Do you know where this figure came from 21 days? Quite interesting ...
Task number 1
Si + HNO 3 + HF → H 2 SIF 6 + NO + ...
N +5 + 3E → N +2 │4 Recovery Reaction
Si 0 - 4E → Si +4 │3 oxidation reaction
N +5 (HNO 3) - Oxidizer, Si - Restore
3SI + 4HNO 3 + 18HF → 3H 2 SIF 6 + 4NO + 8H 2 O
Task number 2.
Using the electronic balance method, make the reaction equation:
B + HNO 3 + HF → HBF 4 + NO 2 + ...
Determine the oxidizing agent and reducing agent.
N +5 + 1e → n +4 │3 Recovery Reaction
B 0 -3E → B +3 │1 oxidation reaction
N +5 (HNO 3) - Oxidizer, B 0 - Restore
B + 3HNO 3 + 4HF → HBF 4 + 3NO 2 + 3H 2 O
Task number 3.
Using the electronic balance method, make the reaction equation:
K 2 Cr 2 O 7 + HCl → Cl 2 + KCl + ... + ...
Determine the oxidizing agent and reducing agent.
2Cl -1 -2E → Cl 2 0 │3 Oxidation reaction
CR +6 (K 2 CR 2 O 7) - Oxidizer, Cl -1 (HCl) - Restore
K 2 Cr 2 O 7 + 14hCl → 3Cl 2 + 2KCL + 2CRCL 3 + 7H 2 O
Task number 4.
Using the electronic balance method, make the reaction equation:
CR 2 (SO 4) 3 + ... + NaOH → Na 2 Cro 4 + NaBr + ... + H 2 O
Determine the oxidizing agent and reducing agent.
BR 2 0 + 2E → 2Br -1 │3 Recovery Reaction
2CR +3 - 6E → 2Cr +6 │1 oxidation reaction
Br 2 - Oxidizer, Cr +3 (CR 2 (SO 4) 3) - Restore
CR 2 (SO 4) 3 + 3Br 2 + 16NAOH → 2NA 2 CRO 4 + 6NABR + 3NA 2 SO 4 + 8H 2 O
Task number 5.
Using the electronic balance method, make the reaction equation:
K 2 Cr 2 O 7 + ... + H 2 SO 4 → L 2 + CR 2 (SO 4) 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 Recovery Reaction
2i -1 -2E → L 2 0 │3 oxidation reaction
CR +6 (K 2 Cr 2 O 7) - Oxidizer, L -1 (HL) - Restore
K 2 Cr 2 O 7 + 6HI + 4H 2 SO 4 → 3L 2 + CR 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O
Task number 6.
Using the electronic balance method, make the reaction equation:
H 2 S + HMNO 4 → S + MNO 2 + ...
Determine the oxidizing agent and reducing agent.
3H 2 S + 2HMNO 4 → 3S + 2MNO 2 + 4H 2 O
Task number 7.
Using the electronic balance method, make the reaction equation:
H 2 S + HCLO 3 → S + HCl + ...
Determine the oxidizing agent and reducing agent.
S -2 -2E → S 0 │3 oxidation reaction
Mn +7 (HMNO 4) - Oxidizer, S -2 (H 2 S) - Restore
3H 2 S + HCLO 3 → 3S + HCl + 3H 2 O
Task number 8.
Using the electronic balance method, make the reaction equation:
NO + HCLO 4 + ... → HNO 3 + HCl
Determine the oxidizing agent and reducing agent.
Cl +7 + 8E → CL -1 │3 Recovery Reaction
N +2 -3E → N +5 │8 oxidation reaction
Cl +7 (HCLO 4) - Oxidizer, N +2 (NO) - Restore
8NO + 3HCLO 4 + 4H 2 O → 8HNO 3 + 3HCL
Task number 9.
Using the electronic balance method, make the reaction equation:
KMNO 4 + H 2 S + H 2 SO 4 → MNSO 4 + S + ... + ...
Determine the oxidizing agent and reducing agent.
S -2 -2E → S 0 │5 oxidation reaction
Mn +7 (KMNO 4) - Oxidator, S -2 (H 2 S) - Restore
2kmno 4 + 5h 2 s + 3h 2 SO 4 → 2mnso 4 + 5s + k 2 SO 4 + 8H 2 O
Task number 10.
Using the electronic balance method, make the reaction equation:
KMNO 4 + KBR + H 2 SO 4 → MNSO 4 + BR 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 Recovery Reaction
2Br -1 -2E → BR 2 0 │5 Oxidation Reaction
Mn +7 (KMNO 4) - Oxidizer, BR -1 (KBR) - Restore
2kmno 4 + 10kbr + 8H 2 SO 4 → 2MNSO 4 + 5Br 2 + 6K 2 SO 4 + 8H 2 O
Task number 11.
Using the electronic balance method, make the reaction equation:
PH 3 + HCLO 3 → HCl + ...
Determine the oxidizing agent and reducing agent.
Cl +5 + 6e → Cl -1 │4 Recovery Reaction
Cl +5 (HCLO 3) - Oxidizer, P -3 (H 3 PO 4) - Restore
3PH 3 + 4HCLO 3 → 4HCl + 3H 3 PO 4
Task number 12.
Using the electronic balance method, make the reaction equation:
PH 3 + HMNO 4 → MNO 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │8 Recovery Reaction
P -3 - 8E → P +5 │3 oxidation reaction
Mn +7 (HMNO 4) - Oxidizer, P -3 (H 3 PO 4) - Restore
3PH 3 + 8HMNO 4 → 8mno 2 + 3H 3 PO 4 + 4H 2 O
Task number 13.
Using the electronic balance method, make the reaction equation:
NO + KCLO + ... → Kno 3 + KCl + ...
Determine the oxidizing agent and reducing agent.
