Genetic series of non-metals examples. Genetic relationship between classes of substances - Knowledge Hypermarket

Amphoteric

hydroxides

Genetic scheme of the main classes of inorganic compounds

With further complication of the composition of substances, hydroxides are formed, the base oxides corresponding to the base, and the acid oxides corresponding to the acid. The bases opposite in properties and acid actively react with each other, forming salts. The interaction of opposites is the driving force of reaction. Therefore, basic and acidic oxides, bases and acids actively interact with each other, and two acidic oxides or two basic oxides do not interact, since their properties are close.

Thus, the properties of a complex compound are determined based on the properties of its constituent elements. The main laws of change of these properties are summarized in the following applications (Table 6).

1. In periods with an increase in the sequence number, the properties of the elements change from metallic to nonmetallic. The number of electrons at the external level increases, the degree of oxidation of the element increases, the radius of the atom and ion decreases, the ionization energy and affinity for the electron increase. In accordance with this, the basic decreases and the acidic properties of oxides and hydroxides increase.

2. In the main subgroups, the basic properties of oxides and hydroxides increase with increasing atomic number. For elements of side groups with an increase in the sequence number, a more complex change in properties is characteristic. First, the metallic properties are enhanced and then reduced.

3. Active metals correspond to oxides and hydroxides with strongly pronounced basic properties. The most active metals are alkali and alkaline-earth. They form water-soluble oxides and strong soluble bases - alkalis.

4. Low-active metals (all but alkaline and alkaline-earth) form weak bases that are difficult to dissolve in water:

Cu (OH) 2, Fe (OH) 3.

5. Oxides and hydroxides with strongly pronounced acidic properties correspond to active non-metals.

6. Amphoteric metals form amphoteric oxides and hydroxides.

7. If an element exhibits different degrees of oxidation, then oxides and hydroxides with different properties correspond to it.

Topic: GENETIC CONNECTION BETWEEN metals and non-metals and their compounds. 9th grade

Objectives: educational: to consolidate the concept of "genetic series", "genetic connection"; teach how to make genetic series of elements (metals and non-metals), to make up the reaction equations corresponding to the genetic series; check how knowledge is learned chemical properties oxides, acids, salts, bases; developing: to develop the ability to analyze, compare, summarize and draw conclusions, make up the equations of chemical reactions; educational: to promote the formation of a scientific worldview.

Providing classes: tables "Periodic System", "Solubility Table", "Metals Activity Series", instructions for students, tasks for testing knowledge.

Progress: 1) Org. moment

2) Check d / s

3) Study new material

4) Fastening

5) D / C

1) Org. moment. Greeting.

2) Check d / s.

Genetic connections are connections between different classes based on their interconversions.
Knowing the classes of inorganic substances, it is possible to make the genetic series of metals and non-metals. The basis of these series is the same element.

Among metals there are two types of rows:

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

For example, K → K 2 O → KOH → KCl

2 . The genetic series, where the base is an insoluble base, then the series can be represented as a chain of transformations:

metal → base oxide → salt → insoluble base →

→basic oxide → metal

For example, Cu → CuO → CuCl 2 → Cu (OH) 2 → CuO → Cu

1 . The genetic series of non-metals, where soluble acid acts as a link in the series. 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 → Na 3 PO 4

2 . The genetic series of non-metals, where insoluble acid acts as a link in the series:

non-metal → acid oxide → salt → acid →

→acid oxide → non-metal

For example,Si→ Sio 2 → Na 2 Sio 3 → H 2 Sio 3 → Sio 2 → Si

Frontal conversation on:

What is a genetic connection? Genetic connections are connections between different classes based on their interconversions. What is a genetic series?

Genetic series - a series of substances - representatives of different classes, which are compounds of one chemical element, connected by interconversions and reflecting the transformations of these substances. The basis of these series is the same element.

What types of genetic series can be distinguished? Among the metals, two types of series can be distinguished:

a) A genetic series in which alkali acts as a base. This series can be represented using the following transformations:

metal → basic oxide → alkali → salt

for example, the genetic series of potassium K → K 2 O → KOH → KCl

b) The genetic series, where the base is an insoluble base, then the series can be represented as a chain of transformations:

metal → base oxide → salt → insoluble base → base oxide → metal

for example: Cu → CuO → CuCl 2   → Cu (OH) 2   → CuO → Cu

Among non-metals it is also possible to distinguish two types of rows:

a) The genetic series of non-metals, where soluble acid acts as a link in the series. 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   → Na 3 PO 4

b) Genetic row of non-metals, where insoluble acid: non-metal → acid oxide → salt → acid → acid oxide → non-metal acts as a link in the series

For example: Si → SiO 2   → Na 2 Sio 3   → H 2 Sio 3 → SiO 2   → Si

Execution of tasks for options:

1. Choose the oxide formulas in your version, explain your choice, based on the knowledge of the characteristics of the composition of this class of compounds. Call them.

2. In the formula column of your variant, find the acid formulas and explain your choice based on the analysis of the composition of these compounds.

3. Determine the valencies of the acid residues in the acid composition.

4. Select salt formulas and name them.

5. Make up the formulas of salts, which can be formed with magnesium and acids of your variant. Write them down, name it.

6. In the formula column of your variation, find the base formulas and explain your choice based on the analysis of the composition of these compounds.

7. In your variant, choose the formulas of substances with which a solution of orthophosphoric acid (hydrochloric, sulfuric) can react. Make the appropriate reaction equations.

9. Among the formulas of your option, select formulas of substances that can interact with each other. Make the appropriate reaction equations.

10. Make a chain of genetic bonds of inorganic compounds, which will include a substance, the formula of which is given in your version at number one.

Option 1

Option 2

Cao

Hno 3

Fe (OH) 3

N 2 O

Zn (NO 3 ) 2

Cr (OH) 3

H 2 SO 3

H 2 S

Pbo

LiOH

Ag 3 PO 4

P 2 O 5

NaOH

Zno

CO 2

BaCl 2

HCl

H 2 CO 3

H 2 SO 4

Cuso 4

From these substances make a genetic line using all the formulas. Write the reaction equations with which you can accomplish this chain of transformations:

I option: ZnSO 4, Zn, ZnO, Zn, Zn (OH) 2 : II  option:Na 2 SO 4,   NaOH, Na, Na 2 O 2 , Na 2 O

4) Fastening1.Al→ Al 2 O 3 → AlCl 3 → Al( OH) 3 → Al 2 O 3

2. P→ P 2 O 5 → H 3 PO 4 → Na 3 PO 4 → Ca 3 ( PO 4 ) 2

3. Zn → ZnCl 2 → Zn (OH) 2 → ZnO → Zn (NO 3 ) 2

4. Cu → CuO → CuCl 2 → Cu (OH) 2 → CuO → Cu

5.N 2 O 5 → HNO 3 → Fe (NO 3 ) 2 → Fe (OH) 2 → FeS → FeSO 4

5) Homework: chart a gradual transition from calcium to calcium carbonate and prepare a report on the use of any salt in medicine (using additional literature).

Among metals, two types of rows can be distinguished: 1. Genetic series, in which alkali acts as a base. This series can be represented by the following transformations: metal-- basic oxide - alkali - salt, for example, the genetic series of potassium K - K 2 O - KOH - KCl.

2. A genetic series, where an insoluble base acts as a base, then the series can be represented as a chain of transformations: metal - basic oxide - salt - insoluble base - basic oxide - metal. Cu-- CuO-- CuCl 2 - Cu (OH) 2 - CuO -\u003e Cu genetic series of copper    "\u003e Cu is a genetic series of copper"\u003e "title =" (! LANG: 2. Genetic series, where the base is an insoluble base, then the series can be represented as a chain of transformations: metal - base oxide - salt - insoluble base-- the main oxide is metal. Cu-- CuO-- CuCl 2 - Cu (OH) 2 - CuO -\u003e"> title="2. A genetic series, where an insoluble base acts as a base, then the series can be represented as a chain of transformations: metal - basic oxide - salt - insoluble base - basic oxide - metal. Cu-- CuO-- CuCl 2 - Cu (OH) 2 - CuO -\u003e"> !}

Among non-metals it is also possible to distinguish two types of rows: 1. The genetic series of non-metals, where soluble acid acts as a link in the series. The chain of transformations can be represented as follows: non-metal - acid oxide - soluble acid - -sol. P-- P 2 O 5 - H 3 PO 4 - Na 3 PO 4. Phosphorus Genetic Range

There is a genetic link between the classes of inorganic compounds. From simple substances you can get complex and vice versa. From compounds of one class, you can get compounds of another class.

