Communication on the topic of the genetic relationship of inorganic compounds. Genetic relationship between classes of inorganic substances

There is a genetic relationship between the classes of inorganic compounds. Complex substances can be obtained from simple substances and vice versa. Compounds of one class can be used to obtain compounds of another class.

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

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

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

(CaOH) 3 PO 4

For typical metals, the following chain of transformations can be carried out:

Ba → BaO → Ba (OH) 2 → BaSO 4

For metals, oxides and hydroxides of which 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.

Relationships between classes:

1. Metals, non-metals→ salt.

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

2Na + C1 2 = 2NaCl

These compounds are stable and usually do not decompose when heated.

2. Basic oxides, acidic oxides→salt.

CaO + CO 2 = CaCO 3;

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

3. Bases, acids→salt.

It is carried out by means of a neutralization reaction:

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

OH - + 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 to form oxides:

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

S + O 2 = SO 2.

6. Basic oxides→ grounds.

Only hydroxides of alkali and alkaline earth metals (alkali) can be obtained by direct interaction with water:

Na 2 O + H 2 O = 2NaOH;

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

The rest of the bases are obtained indirectly.

7. Acidic oxides→acid.

Acidic oxides interact with water to form the corresponding acids:

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

P 2 O 5 + 3H 2 O = 2H 3 PO 4.

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

8. Bases, acidic 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 + CO 2 = CaCO 3 ↓ + H 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 + 2HC1 = 2NaCl + H 2 O,

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

10. Foundations→basic oxides.

Insoluble bases and LiOH decompose on heating:

2LiOH = Li 2 O + H 2 O;

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

11. Acids→acid oxides.

Unstable oxygen-containing acids decompose on heating (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 using 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. Only some unstable oxides (halogen oxides) decompose into non-metal and oxygen.

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 interact with acids (in accordance with the displacement 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 (solid) + H 2 SO 4 (c) = NaHSO 4 + HCl.

16. Salts→ basic oxides, acidic oxides.

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 RELATIONSHIP BETWEEN CLASSES OF INORGANIC COMPOUNDS"

1. Name the substances listed below, distribute them according to 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 which of the following substances can a hydroxide (acid or base) be obtained in one stage: copper, iron oxide (P), barium oxide, nitric oxide (P), nitric 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) the 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 to which class each substance belongs:

1) copper (II) hydroxycarbonate;

2) nitric 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 connection reactions between:

1) simple non-metal substances;

2) a 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 may react with:

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, HC1, 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 interaction possible between the following substances:

1) carbon monoxide (IV) and potassium hydroxide;

2) potassium hydrogen sulfate and calcium hydroxide;

3) calcium phosphate and sulfuric acid;

4) calcium hydroxide and sulfur (IV) oxide;

5) sulfuric acid and potassium hydroxide;

6) calcium bicarbonate and phosphoric acid;

7) silicon oxide and sulfuric acid;

8) zinc oxide and phosphorus (V) oxide.

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

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

10. One of the ways to obtain soda (sodium carbonate) is the action of water and carbon monoxide (IV) on sodium aluminate. Write the reaction equations.

11. Without changing the coefficients, write the reaction products:

1) MgO + 2H 2 SO 4 →

2) 2SO 2 + Ba (OH) 2 →

3) 3N 2 O 5 + 2Аl (OH) 3 →

4) Р 2 O 5 + 4NaOH →

5) P 2 O 5 + 6NaOH →

6) P 2 O 5 + 2NaOH →

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

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

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

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

4) SO 3 + Ca (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 (short) → (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) Сu → СuО → CuSO 4 → Сu (ОН) 2 → СuС1 2 → Cu (NO 3) 2

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

3) P → P 2 O 5 → H 3 PO 4 → K 3 PO 4 → Ca 3 (PO 4) 2 → H 3 PO 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) Сu → 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 sulfate → aluminum chloride → aluminum nitrate → aluminum oxide → potassium aluminate → aluminum hydroxide → aluminum hydroxychloride → 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 → СО 2 → NaHCO 3 → Na 2 CO 3 → СО 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) P 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) К → 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) KC1 → 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 EXPERIMENTS ON THE TOPIC "BASIC CLASSES OF INORGANIC COMPOUNDS"

EXPERIENCE 1. Neutralization reactions.

a) Interaction of a strong acid and a strong base.