Cl +1 + 2e → Cl -1 │3 Recovery Reaction
N +2 - 3E → N +5 │2 oxidation reaction
Cl +1 (KCLO) - Oxidizer, N +2 (NO) - Restore
2no + 3KCLO + 2KOH → 2KNO 3 + 3KCl + H 2 O
Task number 14.
Using the electronic balance method, make the reaction equation:
PH 3 + AGNO 3 + ... → AG + ... + HNO 3
Determine the oxidizing agent and reducing agent.
AG +1 + 1E → AG 0 │8 Recovery Reaction
P -3 - 8E → P +5 │1 oxidation reaction
AG +1 (AGNO 3) - Oxidizer, P -3 (pH 3) - Restore
PH 3 + 8AGNO 3 + 4H 2 O → 8Ag + H 3 PO 4 + 8HNO 3
Task number 15.
Using the electronic balance method, make the reaction equation:
KNO 2 + ... + H 2 SO 4 → I 2 + NO + ... + ...
Determine the oxidizing agent and reducing agent.
N +3 + 1E → N +2 │ 2 Recovery Reaction
2i -1 - 2e → i 2 0 │ 1 oxidation reaction
N +3 (KNO 2) - Oxidizer, I -1 (HI) - Restore
2kno 2 + 2hi + H 2 SO 4 → I 2 + 2NO + K 2 SO 4 + 2H 2 O
Task number 16.
Using the electronic balance method, make the reaction equation:
Na 2 SO 3 + Cl 2 + ... → Na 2 SO 4 + ...
Determine the oxidizing agent and reducing agent.
Cl 2 0 + 2e → 2Cl -1 │1 Recovery Reaction
Cl 2 0 - Oxidizer, S +4 (NA 2 SO 3) - Restore
Na 2 SO 3 + Cl 2 + H 2 O → Na 2 SO 4 + 2HCl
Task number 17.
Using the electronic balance method, make the reaction equation:
KMNO 4 + MNSO 4 + H 2 O → MnO 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 Recovery Reaction
Mn +2 - 2e → Mn +4 │3 oxidation reaction
Mn +7 (KMNO 4) - Oxidizer, Mn +2 (MNSO 4) - Restore
2kmno 4 + 3mnso 4 + 2H 2 O → 5mno 2 + K 2 SO 4 + 2H 2 SO 4
Task number 18.
Using the electronic balance method, make the reaction equation:
KNO 2 + ... + H 2 O → MnO 2 + ... + Koh
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 Recovery Reaction
N +3 - 2e → n +5 │3 oxidation reaction
Mn +7 (KMNO 4) - Oxidizer, N +3 (KNO 2) - Restore
3kno 2 + 2kmno 4 + H 2 O → 2mno 2 + 3Kno 3 + 2KOH
Task №19.
Using the electronic balance method, make the reaction equation:
CR 2 O 3 + ... + KOH → KNO 2 + K 2 CRO 4 + ...
Determine the oxidizing agent and reducing agent.
N +5 + 2E → N +3 │3 Recovery Reaction
2CR +3 - 6E → 2Cr +6 │1 oxidation reaction
N +5 (KNO 3) - Oxidizer, Cr +3 (CR 2 O 3) - Restore
CR 2 O 3 + 3KNO 3 + 4KOH → 3KNO 2 + 2K 2 CRO 4 + 2H 2 O
Task number 20.
Using the electronic balance method, make the reaction equation:
I 2 + k 2 SO 3 + ... → K 2 SO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
I 2 0 + 2e → 2i -1 │1 Recovery Reaction
S +4 - 2e → s +6 │1 oxidation reaction
I 2 - Oxidizer, S +4 (K 2 SO 3) - Restore
I 2 + K 2 SO 3 + 2KOH → K 2 SO 4 + 2KI + H 2 O
Task number 27.
Using the electronic balance method, make the reaction equation:
KMNO 4 + NH 3 → MNO 2 + N 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 Recovery Reaction
2N -3 - 6E → N 2 0 │1 oxidation reaction
Mn +7 (KMNO 4) - Oxidizer, N -3 (NH 3) - Restore
2kmno 4 + 2NH 3 → 2MNO 2 + N 2 + 2KOH + 2H 2 O
Task №22.
Using the electronic balance method, make the reaction equation:
NO 2 + P 2 O 3 + ... → NO + K 2 HPO 4 + ...
Determine the oxidizing agent and reducing agent.
N +4 + 2E → N +2 │2 Recovery Reaction
2P +3 - 4E → 2p +5 │1 oxidation reaction
N +4 (NO 2) - Oxidizer, P +3 (P 2 O 3) - Restore
2NO 2 + P 2 O 3 + 4KOH → 2NO + 2K 2 HPO 4 + H 2 O
Task number 23.
Using the electronic balance method, make the reaction equation:
Ki + H 2 SO 4 → I 2 + H 2 S + ... + ...
Determine the oxidizing agent and reducing agent.
S +6 + 8E → S -2 │1 Recovery Reaction
2i -1 - 2E → i 2 0 │4 oxidation reaction
S +6 (H 2 SO 4) - Oxidizer, I -1 (Ki) - Restore
8ki + 5h 2 SO 4 → 4i 2 + H 2 S + 4K 2 SO 4 + 4H 2 O
Task number 24.
Using the electronic balance method, make the reaction equation:
FESO 4 + ... + H 2 SO 4 → ... + MNSO 4 + K 2 SO 4 + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 Recovery Reaction
2fe +2 - 2e → 2fe +3 │5 oxidation reaction
Mn +7 (KMNO 4) - Oxidizer, Fe +2 (FESO 4) - Restore
10feso 4 + 2kmno 4 + 8H 2 SO 4 → 5Fe 2 (SO 4) 3 + 2mnso 4 + K 2 SO 4 + 8H 2 O
Task №25
Using the electronic balance method, make the reaction equation:
Na 2 SO 3 + ... + Koh → K 2 MNO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 1e → Mn +6 │2 Recovery Reaction
S +4 - 2e → s +6 │1 oxidation reaction
Mn +7 (KMNO 4) - Oxidizer, S +4 (Na 2 SO 3) - Restore
Na 2 SO 3 + 2KMNO 4 + 2KOH → 2K 2 MNO 4 + Na 2 SO 4 + H 2 O
Task number 26.