Simplified genetic relationship between classes of inorganic compounds can be represented by the following scheme:

The sequence of such transformations for non-metals can be represented by the following scheme: CANPO 4

Р → Р 2 О 5 → Н 3 РО 4 → Са 3 (РО 4) 2

(Saon) 3 PO 4

For typical metals you can make the following chain of transformations:

Ba → BaO → Ba (OH) 2 → BaSO 4

For metals whose oxides and hydroxides are amphoteric (semimetals), the following transformations can be carried out:

Al → Al 2 O 3 → Al (OH) 3 → Na → AlCl 3 → AlOHCl 2 → Al (OH) 3 → Al 2 O 3.

Links between classes:

1. Metals, non-metals→ salt.

With the direct interaction of metals and non-metals, salts of oxygen-free acids (halides, sulfides) are formed:

2Na + C1 2 = 2NaCl

These compounds are stable and do not decompose when heated.

2. Basic oxides, acid oxides→  salt.

CaO + CO 2 = CaCO 3;

Na 2 O + SO 3 = Na 2 SO 4.

3. Bases, acids→  salt.

It is carried out through a neutralization reaction:

2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O,

HE - + H + → H 2 O;

Mg (OH) 2 + 2HC1 = MgCl 2 + 2H 2 O,

Mg (OH) 2 + 2H + → Mg 2+ + 2H 2 O.

4. Metals→  basic oxides.

Most metals interact with oxygen to form oxides:

2Ca + O 2 = 2CaO;

4A1 + 3O 2 = 2A1 2 O 3.

Gold, silver, platinum and other noble metals do not interact with oxygen, oxides of such metals are obtained indirectly.

5. Non-metals→ acid oxides.

Non-metals (with the exception of halogens and noble gases) interact with oxygen, forming oxides:

4P + 5O 2 = 2P 2 O 5;

S + O 2 = SO 2.

6. Basic oxides→ grounds.

Direct interaction with water can be obtained only hydroxides of alkali and alkaline-earth metals (alkali):

Na 2 O + H 2 O = 2NaOH;

CaO + H 2 O = Ca (OH) 2.

The remaining grounds are obtained indirectly.

7. Acid oxides→  acid.

Acid oxides react with water to form the corresponding acids:

SO 3 + H 2 O = H 2 SO 4;

Р 2 O 5 + 3H 2 O = 2H 3 PO 4.

The exception is SiO 2, which does not react with water.

8. Bases, acid oxides→ salt.

Alkalis interact with acidic oxides to form salts:

2NaOH + SO 3 = Na 2 SO 4 + H 2 O,

2OH - + SO 3 = SO 4 2- + H 2 O;

Ca (OH) 2 + СO 2 = CaCO 3 ↓ + Н 2 O,

Ca 2+ + 2OH - + CO 2 → CaCO 3 ↓ + H 2 O.

9. Acids, basic oxides→  salt.

Metal oxides dissolve in acids, forming salts:

CuO + H 2 SO 4 = CuSO 4 + H 2 O,

CuO + 2H + = Cu 2+ + H 2 O;

Na 2 O + 2HS1 = 2NaCl + H 2 O,

Na 2 O + 2H + = 2Na + + H 2 O.

10. Grounds→  basic oxides.

Insoluble bases and LiOH decompose when heated:

2LiOH = Li 2 O + H 2 O;

Cu (OH) 2 = CuO + H 2 O.

11. Acids→  acid oxides.

Unstable oxygen-containing acids decompose when heated (H 2 SiO 3) and even without heating (H 2 CO 3, HClO). At the same time, a number of acids are resistant to heating (H 2 SO 4, H 3 PO 4).

H 2 SiO 3 = H 2 O + SiO 2;

H 2 CO 3 = H 2 O + CO 2.

12. Metal oxides→ metals.

Some heavy metal oxides can decompose to metal and oxygen:

2HgO = 2Hg + O 2.

Also, metals are obtained from the corresponding oxides with the help of reducing agents:

3MnO 2 + 4Al = 3Mn + 2Al 2 O 3;

Fe 2 O 3 + 3H 2 = 2Fe + 3H 2 O.

13. Acid oxides→ non-metals

Most non-metal oxides do not decompose when heated. Non-metal and oxygen decompose only some unstable oxides (halogen oxides).

Some non-metals are obtained by reduction from the corresponding oxides:

SiO 2 + 2Mg = 2MgO + Si.

14. Salts, bases → bases.

Insoluble bases are obtained by the action of alkalis on solutions of the corresponding acids:

CuSO 4 + 2NaOH = Cu (OH) 2 ↓ + Na 2 SO 4,

Cu 2+ + 2OH - → Cu (OH) 2 ↓;

FeCl 2 + 2KOH = Fe (OH) 2 ↓ + 2KCl,

Fe 2+ + 2OH - = Fe (OH) 2 ↓.

15. Salts, acids → acids.

Soluble salts react with acids (in accordance with the pressure series) if the result is a weaker or volatile acid:

Na 2 SiO 3 + 2HCl = 2NaCl + H 2 SiO 3 ↓,

SiO 3 2- + 2H + → H 2 SiO 3 ↓;

NaCl (s.) + H 2 SO 4 (k) = NaHSO 4 + HCl.

16. Salts→ basic oxides, acid oxides.

The salts of some oxygen-containing acids (nitrates, carbonates) decompose when heated:

CaCO 3 = CaO + CO 2;

2Cu (NO 3) 2 = 2CuO + 4NO 2 + O 2.

EXERCISES FOR INDEPENDENT WORK ON THE TOPIC “GENETIC CONNECTION BETWEEN THE CLASSES OF INORGANIC COMPOUNDS”

1. Name the substances listed below, distribute them into the classes of inorganic compounds: Na 3 PO 4, H 2 SiO 3, NO, B 2 O 3, MgS, BaI 2, Ca (OH) 2, KNO 3, HNO 2, Cl 2 O 7, Fe (OH) 2, P 2 O 5, HF, MnO 2.

2. From what substances listed below it is possible to obtain hydroxide (acid or base) in one stage: copper, iron oxide (P), barium oxide, nitrogen oxide (P), nitrogen oxide (V), silicon oxide, copper sulfate, potassium chloride , potassium, magnesium carbonate.

3. From the above list write down the formulas of substances related to: 1) oxides; 2) grounds; 3) acids; 4) salts:

CO 2, NaOH, HCl, SO 3, CuSO 4, NaNO 3, KCl, H 2 SO 4, Ca (OH) 2, P 2 O 5, HNO 3, Al (OH) 3.

4. Name the substances: Zn (OH) 2, MgO, P 2 O 3, NaHCO 3, H 3 PO 3, Fe 2 (SO 4) 3, KOH, (AlOH) 3 (PO 4) 2, Ba (MnO 4 ) 2, CO, HI. Indicate which class each substance belongs to.

5. Write the molecular formulas of the following substances and indicate which class each substance belongs to:

1) copper (II) hydroxocarbonate;

2) nitrogen oxide (V);

3) nickel (II) hydroxide;

4) barium hydrogen phosphate;

5) perchloric acid;

6) chromium (III) hydroxide;

7) potassium chlorate;

8) hydrogen sulfide acid;

9) sodium zincate.

6. Give examples of the reactions of the connection between:

1) simple substances-non-metals;

2) simple substance and oxide;

3) oxides;

4) complex substances that are not oxides;

5) metal and non-metal;

6) three substances.

7. Which of the following substances can react:

1) carbon monoxide (IV): HCl, O 2, NO 2, KOH, H 2 O;

2) magnesium oxide: Ba (OH) 2, HCl, CO 2, O 2, HNO 3;

3) iron (II) hydroxide: KCl, HCl, KOH, O 2, H 2 O, HNO 3;

4) hydrogen chloride: Zn, MgO, ZnCl 2, HNO 3, Ca (OH) 2, Cu, (ZnOH) Cl.

8. Is the interaction between the following substances possible?

1) carbon monoxide (IV) and potassium hydroxide;

2) potassium hydrosulfate and calcium hydroxide;

3) calcium phosphate and sulfuric acid;

4) calcium hydroxide and sulfur oxide (IV);

5) sulfuric acid and potassium hydroxide;

6) calcium bicarbonate and phosphoric acid;

7) silica and sulfuric acid;

8) zinc oxide and phosphorus oxide (V).

Write the equations of possible reactions, specify the conditions in which they occur. If reactions can lead to different substances, then indicate what the difference in the conditions of their implementation.