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

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

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

EXPERIENCE 2. Amphotericity of hydroxides.

Obtain zinc hydroxide precipitate from the reagents available in the laboratory. Shake the resulting precipitate and pour small amounts into 2 test tubes. Add hydrochloric acid solution to one of the tubes, and sodium hydroxide solution (excess) to the other. What is being observed? Write the equations of the corresponding reactions in molecular and ionic form.

EXPERIENCE 3. Chemical properties of salts.

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

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

b) The interaction of the salt solution with acids with the formation of a volatile compound.

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

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

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

G ) Interaction of salt solutions with more active metals than the metal that is part of the salt.

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

EXPERIENCE 4. Obtaining basic and acidic salts.

a) Obtaining lead hydroxocarbonate.

Add a little lead (II) oxide to the lead (II) acetate solution and boil the mixture for several minutes. Drain the cooled solution from the sediment and pass a stream of carbon dioxide through it. What is being observed? Filter the precipitate and dry it between the pieces of filter paper. Note the color and nature of the resulting precipitate of lead hydroxycarbonate. Write the reaction equations. Draw up a graphical formula for the salt obtained.

b) Getting magnesium bicarbonate.

Add a little sodium carbonate solution to a very dilute solution of some magnesium salt. What substance is precipitated? Saturate the solution with the precipitate with carbon dioxide. Observe the gradual dissolution of the precipitate. Why is this happening? Write the reaction equations.

EXPERIENCE 5. Obtaining complex salts.

a) Formation of compounds with a complex cation.

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

b) Formation of compounds with a complex anion.

To 1-2 ml of a solution of mercury (II) nitrate, add dropwise a diluted solution of potassium iodide until a precipitate of HgI 2 is formed. Then add an excess of potassium iodide solution until the precipitate dissolves. Write the reaction equations for obtaining a complex salt.

EXPERIENCE 6. Obtaining double salts (potassium alum).

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

CALCULATION PROBLEMS

1. When passing excess 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 a 291 cm 3 solution of copper (II) sulfate with a mass fraction of 10%, a gas obtained by the interaction of 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 sediment has a mass of 71.7 g. Find the volume of the reacted sulfuric acid solution 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 was added a solution containing 4.68 g of sodium sulfide. The precipitate was filtered off, the filtrate was evaporated. Determine the mass of substances in the filtrate after evaporation and the mass of the copper sulfide precipitate.

5. Some of the iron (II) sulfide was treated with an excess of hydrochloric acid. The resulting gas in 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 a KOH solution with a mass fraction of 40% and a density of 1.4 g / cm 3 is needed to completely neutralize the gas obtained during combustion?

7. When firing 100 g of technical pyrite, a gas was obtained, which completely neutralized 400 cm 3 of a 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 (II) oxide and iron (III) oxide was added 16 cm 3 of HC1 solution with a mass fraction of 20% and a density of 1.09 g / cm 3. To neutralize the excess acid, 10.8 cm 3 of NaOH solution with a mass fraction of 10% and a density of 1.05 g / cm 3 were required. Find the masses of substances in the mixture if the volume of the evolved hydrogen is 224 cm 3 (NU).

9. There is a mixture of Ca (OH) 2, CaCO 3 and BaSO 4 weighing 10.5 g. When the mixture was treated with an excess of hydrochloric acid, 672 cm 3 (n.u.) of gas was released, and 71.2 g of acid with a mass shares of 10%. Determine the masses 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 3.5 g. When 20 g of the initial mixture is calcined, its mass decreases by 5.2 g. Find the mass fractions of substances in the original mixture.

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

12. All the hydrogen chloride obtained from 100 g of a mixture of KCl and KNO 3 was dissolved in 71.8 cm 3 of water. On calcining 100 g of the same mixture of salts, 93.6 g of a solid residue remains. Find the mass fraction of hydrogen chloride in 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 entered 8.96 dm 3 of gas (NU). What mass of CaCO 3 remained in the solid phase?

15. When adding water to CaO, its mass increased by 30%. What part of CaO (in% by 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 obtained lead and unreacted oxide was 18.07 g. What mass of lead oxide 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. How much CO has reacted (the oxide is 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 for hydrogen of 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. Two glasses contain 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 the glasses differ after the reaction?

