO;
11. K → KOH → K 2 CO 3 → KCl → K 2 SO 4 → KOH;
12. S → SO 2 → CaSO 3 → H 2 SO 3 → SO 2 → Na 2 SO 3;
13. S → H 2 S → Na 2 S → H 2 S → SO 2 → K 2 SO 3;
14. Cl 2 → HCl → AlCl 3 → KCl → HCl → H 2 CO 3 → CaCO 3;
15. FeO → Fe(OH) 2 → FeSO 4 → FeCl 2 → Fe(OH) 2 → FeO;
16. CO 2 → K 2 CO 3 → CaCO 3 → CO 2 → BaCO 3 → H 2 CO 3;
17. K 2 O → K 2 SO 4 → KOH → KCl → K 2 SO 4 → KNO 3;
18. P 2 O 5 → H 3 PO 4 → Na 3 PO 4 → Ca 3 (PO 4) 2 → H 3 PO 4 → H 2 SO 3;
19. Al 2 O 3 → AlCl 3 → Al(OH) 3 → Al(NO 3) 3 → Al 2 (SO 4) 3 → AlCl 3;
20. SO 3 → H 2 SO 4 → FeSO 4 → Na 2 SO 4 → NaCl → HCl;
21. KOH → KCl → K 2 SO 4 → KOH → Zn(OH) 2 → ZnO;
22. Fe(OH) 2 → FeCl 2 → Fe(OH) 2 → FeSO 4 → Fe(NO 3) 2 → Fe;
23. Mg(OH) 2 → MgO → Mg(NO 3) 2 → MgSO 4 → Mg(OH) 2 → MgCl 2;
24. Al(OH) 3 → Al 2 O 3 → Al(NO 3) 3 → Al 2 (SO 4) 3 → AlCl 3 → Al(OH) 3;
25. H 2 SO 4 → MgSO 4 → Na 2 SO 4 → NaOH → NaNO 3 → HNO 3;
26. HNO 3 → Ca(NO 3) 2 → CaCO 3 → CaCl 2 → HCl → AlCl 3;
27. CuСO 3 → Cu(NO 3) 2 → Cu(OH) 2 → CuO → CuSO 4 → Cu;
28. MgSO 4 → MgCl 2 → Mg(OH) 2 → MgO → Mg(NO 3) 2 → MgCO 3;
29. K 2 S → H 2 S → Na 2 S → H 2 S → SO 2 → K 2 SO 3;
30. ZnSO 4 → Zn(OH) 2 → ZnCl 2 → HCl → AlCl 3 → Al(OH) 3;
31. Na 2 CO 3 → Na 2 SO 4 → NaOH → Cu(OH) 2 → H 2 O → HNO 3;