9. Give the equations of the reactions for obtaining the following substances: sodium orthophosphate (4 ways), potassium sulfate (7 ways), zinc hydroxide.

10. One of the methods for producing soda (sodium carbonate) is the action of water and carbon monoxide (IV) on sodium aluminate. Make up the reaction equations.

11. He changing the coefficients, write the reaction products:

1) MgO + 2H 2 SO 4 →

2) 2SO 2 + Ba (OH) 2 →

3) 3N 2 O 5 + 2Al (OH) 3 →

4) P 2 O 5 + 4NaOH →

5) P 2 O 5 + 6NaOH →

6) P 2 O 5 + 2NaOH →

12. Make the equations of reactions for receiving different types of salts:

1) SO 2 + Ba (OH) 2 → (medium and acid salts),

2) A1 2 O 3 + H 2 O + HNO 3 → (medium salt, basic salts),

3) Na 2 O + H 2 S → (medium and acid salts),

4) SO 3 + Sa (OH) 2 → (medium and basic salts),

5) CaO + H 2 O + P 2 O 5 → (basic salt, acid salts).

13. Complete the reaction equations:

CaO + A1 2 O 3 → CaHPO 4 + Ca (OH) 2 →

Cr 2 O 3 + H 2 SO 4 → AlOHSO 4 + NaOH →

Cr 2 O 3 + NaOH → CaCO 3 + CO 2 + H 2 O →

A1 2 O 3 + HClO 4 → Ca (HCO 3) 2 + HCl →

Mn 2 O 7 + KOH → ZnS + H 2 S →

NO 2 + Ca (OH) 2 → CaSO 4 + H 2 SO 4 →

Zn (OH) 2 + NaOH → (ZnOH) Cl + HCl →

Zn (OH) 2 + HNO 3 → Bi (OH) 3 + H 2 SO 4 (insufficient) →

AlCl 3 + NaOH (deficient) → (FeOH) Cl + NaHS →

AlCl 3 + NaOH → Na 2 ZnO 2 + H 2 SO 4 (excess) →

AlC1 3 + NaOH (excess.) → Ca (AlO 2) 2 + HCl (excess) →

14. Write down the reaction equations with which you can carry out the following transformations:

1) Cu → CuO → CuSO 4 → Cu (OH) 2 → CuC1 2 → Cu (NO 3) 2

2) Zn → ZnO → ZnSO 4 → Zn (OH) 2 → Na 2 ZnO 2 → ZnCl 2

3) Р → Р 2 O 5 → Н 3 РО 4 → К 3 РО 4 → Са 3 (РО 4) 2 → Н 3 РО 4

4) Mg → MgO → MgCl 2 → Mg (OH) 2 → Mg (HSO 4) 2 → MgSO 4

5) Ca → CaO → Ca (OH) 2 → CaCO 3 → Ca (HCO 3) 2 → CO 2

6) Cr → Cr 2 (SO 4) 3 → Cr (OH) 3 → NaСrO 2 → Cr 2 O 3 → K

7) P → P 2 O 5 → HPO 3 → H 3 PO 4 → NaH 2 PO 4 → Na 3 PO 4

8) CuS → CuO → CuSO 4 → Cu (OH) 2 → CuO → Cu

9) Al → Al 2 O 3 → Al 2 (SO 4) 3 → Al (HSO 4) 3 → Al (OH) 3 → K

10) S → SO 2 → SO 3 → NaHSO 4 → Na 2 SO 4 → BaSO 4

11) Zn → ZnO → ZnCl 2 → Zn → Na 2

12) Zn → ZnSO 4 → ZnCl 2 → Zn (OH) 2 → Na 2 → Zn (NO 3) 2

13) Ca → CaCl 2 → CaCO 3 → Ca (HCO 3) 2 → Ca (NO 3) 2

14) Ca → Ca (OH) 2 → CaCO 3 → CaCl 2 → CaCO 3 → Ca (NO 3) 2

15) CuO → CuCl 2 → Cu (NO 3) 2 → CuO → CuSO 4 → Cu

16) CaO → Ca (OH) 2 → Ca (NO 3) 2 → Ca (NO 2) 2 → HNO 2 → NaNO 2

17) MgO → MgSO 4 → MgCl 2 → Mg (NO 3) 2 → Mg (OH) 2 → MgO

18) SO 2 → H 2 SO 3 → KHSO 3 → K 2 SO 3 → KHSO 3 → SO 2

19) P 2 O 5 → H 3 PO 4 → Ca (H 2 PO 4) 2 → Ca 3 (PO 4) 2 → Ca (H 2 PO 4) 2 → CaHPO 4

20) CO 2 → Ca (HCO 3) 2 → CaCO 3 → CaCl 2 → Ca (NO 3) 2 → CaSO 4

21) PbO → Pb (NO 3) 2 → PbO → Na 2 PbO 2 → Pb (OH) 2 → PbCl 2

22) ZnO → ZnSO 4 → Zn (OH) 2 → Na 2 ZnO 2 → Zn (OH) 2 → K 2

23) Al 2 O 3 → AlCl 3 → Al (OH) 3 → NaAlO 2 → Al (OH) 3 → K

24) ZnSO 4 → Zn (OH) 2 → ZnCl 2 → Zn → ZnO → Zn (NO 3) 2

25) AlCl 3 → Al (NO 3) 3 → Al (OH) 3 → NaAlO 2 → A1C1 3 → Al

26) Pb (NO 3) 2 → Pb (OH) 2 → PbO → Na 2 PbO 2 → Pb (OH) 2 → PbSO 4

27) Fe 2 (SO 4) 3 → FeCl 3 → Fe (OH) 3 → FeOH (NO 3) 2 → Fe (NO 3) 3 → Fe 2 O 3

28) K → KOH → KHSO 4 → K 2 SO 4 → KCl → KNO 3

29) Cu (OH) 2 → CuOHNO 3 → Cu (NO 3) 2 → CuSO 4 → CuCl 2 → Cu (NO 3) 2

30) CaCl 2 → Ca → Ca (OH) 2 → CaCl 2 → Ca (NO 3) 2 → CaSO 4

31) Cu → Cu (NO 3) 2 → Cu (OH) 2 → CuSO 4 → Al 2 (SO 4) 3 → A1C1 3

32) Mg → MgSO 4 → MgCl 2 → MgOHCl → Mg (OH) 2 → MgOHNO 3

33) CuSO 4 → CuCl 2 → ZnCl 2 → Zn (OH) 2 → Na 2 ZnO 2 → Zn (OH) 2

34) Hg (NO 3) 2 → Al (NO 3) 3 → Al 2 O 3 → NaAlO 2 → Al (OH) 3 → AlOHCl 2

35) ZnSO 4 → Zn (OH) 2 → ZnCl 2 → AlCl 3 → Al (OH) 3 → A1 2 O 3

36) CuCl 2 → Cu (OH) 2 → CuSO 4 → ZnSO 4 → Zn (OH) 2 → Na 2 ZnO 2

37) Fe (NO 3) 3 → FeOH (NO 3) 2 → Fe (OH) 3 → FeCl 3 → Fe (NO 3) 3 → Fe

38) Al 2 O 3 → AlCl 3 → Al (OH) 3 → NaAlO 2 → NaNO 3 → HNO 3

39) Mg (OH) 2 → MgSO 4 → MgCl 2 → Mg (NO 3) 2 → Mg (OH) 2 → MgO

40) aluminum sulphate → aluminum chloride → aluminum nitrate → aluminum oxide → potassium aluminate → aluminum hydroxide → aluminum hydroxide → aluminum chloride.