20. What volume (n.o.) of sulfur dioxide must be passed through 200 cm 3 of a solution with a mass fraction of NaOH of 0.1% and a density of 1 g / cm 3 in order to obtain an acidic salt?

21. What is the maximum volume (n.u.) of carbon dioxide that can be absorbed by 25 cm 3 of a solution with a mass fraction of NaOH of 25% and a density of 1.1 g / cm 3?

22. What is the minimum volume of a solution with a mass fraction of KOH of 20% and a density of 1.19 g / cm 3 it is possible to absorb all the carbon dioxide obtained in the complete 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 be added 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 Н 3 РО 4, so that after that the solution contains only potassium dihydrogen phosphate?

26. The solution in the form of a suspension contained 56.1 g of a mixture of calcium and magnesium carbonates. To convert them into bicarbonates, we spent all the carbon dioxide obtained by burning 7 dm 3 (NU) ethane. Find the mass of calcium carbonate in your original mixture.

27. To convert 9.5 g of a mixture of sodium hydro- and dihydrogen phosphate into an average salt, you need 10 cm 3 of a solution with a mass fraction of NaOH of 27.7% and a density of 1.3 g / cm 3. 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 acidic and medium salts were 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 acidic and medium salts were obtained. Find the mass of medium salt.

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

31. NaOH was added to the sulfuric acid solution and 3.6 g of hydrogen sulfate and 2.84 g of sodium sulfate were obtained. Determine the chemical quantities of 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. The dissolution of 14.4 g of a mixture of copper and its oxide (II) consumed 48.5 g of a solution with a mass fraction of HNO 3 80%. Find the mass fractions of copper and oxide in the original mixture.

34. Sodium oxide with a mass of 6.2 g was dissolved in 100 cm 3 of water and received solution No. 1. 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. :

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

2) the mass of the HC1 solution, which was used to neutralize the solution No. 1.

35. 3 g of zinc react with 18.69 cm 3 of HC1 solution with a mass fraction of 14.6% and a density of 1.07 g / cm 3. The resulting gas, when heated, is passed over an incandescent CuO weighing 4 g. What is the mass of copper obtained?

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

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

38. Decomposed copper nitrate, and the resulting copper (II) oxide was completely reduced with hydrogen. The resulting products were passed through a tube with P 2 O 5, and the tube mass then 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 to dissolve the experiment copper and what is the mass of decomposed salt?

39. The absorption of nitrogen oxide (IV) with an excess of KOH solution in the cold in the absence of oxygen yielded 40.4 g of KNO 3. What substance is still formed and what is its mass?

40. To neutralize 400 g of a solution containing hydrochloric and sulfuric acids, 287 cm 3 of sodium hydroxide solution with a mass fraction of 10% and a density of 1.115 g / cm 3 were consumed. If an excess of barium chloride solution is added to 100 g of the original solution, 5.825 g of precipitate will fall out. Determine the mass fraction of acids in the original solution.

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

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

43. To neutralize 1 dm 3 of NaOH solution, 66.66 cm 3 of HNO 3 solution with a mass fraction of 63% and a density of 1.5 g / cm 3 were consumed. What volume of sulfuric acid solution with a mass fraction of 24.5% and a density of 1.2 g / cm 3 would be required 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? Present the answer as the quotient of dividing the volume of an alkali solution by an 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 from the decomposition of 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 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. A certain amount of sodium hydroxide solution with a mass fraction of 16% was added to 25 g of a solution of copper (II) sulfate with a mass fraction of 16%. The formed precipitate was filtered off, after which the filtrate was alkaline. For complete neutralization of the filtrate, 25 cm 3 of sulfuric acid solution with a molar concentration of 0.1 mol / dm 3 of solution was required. Calculate the mass of sodium hydroxide solution added.

48. The substance obtained with the complete reduction of CuO weighing 15.8 g with hydrogen of 11.2 dm 3 (NU) was dissolved with heating in concentrated sulfuric acid. What volume of gas (n.o.) was 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 20% and a density of 1.10 g / cm 3, a solution of potassium hydroxide with a mass fraction of KOH 20% was used. What is the chemical amount of water in the resulting solution?