Using the electronic balance method, make the reaction equation:
H 2 O 2 + ... + H 2 SO 4 → O 2 + MNSO 4 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 Recovery Reaction
2O -1 - 2E → O 2 0 │5 oxidation reaction
Mn +7 (KMNO 4) - Oxidizer, O -1 (H 2 O 2) - Restore
5H 2 O 2 + 2KMNO 4 + 3H 2 SO 4 → 5O 2 + 2MNSO 4 + K 2 SO 4 + 8H 2 O
Task number 27.
Using the electronic balance method, make the reaction equation:
K 2 CR 2 O 7 + H 2 S + H 2 SO 4 → CR 2 (SO 4) 3 + K 2 SO 4 + ... + ...
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 Recovery Reaction
S -2 - 2E → S 0 │3 Oxidation reaction
CR +6 (K 2 Cr 2 O 7) - Oxidizer, S -2 (H 2 S) - Restore
K 2 CR 2 O 7 + 3H 2 S + 4H 2 SO 4 → CR 2 (SO 4) 3 + K 2 SO 4 + 3S + 7H 2 O
Task №28.
Using the electronic balance method, make the reaction equation:
KMNO 4 + HCl → MnCl 2 + Cl 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 Recovery Reaction
2CL -1 - 2E → CL 2 0 │5 oxidation reaction
Mn +7 (KMNO 4) - Oxidizer, Cl -1 (HCl) - Restore
2kmno 4 + 16hcl → 2mnCl 2 + 5Cl 2 + 2KCl + 8H 2 O
Task number 29.
Using the electronic balance method, make the reaction equation:
CRCL 2 + K 2 Cr 2 O 7 + ... → CRCL 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 Recovery Reaction
CR +2 - 1E → CR +3 │6 Oxidation Reaction
CR +6 (K 2 Cr 2 O 7) - Oxidizer, Cr +2 (CRCL 2) - Restore
6CrCl 2 + K 2 CR 2 O 7 + 14HCl → 8CrCl 3 + 2KCl + 7H 2 O
Task number 30.
Using the electronic balance method, make the reaction equation:
K 2 Cro 4 + HCl → CRCL 3 + ... + ... + H 2 O
Determine the oxidizing agent and reducing agent.
CR +6 + 3E → CR +3 │2 Recovery Reaction
2CL -1 - 2E → CL 2 0 │3 Oxidation reaction
CR +6 (K 2 CRO 4) - Oxidizer, CL -1 (HCl) - Restore
2K 2 CRO 4 + 16HCl → 2CrCl 3 + 3Cl 2 + 4KCl + 8H 2 O
Task number 31.
Using the electronic balance method, make the reaction equation:
Ki + ... + H 2 SO 4 → I 2 + MNSO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 Recovery Reaction
2L -1 - 2E → L 2 0 │5 oxidation reaction
Mn +7 (KMNO 4) - Oxidizer, L -1 (KL) - Restore
10ki + 2kmno 4 + 8h 2 SO 4 → 5i 2 + 2mnso 4 + 6k 2 SO 4 + 8H 2 O
Task number 32.
Using the electronic balance method, make the reaction equation:
FESO 4 + KCLO 3 + KOH → K 2 FeO 4 + KCl + K 2 SO 4 + H 2 O
Determine the oxidizing agent and reducing agent.
CL +5 + 6E → CL -1 │2 Recovery Reaction
Fe +2 - 4E → Fe +6 │3 oxidation reaction
3Feso 4 + 2KCLO 3 + 12KOH → 3K 2 FeO 4 + 2KCl + 3K 2 SO 4 + 6H 2 O
Task number 33.
Using the electronic balance method, make the reaction equation:
FESO 4 + KCLO 3 + ... → Fe 2 (SO 4) 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Cl +5 + 6e → Cl -1 │1 Recovery Reaction
2fe +2 - 2E → 2fe +3 │3 oxidation reaction
Cl +5 (KCLO 3) - Oxidizer, Fe +2 (FESO 4) - Restore
6Feso 4 + KCLO 3 + 3H 2 SO 4 → 3FE 2 (SO 4) 3 + KCl + 3H 2 O
Task number 34.
Using the electronic balance method, make the reaction equation.
| 37 |
Iron moisture was dissolved in concentrated nitric acid.
Sodium hydroxide solution was added to the resulting solution. The separated sediment was separated and calcined. The resulting solid residue was sparkled with iron.
Write the equations of four described reactions.
| 38 |
Write the reaction equations with which the following transformations can be carried out:
When writing the reaction equations, use structural formulas for organic substances.
Note. It is permissible to use structural formulas of different types (deployed, abbreviated, skeletal), unambiguously reflecting the order of communication of atoms and the relative position of substituents and functional groups in the organic matter molecule.
| 39 |
Phosphorus chloride (V) weighing 4.17 g completely reacted with water. What volume of potassium hydroxide solution with a mass fraction of 10% (1.07 g / ml density) is necessary for the complete neutralization of the resulting solution?