41) Na → NaOH → Na 3 PO 4 → NaNO 3 → HNO 3 → N 2 O 5

42) BaCO 3 → Ba (HCO 3) 2 → BaCO 3 → (BaOH) 2 CO 3 → BaO → BaSO 4

43) Cu → CuSO 4 → (CuOH) 2 SO 4 → Cu (OH) 2 → Cu (HSO 4) 2 → CuSO 4

44) barium → barium hydroxide → barium bicarbonate → barium chloride → barium carbonate → barium chloride → barium hydroxide

45) P → P 2 O 5 → H 3 PO 4 → Ca (H 2 PO 4) 2 → CaHPO 4 → Ca 3 (PO 4) 2

46) Cr → CrO → Cr 2 O 3 → NaCrO 2 → CrCl 3 → Cr (OH) 3 → Cr 2 O 3 → Cr

47) Cr 2 O 3 → CrCl 3 → Cr (OH) 3 → Na 3 → Cr 2 (SO 4) 3 → CrCl 3

48) K → KOH → KCl → KOH → K 2 SO 4 → KNO 3 → KNO 2

49) S → FeS → H 2 S → SO 2 → S → ZnS → ZnO → ZnCl 2 → Zn (OH) 2 → K 2

50) C → CO 2 → CO → CO 2 → Ca (HCO 3) 2 → CaCO 3 → CaCl 2

51) С → CO 2 → NaHCO 3 → Na 2 CO 3 → CO 2

52) S → SO 2 → K 2 SO 3 → KHSO 3 → K 2 SO 3

53) Cu → Cu (OH) 2 → Cu (NO 3) 2 → CuO → Cu

54) Р 2 O 5 → H 3 PO 4 → CaHPO 4 → Ca (H 2 PO 4) 2 → Ca 3 (PO 4) 2

55) Fe → FeCl 2 → Fe (OH) 2 → FeSO 4 → Fe

56) Zn → ZnO → Zn (OH) 2 → Zn (NO 3) 2 → ZnO

57) CuS → SO 2 → KHSO 3 → CaSO 3 → SO 2

58) SO 2 → H 2 SO 4 → CuSO 4 → CuO → Cu (NO 3) 2

59) KHSO 3 → CaSO 3 → Ca (HSO 3) 2 → SO 2 → K 2 SO 4

60) SO 2 → CaSO 3 → SO 2 → NaHSO 3 → SO 2

61) NaHCO 3 → Na 2 CO 3 → NaCl → NaHSO 4 → Na 2 SO 4

62) K → KOH → KCl → KNO 3 → K 2 SO 4 → KCl

63) NaCl → Na → NaOH → Na 2 SO 4 → NaCl

64) Al → AlCl 3 → Al (OH) 3 → A1 2 O 3 → Al (OH) 3

65) CuO → Cu → CuCl 2 → CuSO 4 → CuS

66) Fe → FeSO 4 → Fe (OH) 2 → Fe → Fe (OH) 3

67) Fe → Fe (OH) 2 → FeCl 2 → Fe (NO 3) 2 → Fe

68) Fe (NO 3) 3 → Fe 2 O 3 → FeCl 3 → Fe (NO 3) 3 → Fe

69) CuO → CuSO 4 → Cu (OH) 2 → CuO → Cu

70) MgCO 3 → MgO → MgCl 2 → Mg (OH) 2 → Mg (NO 3) 2

71) Mg → Mg (OH) 2 → MgSO 4 → MgCO 3 → Mg (HCO 3) 2

72) CaO → Ca (OH) 2 → CaCl 2 → CaCO 3 → CO 2

73) CaCO 3 → Ca (HCO 3) 2 → CaCl 2 → Ca (NO 3) 2 → O 2

74) FeS → Fe 2 O 3 → Fe (OH) 3 → Fe 2 (SO 4) 3 → FeCl 3

75) KS1 → K 2 SO 4 → KOH → K 2 CO 3 → KOH

76) CuS → CuO → Cu (OH) 2 → CuSO 4 → Cu

77) Fe → Fe (OH) 3 → Fe (NO 3) 3 → FeCl 3 → Fe 2 (SO 4) 3

78) CuSO 4 → CuO → Cu (NO 3) 2 → CuO → CuS

79) ZnS → H 2 S → SO 2 → Na 2 SO 4 → NaOH

80) Al → Al (OH) 3 → A1 2 (SO 4) 3 → A1 2 O 3 → Al (OH) 3

81) CaCl 2 → CaCO 3 → Ca (HCO 3) 2 → CaCO 3 → CaSiO 3

82) S → ZnS → H 2 S → Ca (HSO 3) 2 → SO 2

83) Na 2 SO 4 → NaCl → HCl → CaCl 2 → Ca (NO 3) 2

84) Na 2 SO 3 → SO 2 → H 2 SO 4 → HCl → FeCl 2

85) С → Na 2 CO 3 → CaCO 3 → CaSiO 3 → H 2 SiO 3

86) P → P 2 O 5 → Ca (H 2 PO 4) 2 → CaHPO 4 → H 3 PO 4

87) Al → A1 2 O 3 → Al (OH) 3 → A1C1 3 → A1 (NO 3) 3

88) HCl → CuCl 2 → Cl 2 → HCl → H 2

89) P 2 O 5 → Na 2 HPO 4 → Na 3 PO 4 → Ca 3 (PO 4) 2 → CaSO 4

90) NH 3 → NH 4 C1 → NH 3 ∙ H 2 O → NH 4 HCO 3 → NH 3

91) NH 4 C1 → KC1 → HCl → CuCl 2 → Cu (OH) 2

92) NH 3 → NH 4 H 2 PO 4 → (NH 4) 2 HPO 4 → NH 3 → NH 4 NO 3

93) KOH → KHCO 3 → K 2 CO 3 → CO 2 → Ca (HCO 3) 2

94) Na → NaOH → NaHCO 3 → Na 2 SO 4 → NaOH

95) KNO 3 → K 2 SO 4 → KC1 → KNO 3 → KNO 2

96) Cl 2 → KC1 → K 2 SO 4 → KNO 3 → KHSO 4

97) FeSO 4 → FeS → SO 2 → KHSO 3 → K 2 SO 4

98) KOH → Cu (OH) 2 → CuSO 4 → Cu (OH) 2 → Cu

99) Fe 2 O 3 → FeCl 3 → Fe (OH) 3 → Fe (NO 3) 3 → Fe 2 O 3

100) Al → A1 2 O 3 → A1 (NO 3) 3 → A1 2 O 3 → Al (OH) 3

101) CaO → CaCO 3 → CaSiO 3 → Ca (NO 3) 2 → O 2

102) Cu → Cu (OH) 2 → Cu → CuSO 4 → CuCl 2

103) H 2 S → SO 2 → ZnSO 4 → ZnS → ZnO

104) Cl 2 → NaCl → HCl → CuCl 2 → CuO

105) Cl 2 → FeCl 3 → Fe 2 O 3 → Fe (OH) 3 → Fe (NO 3) 3

106) P 2 O 5 → Ca 3 (PO 4) 2 → H 3 PO 4 → CaHPO 4 → Ca (H 2 PO 4) 2

107) ZnS → ZnO → Zn → ZnCl 2 → Zn (NO 3) 2

108) ZnO → ZnSO 4 → Zn (NO 3) 2 → ZnO → Zn (OH) 2

109) H 3 PO 4 → NH 4 H 2 PO 4 → (NH 4) 2 HPO 4 → Na 3 PO 4 → Ca 3 (PO 4) 2

110) CaCO 3 → Na 2 CO 3 → Na 3 PO 4 → NaH 2 PO 4 → Ca 3 (PO 4) 2

111) CaCl 2 → CaSO 3 → Ca (OH) 2 → CaCl 2 → Ca (NO 3) 2

112) NaOH → Na 2 CO 3 → NaHSO 4 → NaNO 3 → NaHSO 4

113) Na 2 SiO 3 → Na 2 CO 3 → Na 2 SO 4 → NaCl → Na 2 SO 4

114) KNO 3 → KHSO 4 → K 2 SO 4 → KCl → Na 2 SO 4

115) SiO 2 → K 2 SiO 3 → H 2 SiO 3 → SiO 2 → CaSiO 3

116) Cu → CuCl 2 → Cu (NO 3) 2 → NO 2 → HNO 3

117) Ca (NO 3) 2 → O 2 → SiO 2 → H 2 SiO 3 → SiO 2

118) P → H 3 PO 4 → Ca 3 (PO 4) 2 → CaHPO 4 → Ca (H 2 PO 4) 2

119) CuSO 4 → Cu → CuS → CuO → CuCl 2

120) Al → A1 2 (SO 4) 3 → Al (OH) 3 → A1C1 3 → A1 (NO 3) 3

121) S → SO 3 → H 2 SO 4 → KHSO 4 → BaSO 4

122) N 2 O 5 → HNO 3 → Cu (NO 3) 2 → CuO → Cu (OH) 2

123) Al → A1 2 O 3 → Al (OH) 3 → A1 2 (SO 4) 3 → A1 (NO 3) 3

124) Ca → Ca (OH) 2 → Ca (HCO 3) 2 → CaO → CaCl 2

125) NH 3 ∙ H 2 O → NH 4 C1 → NH 3 → NH 4 HCO 3 → (NH 4) 2 CO 3

126) Cu (OH) 2 → H 2 O → HNO 3 → Fe (NO 3) 3 → Fe

127) SO 2 → Ca (HSO 3) 2 → CaCl 2 → Ca (OH) 2 → Ca (HCO 3) 2

128) NH 3 ∙ H 2 O → NH 4 HCO 3 → CaCO 3 → CaSiO 3 → CaCl 2

129) CuSO 4 → Cu → CuO → Cu (OH) 2 → Cu

130) Fe (OH) 3 → Fe → FeCl 3 → Fe (NO 3) 3 → Fe

131) Zn → Zn (OH) 2 → Na 2 → Zn (OH) 2 → Na 2 ZnO 2 → Zn

132) Zn → ZnO → Na 2 ZnO 2 → Zn (OH) 2 → Na 2 → ZnCl 2

133) Zn → K 2 ZnO 2 → ZnSO 4 → K 2 → Zn (NO 3) 2 → ZnO

134) ZnO → Zn (OH) 2 → K 2 ZnO 2 → ZnSO 4 → ZnCl 2 → ZnO

135) Zn → Na 2 → Na 2 ZnO 2 → Zn (NO 3) 2 → ZnO → Zn

136) Al → K 3 → Al (OH) 3 → Na 3 → A1C1 3 → Al (OH) 3

137) Al 2 O 3 → KAlO 2 → Al (OH) 3 → Al 2 O 3 → Na 3 → Al 2 O 3

138) Al (OH) 3 → A1 2 O 3 → K 3 → Al 2 (SO 4) 3 → A1 (NO 3) 3

139) A1C1 3 → K 3 → Al (NO 3) 3 → NaAlO 2 → Al 2 O 3

140) Be → Na 2 → Be (OH) 2 → Na 2 BeO 2 → BaBeO 2

EXPERIMENTAL EXPERIENCES ON THE TOPIC “BASIC CLASSES OF INORGANIC COMPOUNDS”

EXPERIENCE 1.Neutralization Reactions.

a) The interaction of a strong acid and a strong base.

Pour 5 ml of 2N hydrochloric acid solution into a porcelain dish and add 2 n sodium hydroxide solution dropwise to it. Mix the solution with a glass rod and test its effect on the litmus, transferring a drop of the solution to the litmus test. It is necessary to achieve a neutral reaction (blue and red litmus paper does not change color). Evaporate the resulting solution to dryness. What was formed? Write molecular and ionic reaction equations.

b) The interaction of a weak acid and a strong base.

Pour into a test tube 2 ml of 2N alkali solution and add a solution of acetic acid until the solution is neutral. Write molecular and ionic reaction equations. Explain why the equilibrium ion reaction, in which a weak electrolyte (acetic acid) takes part, shifts towards the formation of water molecules.