50. The gas obtained by passing excess CO 2 over 0.84 g of hot coal is reacted with 14.0 g of heated copper (II) oxide. What volume of a nitric acid solution 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 a constant mass of copper (II) nitrate, the mass of the salt decreased by 6.5 g. What mass of the salt has undergone decomposition?

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

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

54. Upon interaction of sulfur (VI) oxide with water, a solution with a mass fraction of sulfuric acid of 25% was obtained. When an excess of Ba (OH) 2 was added to this solution, a precipitate with a mass of 29.13 g formed. What masses of SO 3 and H 2 O were spent on the formation of the 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 mass of sulfur containing 4.5% impurities by mass was used to obtain SO 2? What is the mass of medium salt?

56. Interaction with 80 g of Ca (NO 3) 2 solution required 50 g of Na 2 CO 3 solution. The precipitate that formed was separated; when it was treated 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 separation of the precipitate?

57. The interaction of zinc with sulfuric acid formed 10 dm 3 (NU) of a mixture of SO 2 and H 2 S with a relative density of 1.51 for argon. What chemical amount of zinc was 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 weight of 11 g was dissolved in an excess of alkali solution. Determine the volume (n.o.) of the evolved gas, if the mass fraction of zinc in the mixture is equal to 30%.

59. Sodium hydroxide with a mass of 4.0 g was fused with aluminum hydroxide with a mass of 9.8 g. Calculate the mass of the obtained sodium meta-aluminate.

60. When treating 10 g of a mixture of copper and aluminum with concentrated nitric acid at room temperature, 2.24 dm 3 of gas (NU) was released. What volume (n.o.) of gas will be released when processing the same mass of a mixture 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 thus obtained was isolated, calcined to constant weight, and 8 g of a solid residue was obtained. Determine the volume of the consumed 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) (oxide is not amphoteric) weighing 39 g was treated with an excess of KOH solution, the evolved gas was burned and 27 g of water was obtained. The insoluble residue was completely dissolved in 25.2 cm 3 of a solution with a mass fraction of HCl 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 with a volume of 2.24 dm 3 (NU) was evolved. For the complete chlorination of the same sample of metals, 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 iron, aluminum and magnesium sawdust weighing 49 g was treated with an excess of dilute 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, and 1.68 dm 3 (NU) of gas was obtained. Find the masses of metals in the mixture.

65. What is the mass of the precipitate formed when the solutions containing 10 g of NaOH and 13.6 g of ZnCl 2 are merged?

66. There are two portions of a mixture of Al, Mg, Fe, Zn of the same molar composition, each weighing 7.4 g. One portion was dissolved in hydrochloric acid and received 3.584 dm 3 of gas (NU), the other in an alkali solution and received 2.016 dm 3 of gas (NU). It is known that in both mixtures there are 3 Zn atoms per A1 atom. Find the masses 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 resulting precipitate was separated and calcined to constant weight equal to 0.2 g. The residue, which did not dissolve after treatment with hydrochloric acid, was calcined in air to obtain 0.8 g of a black substance. Find the mass fraction of aluminum in the mixture.

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

69. Alloy of copper with 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 resulting salt 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 insoluble residue was dissolved in concentrated nitric acid, the resulting solution was evaporated, the dry residue was calcined to constant weight, and 4 g of a solid was obtained. Determine the mass fraction of copper in the alloy.

71. In what mass ratio should two portions of A1 be taken so that when one is introduced into an alkali solution and the other into hydrochloric acid, equal volumes of hydrogen are released?

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

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

74. With the complete oxidation of 7.83 g of an alloy of two metals, 14.23 g of oxides were formed, during the treatment of which with an excess of alkali, 4.03 g of sediment remained undissolved. 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 (consider that a metal oxide with an oxidation state of +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. What is the relationship between the volumes of evolved gases (n.o.)?

76. An alloy of copper with aluminum weighing 1,000 g was treated with an excess of alkali solution, the insoluble residue 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 masses 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 concentrated HNO 3 solution. The resulting salt 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 consumed volume of alkali solution.

78. An alloy of copper and aluminum weighing 2 g was treated with an excess of alkali solution. The residue was filtered off, 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 (n.u.), and the interaction of the same sample with hydrogen chloride requires 5.6 dm 3 (n.u.). When the same mass of a mixture of metals interacts with alkali, 1.68 dm 3 (n.u.) of gas is released. Find the chemical amounts of metals in the mixture.