| The content of the right answer and evaluation guidelines (other formulations of the answer, not distorting its meaning) | Point |
| Response elements: 1) The equation of reactions of hydrolysis of phosphorus chloride and neutralization of two acids are recorded: PCL 5 + 4H 2 O \u003d H 3 PO 4 + 5HCl H 3 PO 4 + 3KOH \u003d K 3 PO 4 + 3H 2 O HCl + KOH \u003d KCl + H 2 O 2) The amount of substance of phosphorus chloride (V) and acids formed during hydrolysis are calculated: n (PCl 5) \u003d 4.17 / 208.5 \u003d 0.02 mol n (H 3 PO 4) \u003d N (PCL 5 ) \u003d 0.02 mol n (HCl) \u003d 5N (PCL 5) \u003d 0.1 mol 3) required required amount of substance and rubber weight: n (koh) \u003d n (HCl) + 3N (H 3 PO 4) \u003d 0 , 1 + 0.06 \u003d 0.16 mol m (KOH) \u003d 0.16 ∙ 56 \u003d 8.96 g 4) The mass and volume of the pitching solution are calculated: M (P-PA) (KOH) \u003d M (KOH) / W (KOH) \u003d 8.96 / 0.10 \u003d 89.6 g V (P-PA) (KOH) \u003d M (P-PA) (KOH) / ρ \u003d 89.6 / 1.07 \u003d 83.7 ml | |
| The answer is correct and complete, includes all the elements named above. | |
| In response, an error is made in one of the above elements. | |
| In response, errors are made in two of the above elements | |
| In response, errors in three of the above elements are allowed | |
| All response elements are recorded incorrectly. | |
| Maximum score | 4 |
Note.In the case when the response contains an error in calculations in one of the elements (second, third or fourth), which led to an incorrect answer, the evaluation of the task is reduced only by 1 score.
| 40 |
With a combustion of 18.8 g of organic matter, 26.88 liters (N.O.) of carbon dioxide and 10.8 ml of water were obtained. It is known that this substance reacts like
Sodium hydroxide and with bromine water.
On the basis of these conditions of the problem:
1) Calculates needed to establish the molecular formula of the organic matter;
2) write down the molecular formula of the organic matter;
3) make a structural formula of the source substance, which uniquely reflects the order of communication of atoms in its molecule;
4) Write the reaction equation of this substance with bromine water.
| The content of the right answer and evaluation guidelines (other formulations of the answer, not distorting its meaning) | Point |
Response elements: General formula of substance - C x H y o z 1) Found the amount of carbon dioxide and water substance: N (CO 2) \u003d 26.88 / 22.4 \u003d 1.2 mol n (H 2 O) \u003d 10, 8/18 \u003d 0.6 mol 2) The molecular formula of the substance is determined: n (C) \u003d n (CO 2) \u003d 1.2 mol m (c) \u003d 1,2 · 12 \u003d 14.4 g n (H) \u003d 2n (H 2 O) \u003d 1.2 mol m (H) \u003d 1.2 g m (o) \u003d 18.8 - 14.4 - 1.2 \u003d 3.2 g N (O) \u003d 3.2 / 16 \u003d 0.2 mol X: Y: Z \u003d 1.2: 1.2: 0.2 \u003d 6: 6: 1 Molecular formula - C 6 H 6 O 3) Composed of a structural formula of substance: 4) The equation of the reaction of this Substances with bromine water:  | |
| The answer is correct and complete, contains all the elements named above. | |
| An error is assumed in one answer element | |
| Answer an error in two response elements is made. | |
| In response, an error in three answers is allowed | |
| All response elements are recorded incorrectly. | |
| Maximum score | 4 |
Work 1.
Using the electronic balance method, make the reaction equation. Determine the oxidizing agent and reducing agent.
Often students are confident that this task does not require special training. However, experience shows that it contains underwater stones that interfere with getting a full score for it.
Let's figure out how to act when preparing to solve the tasks of this type, to pay attention to.
Theoretical information.
Potassium permanganate as an oxidizing agent.
| KMNO. 4 + Restorers→ |
||
| in an acidic environment MN. +2 | in a neutral environment MN. +4 | in an alkaline environment MN. +6 |
| (Salt of the acid that participates in the reaction) | Manganat (K 2 MNO 4 or Knamno 4, Na 2 MNO 4) - |
|
Dichromate and chromate as oxidizing agents.
| K. 2 CR 2 O. 7 (acid and neutral environment), k 2 CRO. 4 (alkaline environment) + reducing agents→ It always works CR +3 |
||
| aclest medium | neutral environment | alkaline environment |
| Salts of those acids that are involved in the reaction: CRCL 3, CR 2 (SO 4) 3 | K 3 in solution, K 3 CRO 3 or KCRO 2 in the melt |
|
Increase the degrees of chromium oxidation and manganese.
| CR +3 + very strong oxidizers→ CR +6 (always independently of the environment!) |
||
| CR 2 O 3, CR (OH) 3, salt, hydrox complexes | Very strong oxidizers: | Alkaline environment: forms chromat K 2 CRO 4 |
| CR (OH) 3, salt | Very strong oxidizers in an acidic environment (HNO 3 or CH 3 COOH): PBO 2, KBIO 3 | Aclement medium: forms dichromat. K 2 Cr 2 O 7 or dichrome acid H 2 CR 2 O 7 |
| Mn + 2, + 4 - oxide, hydroxide, salt | Very strong oxidizers: | Alkaline environment: Mn +6 K 2 MNO 4 - Manganat |
| Mn +2 - Salt | Very strong oxidizers in an acidic environment (HNO 3 or CH 3 COOH): | Aclest Wednesday: MN +7 KMNO 4 - Permanganate |
Nitric acid with metals.
- hydrogen is not highlightedThe nitrogen restoration products are formed.
| The more active the metal and the smaller the concentration of the acid, the further restores nitrogen |
||||
| No. 2 | N. 2 O. | N. 2 | NH. 4 No. 3 |
|
|
| Inactive metals (right of iron) + sample. acid | Active metals (alkaline, alkaline earth, zinc) + conc. acid | Active metals (alkaline, alkaline earth, zinc) + medium dilution acid | Active metals (alkaline, alkaline earth, zinc) + very scan. acid |
| Passivation: With cold concentrated nitric acid do not react: |
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| Do not react with nitric acid at no concentration: |
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Sulfuric acid with metals.