EXPERIENCE 2.  Amphoteric hydroxides.

From the reagents available in the laboratory, obtain a precipitate of zinc hydroxide. Shake the precipitate obtained and pour small amounts of it into 2 tubes. In one of the tubes add a solution of hydrochloric acid, in the other - a solution of sodium hydroxide (excess). What is observed? Write the equations of the corresponding reactions in molecular and ionic form.

EXPERIENCE 3.Chemical properties of salts.

a) The interaction of salt solutions with the formation of a sparingly soluble substance.

Pour 2 ml of sodium carbonate solution into the tube and add barium chloride solution until a white precipitate appears. Write the chemical reaction equation in ionic and molecular form. The precipitate is divided into two parts. Pour sulfuric acid solution into one of the tubes, and sodium hydroxide into the other. Make a conclusion about the solubility of the precipitate in acids and alkalis.

b) Interaction of the salt solution with acids to form a volatile compound.

Pour 2 ml of sodium carbonate solution into the tube and add a small volume of hydrochloric acid solution. What is observed? Write the chemical reaction equations in ionic and molecular form.

c) Interaction of salt solutions with alkalis with the formation of a volatile compound.

Pour a little solution of some ammonium salt into the tube, add 1-2 ml of sodium hydroxide solution and heat to boiling. In a test tube with the reaction mixture to make a wet red litmus test. What is observed? Give an explanation. Write the reaction equations.

g ) Interactions of salts with more active metals than the metal that is part of the salt.

Clean the iron (steel) nail with fine sandpaper. Then dip it in copper sulphate solution. After some time, observe the release of copper on the surface of the nail. Write the corresponding reaction equation in ionic and molecular form.

EXPERIENCE 4.Getting basic and acidic salts.

a) Preparation of lead hydroxocarbonate.

Add a little lead (II) oxide to the lead (II) acetate solution and boil the mixture for a few minutes. The cooled solution is drained from the precipitate and a stream of carbon dioxide is passed through it. What is observed? Filter the precipitate and dry between the sheets of filter paper. Mark the color and nature of the resulting precipitate of lead hydroxocarbonate. Write the reaction equations. Make a graphic formula of the obtained salt.

b) Preparation of magnesium bicarbonate.

To a very highly diluted solution of some magnesium salt add a little solution of sodium carbonate. What substance precipitates? Saturated solution with precipitate with carbon dioxide. Observe the gradual dissolution of the precipitate. Why is this happening? Write the reaction equations.

EXPERIENCE 5.Getting complex salts.

a) Formation of compounds with complex cation.

Add a solution of ammonia dropwise to a tube with 2-3 ml of copper (II) chloride solution until a copper (II) hydroxide precipitate forms, and then add an excess of ammonia solution to dissolve the precipitate. Compare the color of the ions Cu 2+ with the color of the resulting solution. What ions are present in the solution? Write the reaction equation for obtaining a complex compound.

b) Formation of compounds with a complex anion.

To a 1-2 ml solution of mercury (II) nitrate add dropwise a diluted solution of potassium iodide to form a precipitate of HgI 2. Then pour in an excess of potassium iodide solution until the precipitate is dissolved. Write the reactions of obtaining a complex salt.

EXPERIENCE 6.Preparation of double salts (potassium alum).

Weigh 7.5 g of Al 2 (SO 4) 3 ∙ 18H 2 O and dissolve in 50 ml of water, taking a sufficiently large porcelain dish for this purpose. Calculate the reaction equation and weigh the mass of potassium sulfate necessary for the reaction. Prepare a hot saturated solution of potassium sulfate and pour it into a porcelain dish with a solution of aluminum sulfate with stirring. Observe after a while the precipitation of aluminum potassium alum crystals. After cooling and crystallization, pour off the mother liquor, dry the alum crystals between the sheets of filter paper and weigh the resulting crystals. Calculate the yield percentage.

SETTLEMENT TASKS

1. By passing an excess of hydrogen sulfide through 16 g of copper (II) sulfate solution, 1.92 g of precipitate was obtained. Find the mass fraction of copper sulfate in the used solution and the volume of consumed hydrogen sulfide.

2. For the complete precipitation of copper in the form of sulfide from 291 cm 3 of copper (II) sulfate solution with a mass fraction of 10%, gas obtained by reacting 17.6 g of iron (II) sulfide with an excess of hydrochloric acid was used. Find the density of the original copper sulfate solution.

3. The gas released during the interaction of the K 2 S solution with dilute sulfuric acid is passed through an excess of lead (II) nitrate solution. The resulting precipitate has a mass of 71.7 g. Find the volume of the reacted solution of sulfuric acid, if its density is 1.176 g / cm 3, and the mass fraction is 25%.

4. To a solution containing 8 g of copper (II) sulfate, a solution containing 4.68 g of sodium sulfide was added. The precipitate was filtered, the filtrate was evaporated. Determine the mass of substances in the filtrate after evaporation and the mass of copper sulfide precipitate.

5. Some of the iron (II) sulfide was treated with an excess of hydrochloric acid. The resulting gas in the reaction with 12.5 cm 3 of NaOH solution with a mass fraction of 25% and a density of 1.28 g / cm 3 formed an acid salt. Find the mass of the original iron sulfide.

6. Iron (II) sulfide weighing 176 g was treated with an excess of hydrochloric acid, and the resulting gas was burned in excess air. What volume of KOH solution with a mass fraction of 40% and a density of 1.4 g / cm 3 is needed to fully neutralize the gas produced during combustion?

7. When roasting 100 g of technical pyrite, they obtained a gas, which was completely neutralized with 400 cm 3 of NaOH solution with a mass fraction of 25% and a density of 1.28 g / cm 3. Determine the mass fraction of impurities in pyrite.