80. Potassium hydride with a mass of 5.0 g was dissolved in water with a volume of 80 cm 3 and aluminum with a mass of 0.81 g was added to the resulting solution. Find the mass fractions of substances in the resulting solution with an accuracy of 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. - t. XXVIII. - S. 9-16.

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

3. Glinka, N.L. Tasks and exercises in general chemistry: textbook. 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: textbook. manual for students of higher. study. institutions / R.A. Lidin, V.A. Molochko, L.L. Andreeva; ed. R.A. Lidina. - M.: VLADOS, 2004 .-- 383 p.

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

6. Stepin, B.D. Application of the 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. - t. XXVIII. - S. 17–20.

Introduction ……………………………………………………………… General rules for the nomenclature of inorganic substances ………… .. Oxides ………………………………… ……………………………… Basic oxides ………………………………………………… ... Acidic oxides ……………………………… ………………………. Amphoteric oxides ………………………………………………… .. Obtaining oxides …………………………………………………… Exercises for independent work on topic "Oxides" …………………………………………………. Acids ……………………………………………………………. Exercises for independent work on the topic "Acids" ………………………………………………… Bases ……………………………………………………… ……… .. Exercises for independent work on the topic "Foundations" …………………………………………………. Salts …………………………………………………………………. Exercises for independent work on the topic "Salts" …………………………………………………… ... Genetic relationship between classes of inorganic compounds ... …………… .. Exercises for independent work on the topic “Genetic relationship between classes of inorganic compounds” ………………………………………… Experimental experiments on the topic “Basic classes of inorganic compounds …………… ……… .. Design tasks ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

Similar information.

>> Chemistry: Genetic relationship between classes of organic and inorganic substances

Material world. in which we live and a tiny part of which we are, is one and at the same time infinitely diverse. The unity and diversity of chemicals in this world is most clearly manifested in the genetic connection of substances, which is reflected in the so-called genetic series. Let's highlight the most characteristic features of such series:

1. All substances of this series must be formed by one chemical element.

2. Substances formed by the same element must belong to different classes, that is, reflect different forms of its existence.

3. Substances that form the genetic line of one element must be linked by interconversions. On this basis, complete and incomplete genetic series can be distinguished.

Summarizing the above, we can give the following definition of the genetic series:
A number of substances of representatives of different classes are called genetic, which are compounds of the same chemical element, interconverted and reflecting the common origin of these substances or their genesis.

Genetic link - the concept is more general than the genetic line. which is, albeit a bright, but private manifestation of this connection, which is realized in any mutual transformations of substances. Then, obviously, the first targeted series of substances in the text of the paragraph fits this definition.

To characterize the genetic relationship of inorganic substances, we will consider three types of genetic series:

II. Genetic range of a non-metal. Similar to the metal series, a number of non-metal with different oxidation states is richer in bonds, for example, the genetic series of sulfur with oxidation states +4 and +6.

Only the last transition can cause difficulty. If you perform tasks of this type, then be guided by the rule: in order to get a simple substance from an elementary compound, you need to take its most reduced compound for this purpose, for example, a volatile hydrogen compound of a non-metal.

III. The genetic line of the metal, which corresponds to amphoteric oxide and hydroxide, is very rich in soyaz. since they exhibit, depending on the conditions, the properties of an acid, then the properties of a base. For example, consider the genetic makeup of zinc:

In organic chemistry, it is also necessary to distinguish between a more general concept - a genetic link and a more particular concept of a genetic row. If the basis of the genetic series in inorganic chemistry is formed by substances formed by one chemical element, then the basis of the genetic series in organic chemistry (the chemistry of carbon compounds) is made up of substances with the same number of carbon atoms in the molecule. Consider the genetic series of organic substances, in which we include the largest number of classes of compounds:

Each number above the arrow corresponds to a certain reaction urn (the reaction equation is indicated by a number with a prime):

Iodine, the definition of the genetic series does not fit the last transition - a product is formed not with two, but with many carbon atoms, but with its help the genetic connections are most diversely represented. And finally, we will give examples of the genetic relationship between the classes of organic and inorganic compounds, which prove the unity of the world of substances, where there is no division into organic and inorganic substances.