- diluted Sulfuric acid reacts as ordinary mineral acid with metal left in a row of stresses, while hydrogen is distinguished;
- when reactions with metals concentrated Sulfuric acid hydrogen is not highlightedThe sulfur restoration products are formed.
| SO. 2 | H. 2 S. | H. 2 |
|
| Inactive metals (right of iron) + conc. acid | Alkaline earth metals + conc. acid | Alkali metals and zinc + concentrated acid. | Diluted sulfuric acid behaves like ordinary mineral acid (for example, hydrochloric) |
| Passivation: With cold concentrated sulfuric acid do not react: |
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| Do not react with sulfuric acid at no concentration: |
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Disproportionation.
Disproportionation reactions - these are reactions in which the same The element is both the oxidizing agent, and the reducing agent, at the same time increasing, and lowering its degree of oxidation:
| 5KCl + KCLO 3 + 3H 2 O |
||
Disproportionation of non-metals - sulfur, phosphorus, halogen (except fluorine).
| Sulfur + cheek → 2 salts, sulfide and metal sulfite (reaction is boiling) | S 0 → S -2 and S +4 |
| Phosphorus + alkali → phosphine pH 3 and salt hypophosphitis KN 2 PO 2 (the reaction is boiling) | P 0 → P -3 and P +1 |
| Chlorine, bromine, iodine + water (without heating) → 2 Acids, HCl, HCLO | Cl 2 0 → Cl - and Cl + |
| Brom, iodine + water (when heated) → 2 acids, HBR, HBRO 3 | Cl 2 0 → Cl - and Cl +5 |
Disproportionation of nitrogen oxide (IV) and salts.
| NO 2 + Water → 2 Acids, nitrogen and nitrogenous | N +4 → N +3 and N +5 |
| S +4 → S -2 and S +6 |
|
| Cl +5 → Cl - and Cl +7 |
The activity of metals and non-metals.
To analyze the activity of metals, either an electrochemical series of metals voltage, or their position in the periodic table is used, or their position in the periodic table. The more active the metal, the easier it will give electrons and the more good reducing agent it will be in oxidative reaction reactions.
Electrochemical row of metal voltages.
| Li RB K BA SR CA NA MG AL MN ZN CR FE CD CO NI SN PB H SB BI CU HG AG PD PT AU |
The activity of non-metals can also be determined by their position in the Mendeleev table.
Remember! Nitrogen - more active nonmetall than chlorine!
More active nonmetall will be an oxidizing agent, and less active will be content with the role of the reducing agent, if they react with each other.
Number of electronegability of non-metals:
Features of the behavior of some oxidizing agents and reducing agents.
a) Oxygen-containing salts and chlorine acids in reactions with reducing agents usually go to chlorides: KCLO 3 + P \u003d P 2 O 5 + KCL
b) If substances are involved in the reaction, in which the same element has a negative and positive degree of oxidation - they are found in the zero degree of oxidation (there is a simple substance). H 2 S -2 + S (+4) O 2 \u003d S 0 + H 2 O
Necessary skills.
Alignment of degrees of oxidation.
It must be remembered that the degree of oxidation is hypothetical Atom charge (i.e. conditional, imaginary), but it should not go beyond common sense. It can be integer, fractional or equal to zero.
Exercise 1:Arrange the degrees of oxidation in substances:
Nonfes 2 Ca (OCL) CLH 2 S 2 O 8
Settlement of oxidation degrees in organic substances.
Remember that we are interested in the degree of oxidation of only those carbon atoms that change their surroundings in the process of the OSR, while the total charge of the carbon atom and its non-harmonic environment is taken for 0.
Task 2:Determine the degree of oxidation of carbon atoms circled by the frame together with the non-harmonic environment:
2-methylbutene-2: CH 3 -CH \u003d C (CH 3) -CH 3
acetone: (CH 3) 2 C \u003d O
acetic acid: CH3-coxy
Do not forget to ask yourself the main question: who in this reaction gives the electrons, and who takes them, and what do they turn? In order not to succeed, the electrons arrive from nowhere or fly away.
Example: KNO. 2 + Ki + h 2 SO. 4 → … + … + … + …
In this reaction, it is necessary to see that Kilia Ki iodide may be only a reducing agent, therefore, Kalistry Nitrite KNO 2 will receive electrons lowing Its degree of oxidation.
And in these conditions (diluted solution) nitrogen goes from +3 to the nearest oxidation degree +2.
KNO 2 + KI + H 2 SO 4 → I 2 + NO + K 2 SO 4 + H 2 O
The compilation of the electronic balance is more difficult if the formula unit of the substance contains several oxidizing agent atoms or reducing agent.
In this case, this must be taken into account in the semi-resource, calculating the number of electrons.
The most frequent problem is with potassium dichromate K 2 CR 2 O 7, when it turns into +3 in the role of the oxidant:
2cr +6 + 6e → 2Cr +3
The same two can not be forgotten when equalizing, because they indicate the number of atoms of this species in the equation.
Task 3:What coefficient you need to put before FESO 4 And before FE. 2 (So. 4 ) 3 ?
FESO 4 + K 2 Cr 2 O 7 + H 2 SO 4 → Fe 2 (SO 4) 3 + CR 2 (SO 4) 3 + K 2 SO 4 + H 2 O
Fe +2 - 1e → Fe +3
2Cr +6 + ... E → 2Cr +3
Task 4:What coefficient in the reaction equation will stand in front of magnesium?
HNO 3 + MG → Mg (NO 3) 2 + N 2 O + H 2 O
Determine in which medium (acidic, neutral or alkaline) reaction flows.