8. To 2 g of a mixture of iron, iron oxide (II) and iron oxide (III) was added 16 cm 3 of the HC1 solution with a mass fraction of 20% and a density of 1.09 g / cm 3. To neutralize the excess acid, it took 10.8 cm 3 of NaOH solution with a mass fraction of 10% density of 1.05 g / cm 3. Find the masses of the substances in the mixture if the volume of hydrogen released is 224 cm 3 (NU).

9. There is a mixture of Ca (OH) 2, CaCO 3 and BaSO 4 with a mass of 10.5 g. When the mixture was treated with an excess of hydrochloric acid, 672 cm 3 (IU) of gas was released, and 71.2 g of the acid reacted with a mass share of 10%. Determine the mass of substances in the mixture.

10. There is a mixture of barium chloride, calcium carbonate and sodium bicarbonate. When 10 g of this mixture is dissolved in water, the insoluble residue is equal to 3.5 g. When calcining 20 g of the initial mixture, its mass decreases by 5.2 g. Find the mass fractions of substances in the initial mixture.

11. There is a solution containing both sulfuric and nitric acids. To completely neutralize 10 g of this solution, 12.5 cm 3 of KOH solution is consumed with a mass fraction of 19% and a density of 1.18 g / cm 3. When an excess of barium chloride is added to 20 g of the same mixture of acid solution, 4.66 g of precipitate fall out. Find the mass fractions of acids in the mixture.

12. All hydrogen chloride obtained from 100 g of a mixture of KC1 and KNO 3 was dissolved in 71.8 cm 3 of water. When calcining 100 g of the same mixture of salts, 93.6 g of solid residue remain. Find the mass fraction of hydrogen chloride in the water.

13. By passing 2 m 3 of air (NU) through a solution of Ca (OH) 2, 3 g of carbonic acid salt precipitate was obtained. Find the volume and mass fraction of CO 2 in the air.

14. Carbon dioxide is passed through a suspension containing 50 g of CaCO 3. The reaction came 8.96 dm 3 gas (NU). What is the mass of CaCO 3 left in the solid phase?

15. When adding water to CaO, its mass increased by 30%. What part of CaO (in mass%) was extinguished?

16. Lead (II) oxide weighing 18.47 g was heated in a stream of hydrogen. After the reaction, the mass of the resulting lead and unreacted oxide was 18.07 g. What is the mass of lead oxide that did not react?

17. Carbon monoxide is passed through iron (III) oxide when heated. The mass of the solid residue after the reaction is 2 g less than the initial mass of iron oxide. What volume of CO did the reaction take (is the oxide completely reduced)?

18. There is 8.96 dm 3 (NU) of a mixture of N 2, CO 2 and SO 2 with a relative density of hydrogen 25. After passing it through an excess of KOH solution, the volume of the mixture decreased by 4 times. Find the volumes of gases in the original mixture.

19. In two glasses there is 100 g of HC1 solution with a mass fraction of 2.5%. In one glass was added 10 g of CaCO 3, in the other - 8.4 g of MgCO 3. How will the mass of glasses differ after the reaction?

20. What volume (NU) of sulfur dioxide must be passed through 200 cm 3 of a solution with a 0.1% mass fraction of NaOH and a density of 1 g / cm 3 to get an acid salt?

21. What is the maximum amount (carbon dioxide) of carbon dioxide can absorb 25 cm 3 of a solution with a 25% mass fraction of NaOH and a density of 1.1 g / cm 3?

22. What is the minimum volume of the solution with a mass fraction of 20% KOH and a density of 1.19 g / cm 3 that can absorb all the carbon dioxide produced with the full reduction of 23.2 g of magnetite with carbon monoxide?

23. What is the minimum mass of KOH that must be reacted with 24.5 g of orthophosphoric acid so that only potassium dihydrogen phosphate is the product?

24. What is the minimum mass of Ca (OH) 2 to add to 16 g of calcium bicarbonate solution with a mass fraction of 5% salt to obtain a medium salt?

25. What mass of potassium hydrogen phosphate should be added to a solution containing 12.25 g of H 3 PO 4 so that after this the solution contains only potassium dihydrogen phosphate?

26. The solution contained 56.1 g of a mixture of calcium and magnesium carbonates in suspension. To convert them into hydrocarbonates, they spent all the carbon dioxide produced by burning 7 dm 3 (NU) of ethane. Find the mass of calcium carbonate in the initial mixture.

27. To convert 9.5 g of a mixture of sodium hydro- and dihydrogen phosphate to a medium salt, 10 cm 3 of a solution with a mass fraction of NaOH of 27.7% and a density of 1.3 g / cm 3 is necessary. Find the mass of hydrogen phosphate in the mixture.

28. By passing carbon dioxide through a solution containing 6 g of NaOH, 9.5 g of a mixture of acid and medium salts was obtained. Find the amount of carbon dioxide consumed.

29. After passing 11.2 dm 3 (NU) CO 2 through the KOH solution, 57.6 g of a mixture of acid and medium salts were obtained. Find the mass of medium salt.

30. What mass of orthophosphoric acid must be neutralized in order to obtain 1.2 g of dihydro- and 4.26 g of sodium hydrogen phosphate?

31. NaOH was added to the sulfuric acid solution to obtain 3.6 g of hydrosulfate and 2.84 g of sodium sulfate. Determine the chemical amounts of the acid and alkali that have reacted.

32. After passing hydrogen chloride through 200 cm 3 of NaOH solution with a mass fraction of 10% and a density of 1.1 g / cm 3, the mass fraction of NaOH in the resulting solution decreased by half. Determine the mass fraction of NaCl in the resulting solution.

33. Dissolving 14.4 g of a mixture of copper and its oxide (II) consumed 48.5 g of a solution with an HNO 3 mass fraction of 80%. Find the mass fractions of copper and oxide in the initial mixture.

34. Sodium oxide mass of 6.2 g was dissolved in 100 cm 3 of water and solution No. 1 was obtained. Then hydrochloric acid with a mass fraction of 10% was added to this solution until the medium became neutral, and solution No. 2 was obtained. Determine :

1) mass fractions of substances in solutions No. 1, 2;

2) the mass of the solution of HC1, which went to neutralize the solution number 1.

35. 3 g of zinc interact with 18.69 cm 3 of HC1 solution with a mass fraction of 14.6% and a density of 1.07 g / cm 3. When heated, the resulting gas is passed over hot CuO weighing 4 g. What is the mass of copper that is obtained?

36. The gas released after the treatment of calcium hydride with excess water, missed over FeO. As a result, the oxide mass decreased by 8 g. Find the CaH 2 mass treated with water.

37. During calcination of the CaCO 3 sample, its mass decreased by 35.2%. The solid reaction products were dissolved in an excess of hydrochloric acid to obtain 0.112 dm 3 (NU) of gas. Determine the mass of the original sample of calcium carbonate.

38. Copper nitrate was decomposed, and the resulting copper (II) oxide was completely reduced by hydrogen. The resulting products passed through a tube with P 2 O 5, and the mass of the tube after that increased by 3.6 g. What is the minimum volume of sulfuric acid with a mass fraction of 88% and a density of 1.87 g / cm 3 needed for dissolution obtained in the experiment copper and what is the mass of decomposed salt?

39. Upon absorption of nitric oxide (IV) by an excess of KOH solution in the cold in the absence of oxygen, 40.4 g of KNO 3 were obtained. What substance is still formed and what is its mass?

40. To neutralize 400 g of a solution containing hydrochloric and sulfuric acid, consumed 287 cm 3 of sodium hydroxide solution with a mass fraction of 10% and a density of 1,115 g / cm 3. If an excess solution of barium chloride is added to 100 g of the initial solution, 5.825 g of precipitate will fall out. Determine the mass fractions of acids in the initial solution.

41. After passing carbon dioxide through a solution of sodium hydroxide, 13.7 g of a mixture of medium and acid salts was obtained. To convert them into sodium chloride, 75 g of hydrochloric acid with a mass fraction of HCl of 10% is needed. Find the volume of carbon dioxide absorbed.

42. A mixture of hydrochloric and sulfuric acids with a total mass of 600 g of solution with the same mass fractions of acids was treated with an excess of sodium bicarbonate and 32.1 dm 3 of gas (N o.) Was obtained. Find the mass fraction of acids in the initial mixture.

43. To neutralize 1 dm 3 of NaOH solution, 66.66 cm 3 of HNO 3 solution was consumed with a mass fraction of 63% and a density of 1.5 g / cm 3. What volume of sulfuric acid solution with a mass fraction of 24.5% and a density of 1.2 g / cm 3 would be needed to neutralize the same amount of alkali?