Let's take the opportunity to repeat the names of the reactions corresponding to the proposed transitions:
1. Firing limestone:

1. Write down the reaction equations illustrating the following transitions:

3. In the interaction of 12 g of saturated monohydric alcohol with sodium, 2.24 liters of hydrogen (n.u.) were released. Find the molecular formula for the alcohol and write down the formulas for the possible isomers.

Lesson content lesson outline support frame lesson presentation accelerative methods interactive technologies Practice tasks and exercises self-test workshops, trainings, cases, quests homework discussion questions rhetorical questions from students Illustrations audio, video clips and multimedia photos, pictures, charts, tables, schemes humor, anecdotes, fun, comics parables, sayings, crosswords, quotes Add-ons abstracts articles chips for the curious cheat sheets textbooks basic and additional vocabulary of terms others Improving textbooks and lessonsbug fixes in the tutorial updating a fragment in the textbook elements of innovation in the lesson replacing outdated knowledge with new ones For teachers only perfect lessons calendar plan for the year methodological recommendations of the discussion program Integrated lessons

To form a concept of the genetic link and the genetic series.
Consider the genetic series of metals and non-metals.
Find out the genetic relationship between the classes of inorganic compounds.
To continue to develop the ability to use the solubility table and the periodic system of D.I. Medeleev to predict possible chemical reactions, as well as apply the knowledge gained on the topics of properties of classes of substances.
Review the main classes of inorganic compounds and their classification.
Develop a cognitive interest in the subject, the ability to quickly and clearly answer questions.
Continue to develop the ability to think logically, work with a textbook, and work with the information received.
To consolidate and systematize knowledge on this topic.

Equipment: Periodic table of D.I. Mendeleev, overhead, the table "Acids", the scheme "Genetic connection", cards for the game "Conveyor", "Creative task".

Reagents: In racks there are 3 test tubes with solutions of HCI, NaCI, NaOH, universal indicator paper. On the teacher's table: Na, H 2 O crystallizer, phenolphthalein, H 2 SO 4.

The class is divided into 4 microgroups: “Oxides”, “Acids”, “Salts”, “Bases”.

During the classes

I. Organizational moment.

1. Discipline.
2. The readiness of the class for the lesson.
3. Setting the goal of the lesson, motivation.

II. Main part.

1. Target setting of the lesson

There is nothing else in nature
Neither here nor there in the cosmic depths.
Everything - from small grains of sand - to planets
It consists of single elements.

Like a formula, like a work schedule,
The structure of the Mendeleev system is strict,
A living world is going on around you,
Enter it with your hands.

Today we are gathered here to put the best eighth graders of our school to the test and answer the question: "Are they worthy to become citizens of a great chemical country?" This country is ancient and magical, keeping many mysteries. Not a single person has yet been able to guess many of them. Only the most intelligent, courageous and persistent this country reveals its secrets. So, let's begin!

So, having studied the topic "The most important classes of inorganic compounds" you got the idea that inorganic compounds are diverse and interrelated. In the lesson, we will consider small fragments of interconversions of substances, recall the classification of inorganic substances, talk about the unity and diversity of chemicals.

The task of our lesson is to summarize information about substances, about individual classes of inorganic compounds and their classification as a whole, to consolidate knowledge about genetic series, genetic relationships, the interaction of substances of different classes, to learn the ability to apply knowledge in practice.

Write down the topic of our lesson in notebooks "Genetic relationship between inorganic compounds."

But, first, tell us about what substances are you talking about (name, formula)?

An owl sits on a bitch
Exhales _____________________________
My boots of the one
Pass ___________________________
Everyone knows him
They buy in the store,
You can't cook dinner without him -
In small doses, dishes need ___________
A bottle of substance, usually found in every apartment,
From birth, any child is familiar with him,
As soon as he leaves the hospital with his mother,
She is bathed in a bath with _________
What a miracle look
He rides on the board,
Leaves a trace behind itself. ____________________
If you don't have a baking powder for the test
you instead.
Put in pies. ________________________

Translate from chemical language to

Not all that glitters is aurum.
Strike the ferrum while it's hot.
_____________________________________________________________
The word is argentum, and silence is aurum.
_____________________________________________________________
5. Cooper not worth a dime.
_____________________________________________________________
Steadfast stanum soldier.
_____________________________________________________________
Since then, a lot of H 2 O has flowed under the bridge.
_____________________________________________________________

All these substances belong to some class of inorganic substances. Answer the question:

- How are inorganic substances classified into classes based on composition and properties?
- Name the classes of inorganic compounds you know

By microgroups:

- Give definitions.
Pupils give definitions to substances.