This can be made either about the products of repair of manganese and chromium, or by type of compounds that turned out in the right side of the reaction: for example, if we see in the products acid, acid oxide - It means that it is definitely not an alkaline medium, but if metal hydroxide drops - it is definitely not acidic. Well, of course, if we see metal sulfates in the left side, and in the right - nothing like sulfur compounds - apparently, the reaction is carried out in the presence of sulfuric acid.
Task 5:Determine the medium and substances in each reaction:
PH 3 + ... + ... → K 2 MNO 4 + ... + ...
PH 3 + ... + ... → MNSO 4 + H 3 PO 4 + ... + ...
Remember that water is a free traveler, it can both participate in the reaction and form.
Task 6:Which side of the reaction will be water? What will zinc going?
KNO 3 + Zn + Koh → NH 3 + ...
Task 7:Soft and hard oxidation of alkenes.
Extract and equalize the reaction, pre-laying the degree of oxidation in organic molecules:
CH 3 -CH \u003d CH 2 + KMNO 4 + H 2 O (cold. Rr.) → CH 3 -CHOH-CH 2 OH + ...
Sometimes any product of the reaction can be determined only by making an electronic balance and realizing which particles we have more:
Task 8:What products will turn out? Extract and equalize the reaction:
MNSO 4 + KMNO 4 + H 2 O → MNO 2 + ...
What are the reagents in the reaction?
If the answer to this question does not give us schemes, then you need to analyze what kind of oxidizing agent and the reducing agent - strong or not so? If the average force oxidizer, it can hardly oxidize, for example, sulfur from -2 B +6, usually oxidation goes only to S 0. And vice versa, if Ki is a strong reducing agent and can restore sulfur from +6 to -2, then KBR is only up to +4.
Task 9:What will sulfur go? Extract and equalize the reactions:
H 2 S + KMNO 4 + H 2 O → ...
H 2 S + HNO 3 (conc.) → ...
Check that the reaction is the oxidizer, and the reducing agent.
Task 10:How many more products in this reaction, and what?
KMNO 4 + HCl → MnCl 2 + ...
If both substances can show properties and reducing agent, and the oxidant - it is necessary to think about which one of them more Active oxidizer. Then the second will be a reducing agent.
Task 11:Which of these halogen oxidizers, and who is a reducing agent?
Cl 2 + i 2 + H 2 O → ... + ...
If one of the reagents is a typical oxidizing agent or a reducing agent - then the second will "perform his will", or giving the electrons to the oxidizing agent, or by taking the reducing agent.
Hydrogen peroxide - substance with dual natureIn the role of an oxidizing agent (which is more characteristic) turns into water, and as a reducing agent, it goes into free gas oxygen.
Task 12:What role does hydrogen peroxide in each reaction perform?
H 2 O 2 + Ki + H 2 SO 4 →
H 2 O 2 + K 2 CR 2 O 7 + H 2 SO 4 →
H 2 O 2 + KNO 2 →
Sequence of coefficients in the equation.
First, smear the coefficients obtained from the electronic balance.
Remember that you can double or cut them only together. If any substance acts in the role of the medium, and in the role of an oxidizing agent (reducing agent) - it will be necessary to equalize it later, when almost all coefficients are arranged.
The penultimate equal to hydrogen, and oxygen we only check!
Task 13:Extract and equalize:
HNO 3 + Al → Al (NO 3) 3 + N 2 + H 2 O
Al + Kmno 4 + H 2 SO 4 → Al 2 (SO 4) 3 + ... + K 2 SO 4 + H 2 O
Do not rush, recalculating oxygen atoms! Do not forget to multiply, and not fold indexes and coefficients.
The number of oxygen atoms in the left and right part should go!
If this did not happen (provided that you consider them correctly), it means somewhere a mistake.
Possible mistakes.
Oxidation degrees: Check each substance carefully.
Often mistaken in the following cases:
a) the degree of oxidation in the hydrogen compounds of non-metals: phosphine pH 3 - the degree of oxidation in phosphorus - negative;
b) in organic substances - check again, whether all the environment of the atom with taken into account;
c) ammonia and ammonium salts - nitrogen in them always has a degree of oxidation -3;
d) oxygen salts and chlorine acids - chlorine in them may have a degree of oxidation +1, +3, +5, +7;
e) peroxides and supersides - in them oxygen does not have the degree of oxidation -2, it happens -1, and in KO 2 - even - (½)
e) double oxides: Fe 3 O 4, Pb 3 O 4 - Metals have in them two different The degree of oxidation is usually only one of them participates in the transfer of electrons.
Task 14:Extract and equalize:
Fe 3 O 4 + HNO 3 → Fe (NO 3) 3 + NO + ...
Task 15:Extract and equalize:
KO 2 + KMNO 4 + ... → ... + ... + k 2 SO 4 + H 2 O
The choice of products without taking into account the transfer of electrons is, that is, for example, in the reaction there is only an oxidizing agent without a reducing agent or vice versa.
Example:in the reaction MNO. 2 + HCL.→ Mncl 2 + Cl. 2 + H. 2 O.free chlorine is often lost. It turns out that the electrons to the manganese flew from space ...
Incorrect products from a chemical point of view: there can be no substance that enters into interaction with the environment!
a) in an acidic environment, metal oxide, base, ammonia cannot be obtained;
b) in an alkaline medium will not work or acidic oxide;
c) oxide or all the more metal, violently reactive with water, are not formed in an aqueous solution.
Task 16:Find in reactionserroneous Products explain why they cannot be obtained under these conditions:
BA + HNO 3 → BAO + NO 2 + H 2 O
PH 3 + KMNO 4 + KOH → K 2 MNO 4 + H 3 PO 4 + H 2 O
P + HNO 3 → P 2 O 5 + NO 2 + H 2 O
FESO 4 + KMNO 4 + H 2 SO 4 → Fe (OH) 3 + MNSO 4 + K 2 SO 4 + H 2 O
Answers and solutions to tasks with explanations.