44. In what volume ratio should one take a solution of sulfuric acid with a mass fraction of 5% and a density of 1.03 g / cm 3 and a solution of barium hydroxide with a mass fraction of 5% and a density of 1.1 g / cm 3 for complete neutralization? Imagine the answer as the quotient of the volume of the alkali solution on the acid solution

45. Calculate the minimum volume of ammonia solution with a density of 0.9 g / cm 3 and a mass fraction of 25%, which is necessary for the complete absorption of carbon dioxide obtained by decomposing 0.5 kg of natural limestone with a mass fraction of calcium carbonate equal to 92%.

46. ​​For the complete conversion of 2.92 g of a mixture of sodium hydroxide and sodium carbonate into chloride, 1.344 dm 3 of hydrogen chloride (NU) is required. Find the mass of sodium carbonate in the mixture.

47. To 25 g of copper (II) sulfate solution with a mass fraction of 16% was added a quantity of sodium hydroxide solution with a mass fraction of 16%. The precipitate formed was filtered, after which the filtrate was alkaline. To completely neutralize the filtrate, it took 25 cm 3 of a solution of sulfuric acid with a molar concentration of 0.1 mol / dm 3 of the solution. Calculate the mass of the added sodium hydroxide solution.

48. The substance obtained with the complete reduction of CuO with a mass of 15.8 g of hydrogen with a volume of 11.2 dm 3 (NU) was dissolved by heating in concentrated sulfuric acid. What is the volume of gas (NU) released as a result of the reaction?

49. For the complete neutralization of 50 cm 3 of hydrochloric acid with a mass fraction of HCl of 20% and a density of 1.10 g / cm 3, a solution of potassium hydroxide with a mass fraction of KOH of 20% was used. What is the chemical amount of water contained in the resulting solution?

50. The gas obtained by passing an excess of CO 2 over 0.84 g of hot coal, is sent to the reaction with 14.0 g of heated copper (II) oxide. What volume of a solution of nitric acid with a mass fraction of 63% and a density of 1.4 g / cm 3 is needed to completely dissolve the substance obtained in the last reaction?

51. When calcined to constant weight of copper (II) nitrate, the weight of salt decreased by 6.5 g. What weight of salt was decomposed?

52. When an excess of hydrochloric acid was used, 6.72 dm 3 (NW) of gas was released on a mixture of aluminum with an unknown monovalent metal, and the mass of the mixture was halved. When treating the residue with dilute nitric acid, 0.373 dm 3 (NU) NO was released. Identify unknown metal.

53. The mass of the chalk sample is 105 g, and the chemical quantity of the oxygen element in its composition is 1 mol. Determine the mass fraction of CaCO 3 in the chalk sample (oxygen is included only in the composition of calcium carbonate).

54. In the interaction of sulfur oxide (VI) with water, a solution was obtained with a mass fraction of sulfuric acid of 25%. When an excess of Ba (OH) 2 was added to this solution, a precipitate of 29.13 g was deposited. What masses of SO 3 and H 2 O were spent on the formation of an acid solution?

55. When passing SO 2 through 200 g of a solution with a mass fraction of NaOH of 16%, a mixture of salts was formed, including 41.6 g of an acid salt. What is the mass of sulfur containing 4.5% impurities by weight, was used to obtain SO 2? What is the weight of medium salt?

56. The interaction with 80 g of Ca (NO 3) 2 solution took 50 g of Na 2 CO 3 solution. The precipitation was separated, while processing it with an excess of hydrochloric acid, 2.24 dm 3 (NU) of gas was released. What are the mass fractions of salts in the initial solutions? What is the mass fraction of sodium nitrate in the solution after separating the precipitate?

57. When zinc interacts with sulfuric acid, 10 dm 3 (NU) of a mixture of SO 2 and H 2 S with an argon relative density of 1.51 is formed. What is the chemical amount of zinc dissolved? What is the mass fraction of SO 2 in the gas mixture?

58. A sample of a mixture of zinc and aluminum sawdust with a total mass of 11 g was dissolved in an excess of alkali solution. Determine the volume (NU) of the released gas, if the mass fraction of zinc in the mixture is 30%.

59. Sodium hydroxide weighing 4.0 g was alloyed with aluminum hydroxide weighing 9.8 g. Calculate the mass of the sodium metaaluminate obtained.

60. When processing 10 g of a mixture of copper and aluminum with concentrated nitric acid at room temperature  released 2.24 dm 3 gas (NU). What volume (NU) of gas will be released when processing the same mixture mass with an excess of KOH solution?

61. An alloy of copper and aluminum weighing 20 g was treated with an excess of alkali, the insoluble residue was dissolved in concentrated nitric acid. The salt obtained was isolated, calcined to constant weight, and 8 g of solid residue was obtained. Determine the volume of spent NaOH solution with a mass fraction of 40% and a density of 1.4 g / cm 3).

62. A mixture of aluminum and metal oxide (II) (non-amphoteric oxide) weighing 39 g was treated with an excess of KOH solution, the emitted gas was burned to obtain 27 g of water. The undissolved residue was completely dissolved in 25.2 cm 3 of a solution with a mass fraction of HC1 of 36.5% and a density of 1.19 g / cm 3). Determine the oxide.

63. A mixture of zinc and copper chips was treated with an excess of KOH solution, while a gas of 2.24 dm 3 (NU) was released. For the complete chlorination of the same metal sample, chlorine with a volume of 8.96 dm 3 (NU) was required. Calculate the mass fraction of zinc in the sample.

64. A mixture of sawdust of iron, aluminum and magnesium with a mass of 49 g was treated with excess diluted H 2 SO 4, thus obtaining 1.95 mol of gas. Another portion of the same mixture weighing 4.9 g was treated with an excess of alkali solution; 1.68 dm 3 (NU) of gas was obtained. Find the mass of metals in the mixture.

65. What is the mass of sediment formed when merging solutions containing 10 g of NaOH and 13.6 g of ZnCl 2?

66. There are two portions of the mixture of Al, Mg, Fe, Zn, the same in molar composition, each weighing 7.4 g. One portion was dissolved in hydrochloric acid and 3.584 dm 3 of gas (NU) was obtained, the other in alkali solution and Received 2.016 dm 3 gas (NU). It is known that in both mixtures there are 3 Zn atoms per A1 atom. Find the mass of metals in the mixture.

67. A mixture of copper, magnesium and aluminum weighing 1 g was treated with an excess of hydrochloric acid. The solution was filtered, an excess of NaOH solution was added to the filtrate. The precipitate was separated and calcined to a constant mass of 0.2 g. The residue, which was undissolved after treatment with hydrochloric acid, was calcined in air and 0.8 g of black substance was obtained. Find the mass fraction of aluminum in the mixture.

68. When heated in a stream of oxygen alloy of zinc, magnesium and copper, the mass of the mixture increased by 9.6 g. The product is partially dissolved in alkali, and 40 cm 3 of a solution with a mass fraction of KOH 40% and density of 1.4 g / cm is necessary for dissolution 3 For the reaction with the same portion of the alloy, 0.7 mol HC1 is needed. Find chemical amounts of metals in the alloy.

69. An alloy of copper and zinc weighing 5 g was treated with an excess of NaOH solution. Then the solid residue was separated and treated with concentrated HNO 3, the salt thus obtained was isolated, calcined to constant weight, and 2.5 g of a solid residue was obtained. Determine the mass of metals in the alloy.

70. An alloy of copper and aluminum weighing 12.8 g was treated with an excess of hydrochloric acid. The undissolved residue was dissolved in concentrated nitric acid, the resulting solution was evaporated, the dry residue was calcined to constant weight and 4 g of solid were obtained. Determine the mass fraction of copper in the alloy.

71. In what mass ratio should take two portions of A1, so that when one is added to the solution of alkali and the other to equal amounts of hydrogen are released into the hydrochloric acid?

72. When processing a mixture of aluminum and copper (II) oxide with an excess of KOH solution, 6.72 dm 3 (NU) of gas was released, and by dissolving the same portion of the mixture in concentrated HNO 3 at room temperature, 75.2 g of salt was obtained. Find the mass of the initial mixture of substances.

73. What mass of copper (II) oxide can be restored by hydrogen obtained by reacting an excess of aluminum with 139.87 cm 3 of a solution with 40% NaOH and a density of 1.43 g / cm 3?