– Classification of these classes of substances.
Students provide answers.

On the slide:

From the proposed list of inorganic compounds, select the formulas:
Group 1 - oxides,
Group 2 - acids,
Group 3 - salts.
Group 4 - bases.

Name these substances.

Pupils complete the assignment in notebooks in microgroups.

Correct answer:

Now let's play a game with you "Tic-tac-toe".

Slide 19 . Appendices 1.

Distribute the substances, the formulas of which are given in the table by classes. From the letters corresponding to the correct answers, get the name of the great Russian scientist

Formulas	Oxides	Acids	Foundations	Salt
K 2 O	M	A	NS	A
H 2 CO 3	NS	E	T	R
P 2 O 5	H	AND	M	A
CuSO 4	NS	O	WITH	D
Ca (OH) 2	L	AND	E	WITH
Fe (NO 3) 3	A	H	Have	L
SO 2	E	L	Z	A
H 3 PO 4	H	E	L	WITH
Na 3 PO 4	H	Have	M	V

Answer: Mendeleev.

Problem task.

Can different classes of inorganic compounds interact with each other?

Highlight the signs of the genetic series:

Ca Ca (OH) 2 CaCO 3 CaO CaSO 4 CaCl 2 Ca?

substances of different classes;
different substances are formed by one chemical element;
different substances of one chemical element are interconverted.

There is an important connection between classes, which is called genetic ("genesis" in Greek means "origin"). This connection consists in the fact that substances of other classes can be obtained from substances of one class.

A number of substances are called genetic - representatives of different classes of inorganic compounds, which are compounds of the same chemical element, connected by interconversions and reflecting the common origin of these substances.

The genetic series reflects the relationship of substances of different classes, which are based on the same chemical element.

Genetic link is a link between substances of different classes formed by one chemical element, interconverted and reflecting the unity of their origin.

There are two main pathways for genetic relationships between substances: one of them begins with metals, the other with non-metals.
Among the metals, two types of series can also be distinguished:

1. 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 ... 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.

For example: Cu - CuO - CuCl 2 --Cu (OH) 2 --CuO -> Cu

Among non-metals, two types of series can also be distinguished:
1 ... Genetic series of non-metals, where soluble acid acts as a link in the series.

The transformation chain can be represented as follows:
nonmetal - acidic oxide - soluble acid - salt.

For example:
P - P 2 O 5 --H 3 PO 4 --Na 3 PO 4.
2 ... Genetic series of non-metals, where insoluble acid acts as a link in the series:
nonmetal - acidic oxide - salt - acid - acidic oxide - nonmetal

For example: Si - SiO 2 --Na 2 SiO 3 --H 2 SiO 3 --SiO 2 --Si.

Carry out transformations by microgroups.

Physical education "Red cat".

The solution of the problem.

Once Yukh conducted experiments to measure the electrical conductivity of solutions of different salts. On his laboratory table were beakers with solutions KCl, BaCl 2, K 2 CO 3, Na 2 SO 4 and AgNO 3 ... A label was neatly glued to each glass. There was a parrot in the laboratory whose cage was not very well locked. When Yukh, absorbed in the experiment, looked back at the suspicious rustle, he was horrified to find that the parrot, grossly violating safety rules, was trying to drink from a glass with BaCl 2 solution. Knowing that all soluble barium salts are extremely poisonous, Yukh 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. The glass with which solution was used to save the parrot?

BaCl 2 + Na 2 SO 4 = BaSO 4 (precipitate) + 2NaCl (barium sulfate is so slightly soluble that it cannot be poisonous, like some other barium salts).

Demonstration experiment. Teacher shows samples in test tubes :

1 - a piece of calcium, 2 - quicklime, 3 - slaked lime, 4 - gypsum asks the question:

"What do these samples have in common?" and writes a chain of formulas for the samples presented.

Ca CaO Ca (OH) 2 CaSO 4

Ok guys! Think how, using chemical reactions, you can go from a simple substance to a complex one, from one class of compounds to another. Let's conduct an experiment proving the presence of copper atoms in its various compounds. In the course of the experiment, write down the chain of transformations. What are the types of chemical reactions?