Exercise 1:
H + C 0 O -2 H + Fe +2 S 2 - Ca +2 (O -2 Cl +) Cl - H 2 + S 2 +7 O 8 -2
Task 2:
2-methylbutene-2: CH 3 -C -1 H +1 \u003d C 0 (CH 3) -CH 3
acetone: (CH 3) 2 C +2 \u003d O -2
acetic acid: CH 3 -C +3 O -2 O -2 N +
Task 3:
Since in the dichromate molecule 2 of the chromium atom, then they give electrons 2 times more - i.e. 6.
6Feso 4 + K 2 Cr 2 O 7 + 7H 2 SO 4 → 3FE 2 (SO 4) 3 + Cr 2 (SO 4) 3 + + K 2 SO 4 + 7H 2 O
Task 5:
If the environment is alkaline, then phosphorus +5 will exist in the form of salt - Potassium phosphate.
PH 3 + 8KMNO 4 + 11KOH → 8K 2 MNO 4 + K 3 PO 4 + 7H 2 O
| R -3 - 8E → P +5 | ||
| Mn +7 + 1e → Mn +6 |
If the medium is acidic, then phosphine goes into phosphoric acid.
PH 3 + KMNO 4 + H 2 SO 4 → MNSO 4 + H 3 PO 4 + K 2 SO 4 + H 2 O
| R -3 - 8E → P +5 | ||
| Mn +7 + 5e → Mn +2 |
Task 6:
As zinc - amphoterous Metal, in an alkaline solution it forms hydroxacomplex. As a result of the arrangement of the coefficients, it is found that water must be present in the left side of the reaction:
KNO 3 + 4ZN + 7KOH + 6N 2 O → N -3 H 3 + + 4k 2
| Zn 0 - 2E → Zn 2+ | ||
| N +5 + 8e → N -3 |
Task 7:
Electrons give two atoms S. In the alkene molecule. Therefore, we must take into account general The number of aligned electrons given by the entire molecule:
3CH 3 -C -1 H \u003d C -2 H 2 + 2KMN +7 O 4 + 4H 2 O (Hall. Rr.) → 3CH 3 -C 0 HOH-C -1 H 2 OH + 2MN +4 O 2 + 2Koh
| Mn +7 + 3e → Mn +4 | ||||||
| C -1 - 1e → C 0 | ||||||
| C -2 - 1e → C -1 |
||||||
| 3SH 3 -C -1 H \u003d C -2 H 2 + 10KMN +7 O 4 | 3CH 3 -C +3 Ook + 3K 2 C +4 O 3 + 10mn +4 O 2 + KOH + 4N 2 |
|
| Mn +7 + 3e → Mn +4 | ||||
| C -1 - 4E → C +3 | ||||
| C -2 - 6E → C +4 |
||||
Note that out of 10 potassium ions 9 are distributed between the two salts, so alkali will succeed only one molecule.
The task 8:
3mnso 4 + 2kmno 4 + 2 H 2 O → 5mno 2 + k 2 SO 4 + 2H 2 SO 4
| Mn 2+ - 2E → Mn +4 | ||
| Mn +7 + 3e → Mn +4 |
In the process of drawing up balance, we see that on 2 ions to + There are 3 sulfate ions. So, in addition to sulfate, potassium is formed yet sulfuric acid (2 molecules).
Task 9:
3H 2 S + 2KMNO 4 + (H 2 O) → 3S 0 + 2MNO 2 + 2KOH + 2H 2 O
(Permanganate is not a very strong oxidizing agent in solution; Please note that water transfers In the process of adjusting to the right!)
H 2 S + 8HNO 3 (conc.) → H 2 S +6 O 4 + 8NO 2 + 4H 2 O
(Concentrated nitric acid is a very strong oxidizer)
Task 10:
Do not forget that manganese accepts electrons, wherein chlorine must give them.
Chlorine is allocated as a simple substance.
2kmno 4 + 16hcl → 2mnCl 2 + 5Cl 2 + 2KCl + 8H 2 O
Task 11:
The higher the nonmetall in the subgroup, the more he active oxidizer. Chlorine in this reaction will be an oxidizing agent. The iodine goes into the most stable positive degree of oxidation +5, forming an iodinite acid.
5Cl 2 + I 2 + 6H 2 O → 10HCl + 2HIO 3
Task 12:
H 2 O 2 + 2KI + H 2 SO 4 → I 2 + K 2 SO 4 + 2H 2 O
(peroxide - oxidizing agent, because Restorener - ki)
3H 2 O 2 + K 2 CR 2 O 7 + 4H 2 SO 4 → 3O 2 + CR 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O
(peroxide - reducing agent, because Oxidizer - Permanganate potassium)
H 2 O 2 + KNO 2 → KNO 3 + H 2 O
(peroxide - oxidizing agent, because the role of the reducing agent is more characteristic of potassium nitrite, which seeks to go to nitrate)
The task 13:
36hno 3 + Al → 10Al (NO 3) 3 + 3N 2 + 18H 2 O
10Al + 6kmno 4 + 24h 2 SO 4 → 5Al 2 (SO 4) 3 + 6mnso 4 + 3k 2 SO 4 + 24H 2 O
Task 14:
In the molecule Fe 3 O 4 of the three iron atoms, only one charges +2. Onokissed +3.
(Fe +2 O Fe 2 +3 O 3)
3Fe
3
O.