74. With the complete oxidation of 7.83 g of the alloy of two metals, 14.23 g of oxides were formed, during the treatment of which an excess of alkali remained undissolved 4.03 g of sediment. Determine the qualitative composition of the metals forming the alloy, if their cations have an oxidation state of +2 and +3, and the molar ratio of oxides is 1: 1 (assume that the metal oxide with oxidation state +3 has amphoteric properties).

75. Two portions of aluminum, having the same mass, were dissolved: one in a solution of potassium hydroxide, the other in hydrochloric acid. How do the volumes of evolved gases (nos) relate to each other?

76. An alloy of copper with aluminum weighing 1,000 g was treated with an excess of alkali solution, the undissolved precipitate was dissolved in nitric acid, then the solution was evaporated, the residue was calcined to constant weight. The mass of the new residue is 0.398 g. What are the mass of metals in the alloy?

77. An alloy of zinc and copper weighing 20 g was treated with an excess of NaOH solution with a mass fraction of 30% and a density of 1.33 g / cm 3. The solid residue was isolated and treated with an excess of a concentrated solution of HNO 3. The salt formed during this process was isolated and calcined to constant weight. The mass of the solid residue was 10.016 g. Calculate the mass fractions of metals in the alloy and the volume of alkali solution consumed.

78. An alloy of copper and aluminum weighing 2 g was treated with an excess of alkali solution. The residue was filtered, washed, dissolved in HNO 3, the solution was evaporated and calcined to constant weight. The mass of the residue after calcination was 0.736 g. Calculate the mass fractions of metals in the alloy.

79. Chlorination of a mixture of iron, copper and aluminum requires 8.96 dm 3 of chlorine (NU), and the interaction of the same sample with hydrogen chloride requires 5.6 dm 3 (NU). When interacting with the same mass of a mixture of metals with alkali, 1.68 dm 3 (NU) of gas is released. Find chemical quantities of metals in the mixture.

80. Potassium hydride weighing 5.0 g was dissolved in water with a volume of 80 cm 3 and aluminum weighing 0.81 g was added to the resulting solution. Find the mass fractions of the substances in the resulting solution with an accuracy of up to thousandths of a percent.

BIBLIOGRAPHY

1. Barannik, V.P. Modern Russian nomenclature of inorganic compounds / V.P. Barannik // Journal of the All-Union Chemical Society. DI. Mendeleev. - 1983. - Vol. XXVIII. - p. 9–16.

2. Wroblewski, A.I. Chemistry simulator / A.I. Wroblewski. - 2nd ed., Pererab. and add. - Minsk: Krasiko-Print, 2007. - 624 p.

3. Glinka, N.L. Tasks and exercises in general chemistry: studies. manual for universities / ed. V.A. Rabinovich and H.M. Rubina. - M.: Integral-Press, 2004. - 240 p.

4. Lidin, R.A. Tasks in general and inorganic chemistry: studies. manual for students of higher. studies. institutions / R.A. Lidin, V.A. Milk, L.L. Andreeva; by ed. R.A. Lidina. - M.: VLADOS, 2004. - 383 p.

5. Lidin, R.A. Fundamentals of inorganic substances nomenclature / R.А. Lidin [et al.]; by ed. B.D. Stepina. - M .: Chemistry, 1983. - 112 p.

6. Stepin, B.D. Application of IUPAC rules on the nomenclature of inorganic compounds in Russian / B.D. Stepin, R.A. Lidin // Journal of the All-Union Chemical Society. DI. Mendeleev. - 1983. - Vol. XXVIII. - p. 17–20.

educational: to fix the concept of "genetic series", "genetic connection"; teach how to make genetic series of elements (metals and non-metals), to make up the reaction equations corresponding to the genetic series; check how knowledge of the chemical properties of oxides, acids, salts, bases is assimilated;

developing:   develop the ability to analyze, compare, summarize and draw conclusions, make up the equations of chemical reactions;

educational :   promote the formation of a scientific worldview.

2. Providing classes: the tables “Periodic system”, “Solubility table”, “Metals activity series”, instructions for students, tasks for testing knowledge.

3. Order of performance:

3.1. Frontal survey.

3.2. The solution of tasks.

3.3. Perform verification work on options.

4. Report layout:

4.1. Write the topic and objectives of the practical lesson.

4.2. Record problem solving.

4.3. Solve your own version of independent work, the decision to write in a notebook and pass on to the teacher for review.

Working process

1. Frontal conversation on:

What is a genetic connection?

Genetic connections   - these are connections between different classes based on their interconversions.

What is a genetic series?

Genetic row   - a series of substances - representatives of different classes, which are compounds of one chemical element, connected by interconversions and reflecting the transformations of these substances. The basis of these series is the same element.

What types of genetic series are usually allocated?

Among metals there are two types of rows:

a) A genetic series in which alkali acts as a base. This series can be represented using the following transformations:

metal → basic oxide → alkali → salt

for example, the genetic series of potassium K → K 2 O → KOH → KCl

b) The genetic series, where the base is an insoluble base, then the series can be represented as a chain of transformations:

metal → base oxide → salt → insoluble base → base oxide → metal

for example: Cu → CuO → CuCl 2 → Cu (OH) 2 → CuO → Cu

Among non-metals it is also possible to distinguish two types of rows:

a) The genetic series of non-metals, where soluble acid acts as a link in the series. 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 → Na 3 PO 4

b) Genetic row of non-metals, where insoluble acid: non-metal → acid oxide → salt → acid → acid oxide → non-metal acts as a link in the series

For example: Si → SiO 2 → Na 2 SiO 3 → H 2 SiO 3 → SiO 2 → Si

Execution of tasks for options:

I option: ZnSO 4, Zn, ZnO, Zn, Zn (OH) 2

Option II: Na 2 SO 4, NaOH, Na, Na 2 O 2, Na 2 O

Homework:   chart a gradual transition from calcium to calcium carbonate and prepare a report on the use of any salt in medicine (using additional literature).

Instructions for practical lesson

Genetic relationship between the main classes of inorganic substances.

Goals: to fix the concept of "genetic series", "genetic connection"; learn to make genetic series of elements (metals and non-metals), to make up the reaction equations corresponding to the genetic series; repeat the properties of oxides, acids, salts, bases.

Working process

Write down the definitions of the concepts:

Genetic relationship - __________________________________________

Genetic row - ___________________________________________

A genetic series of metals in which alkali acts as a base.   can be represented in general: metal → basic oxide → alkali → salt. Make this series using potassium. Write the equations of reactions, with the help of which you can accomplish this chain of transformations.

The genetic series of non-metals, where soluble acid acts as a link in the series   can be represented as follows: non-metal → acid oxide → soluble acid → salt. Make this series using phosphorus. Write the equations of reactions, with the help of which you can accomplish this chain of transformations.

Genetic row, where the base is insoluble base can be represented by the chain of transformations: metal → basic oxide → salt → insoluble base → basic oxide → metal. Build this series using copper. Write the equations of reactions, with the help of which you can accomplish this chain of transformations.

The genetic series of non-metals, where insoluble acid acts as a link in the series can be represented by the chain of transformations: non-metal → acid oxide → salt → acid → acid oxide → non-metal. Build this series using silicon. Write the equations of reactions, with the help of which you can accomplish this chain of transformations.

Perform tasks for options:

1. Choose the oxide formulas in your version, explain your choice, based on the knowledge of the characteristics of the composition of this class of compounds. Call them.

2. In the formula column of your variant, find the acid formulas and explain your choice based on the analysis of the composition of these compounds.

3. Determine the valencies of the acid residues in the acid composition.

4. Select salt formulas and name them.

5. Make up the formulas of salts, which can be formed with magnesium and acids of your variant. Write them down, name it.

6. In the formula column of your variation, find the base formulas and explain your choice based on the analysis of the composition of these compounds.

7. In your variant, choose the formulas of substances with which a solution of orthophosphoric acid (hydrochloric, sulfuric) can react. Make the appropriate reaction equations.

9. Among the formulas of your option, select formulas of substances that can interact with each other. Make the appropriate reaction equations.

10. Make a chain of genetic bonds of inorganic compounds, which will include a substance, the formula of which is given in your version at number one.

From these substances make a genetic line using all the formulas. Write the reaction equations with which you can accomplish this chain of transformations:

I  option:  ZnSO 4, Zn, ZnO, Zn, Zn (OH) 2

IIoption:   Na 2 SO 4, NaOH, Na, Na 2 O 2, Na 2 O

Homework: chart a gradual transition from calcium to calcium carbonate and prepare a report on the use of any salt in medicine (using additional literature).