The work is carried out according to the instructional map.

Observe the safety instructions!

Instructional card.

Laboratory work: "Practical implementation of the chain of chemical transformations."

Check the availability of equipment and reagents in the workplace.

Equipment: test tube rack, spirit lamp, matches, test tube clamp, crucible tongs.

Reagents and materials: hydrochloric acid solution (1: 2), copper wire, iron nail or paper clip, threads.

Completing of the work.

Carry out reactions in which chemical transformations are carried out.

Copper wire copper (II) oxide copper (II) chloride copper

Anneal the copper wire, holding it with crucible tongs, in the upper part of the spirit lamp flame (1–2 min). What are you watching?

Carefully remove black deposits from the wire and place in a test tube. Note the color of the substance.

Pour 1 ml of hydrochloric acid solution (1: 2) into a test tube. Heat the contents slightly to speed up the reaction. What are you watching?

Carefully (why?) Immerse an iron nail (paper clip) in the test tube with the solution.

After 2-3 minutes, remove the nail from the mortar and describe the changes that have occurred to it.

What substance are they caused by the formation?

Describe and compare the color of the resulting and original solutions.

Clean up your workplace.

Attention! Heat the solution with copper oxide very carefully, holding the test tube high above the flame of an alcohol lamp.

III. Conclusion.

Teacher. The concepts “oxide”, “acid”, “base”, “salt” form a system that is in close interconnection, it is revealed when substances of one class are obtained from substances of another class. It manifests itself in the process of interaction of substances and is actively used in human practice. Do you guys think we have achieved the goal that we set at the beginning of the lesson?

V. Homework.

Slides 30, 31.

Vi. Summing up the lesson, assessment, reflection.

Teacher. Guys, it's time to take stock. What did you learn today, what did you learn new, what did you do in the lesson?

Students provide answers.

Genetic link- this is a connection between substances of different classes, based on their interconversions and reflecting the unity of their origin, that is, the genesis of substances. A complex substance can be obtained from simple substances, and simple substances can be obtained from a complex substance.

The genetic link is reflected in the genetic series.

Characteristic features of genetic series:

1. All substances of this series must be formed by one chemical element.

2. Substances formed by the same element must belong to different classes, that is, reflect different forms of its existence.

3. Substances that form the genetic line of one element must be linked by interconversions. On this basis, complete and incomplete genetic series can be distinguished.

Among the metals, two types of rows can be distinguished:

1. Genetic series, in which alkali acts as a base. This series can be represented using the following transformations:

metal → basic oxide → alkali → salt (Example: K → K 2 O → KOH → KCl)

2 Genetic range of metals, which correspond to insoluble bases. There are more genetic connections in this row, because it more fully reflects the idea of direct and reverse transformations (reciprocal).

metal → basic oxide → salt → base → basic oxide → metal.

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

Among non-metals, two types of series can also be distinguished:

1. Genetic series of non-metals, where soluble acid acts as a link in the series.

non-metal → acid oxide → soluble acid → salt

(For example: P → P 2 O 5 → H 3 PO 4 → Ca 3 (PO 4) 2)

2. 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

(you can consider the series from one side or the other)

Quantum mechanical model of the atom, de Broglie and Schrödinger's equations, Heisenberg's uncertainty principle. Atomic orbital. quantum numbers

The CMM is based on the quantum theory of the atom, according to which the electron has both the properties of a particle and the properties of a wave. In other words, the position of the electron at a certain point can be judged not exactly, but with a certain degree of probability. Therefore, in KMM orbits Bora was replaced orbitals(a kind of "electron clouds" - the area of space in which there is a probability of the presence of an electron).

Principal quantum number n

Describes:

· The average distance from the orbital to the nucleus;

· Energy state of an electron in an atom.

The larger the value of n, the higher the energy of the electron and the larger the size of the electron cloud. If there are several electrons in an atom with the same n, then they form electron clouds of the same size - electronic shells.

Orbital quantum number l (azimuthal)

Describes the shape of the orbital which depends on n.

The orbital number l can take integer values in the range from 0 to n-1. For example, for n = 2: l = 0 l = 1
Orbitals that have the same n but different l are called energy sublevels and denoted by letters of the Latin alphabet:

Magnetic quantum number m