4
+ 28hno 3 → 9Fe +3 (NO 3) 3 + NO + 14H 2 O
Task 16:
Ba + HNO 3 → Bao + NO 2 + H 2 O (aqueous solution)
BA + HNO 3 → BA (No. 3
)
2
+ NO 2 + H 2 O
PH 3 + KMNO 4 + KOH → K 2 MNO 4 + H 3 PO 4 + H 2 O (alkaline environment)
PH 3 + KMNO 4 + KOH → K 2 MNO 4 + K. 3
PO. 4
+ H 2 O
P + HNO 3 → P 2 O 5 + NO 2 + H 2 O (aqueous solution)
P + HNO 3 → H. 3
PO. 4
+ NO 2 + H 2 O
FESO 4 + KMNO 4 + H 2 SO 4 → FE (OH) 3 + MNSO 4 + K 2 SO 4 + H 2 O (sour Wednesday)
FESO 4 + KMNO 4 + H 2 SO 4 → FE. 2
(So. 4
)
3
+ MNSO 4 + K 2 SO 4 + H 2 O
1.cr2 (SO4) 3 + ... + NaOH → Na2Cro4 + NaBr + ... + H2O
2. Si + HNO3 + HF → H2SIF6 + NO + ...
3. P + HNO3 + ... → NO + ...
4. K2Cr2O7 + ... + H2SO4 → I2 + CR2 (SO4) 3 + ... + H2O
5. P + HNO3 + ... → No2 + ...
6. K2Cr2O7 + HCl → CL2 + KCL + ... + ...
7. B + HNO3 + HF → HBF4 + NO2 + ...
8. KMNO4 + H2S + H2SO4 → MNSO4 + S + ... + ...
9. KMNO4 + ... → CL2 + MNCL2 + ... + ...
10. H2S + HMNO4 → S + MNO2 +
11. KMNO4 + KBR + H2SO4 → MNSO4 + BR2 + ... + ...
12. KCLO + ... → I2 + KCL + ...
13. KnO2 + ... + H2SO4 → NO + I2 + ... + ...
14. Nano2 + ... + H2SO4 → NO + I2 + ... + ...
15. HCOH + KMNO4 → CO2 + K2SO4 + ... + ...
16. PH3 + HMNO4 → MnO2 + ... + ...
17. P2O3 + HNO3 + ... → NO + ...
18. PH3 + HCLO3 → HCl + ...
19. Zn + Kmno4 + ... → ... + MNSO4 + K2SO4 + ...
20. FECL2 + HNO3 (end.) → Fe (NO3) 3 + HCl + ... + ...
Tasks C1.(solutions and answers)
1. CR2 (SO4) 3 + 3Br2 + 16NAOH \u003d 2NA2CRO4 + 6NABR + 3NA2SO4 + 8H2O
Si0 - Restorener, HNO3 (N + 5) - Oxidizer
3. 3p + 5hnO3 + 2H2O \u003d 3H3PO4 + 5NO
Ki (I-) - reducing agent, K2CR2O7 (CR + 6) - Oxidizer
5.
K2CR2O7 (CR + 6) - Oxidizer, HCl (CL-) - Restore
7. B + 3HNO3 + 4HF \u003d HBF4 + 3NO2 + 3H2O
H2S (S-2) - Restorener, KMNO4 (Mn + 7) - Oxidizer
9. 2kmnO4 + 16HCl \u003d 5Cl2 + 2mnCl2 + 2KCl + 8H2O
H2S (S-2) - Restorener, HMNO4 (Mn + 7) - Oxidizer
11. 2kmno4 + 10kbr + 8H2S04 \u003d 2MNSO4 + 5B2 + 6K2SO4 + 8H2O
KCLO (CL + 1) - Oxidizer, Hi (I-) - Restore
13. KNO2 + 2HI + H2SO4 \u003d 2NO + I2 + K2SO4 + 2H2O
NO (N + 2) - Restorener, KCLO (CL + 1) - Oxidizer
15. 5hcoh + 4kmnO4 + 6H2SO4 \u003d 5CO2 + 2K2SO4 + 4MNSO4 + 11H2O
KMNO4 (Mn + 7) - Oxidizer, PH3 (P-3) - Restore
17. 3P2O3 + 4HnO3 + 7H2O \u003d 4NO + 6H3PO4
PH3 (P-3) - Restorener, HCLO3 (CL + 5) - Oxidizer
19. 5ZN + 2KmnO4 + 8H2SO4 \u003d 5ZNSO4 + 2MNSO4 + K2SO4 + 8H2O
FECL2 (Fe + 2) - Restorener, HNO3 (N + 5) - Oxidizer
Tasks C2.
1. Substances are given: Magnesium Ammonia, Nitrogen, Nitric Acid (RSC). Write the equations of four possible reactions between these substances.
2. Substances are given: calcium, phosphorus, nitric acid. Write the equations of four possible reactions between these substances.
3. Substances are given: sodium sulfite, water, potassium hydroxide, potassium permanganate, phosphate acid. Write the equations of four possible reactions between these substances.
4. Substances are given: copper, nitric acid, copper sulphide (II), nitrogen oxide (II).
5. Write the equations of four possible reactions between these substances.
6. Substances are given: sulfur, hydrogen sulfide, nitric acid (conc.), Sulfuric acid (conc.). Write the equations of four possible reactions between these substances.
7. Aqueous solutions are given: iron (III) chloride, sodium iodide, sodium bichromate, sulfuric acid and cesium hydroxide. Write the equations of four possible reactions between these substances.
9. Write the equations of four possible reactions between these substances.
10. Substances are given: carbon, hydrogen, sulfuric acid (conc.), Potassium dichromat. Write the equations of four possible reactions between these substances.
11. Substances are given: silicon, hydrochloric acid, caustic sodium, sodium bicarbonate. Write the equations of four possible reactions between these substances.
12. Substances are given: aluminum, water, diluted nitric acid, concentrated sodium hydroxide solution. Write the equations of four possible reactions.
13. Aqueous solutions are given: sodium sulfide, hydrogen sulfide, aluminum chloride, chlorine. Write the equations of four possible reactions between these substances.
14. Substances are given: sodium oxide, iron oxide (III), iodine hydrogen, carbon dioxide. Write the equations of four possible reactions between these substances.
15. Aqueous solutions are given: potassium hexagidroxaluminate, aluminum chloride, hydrogen sulfide, rubidium hydroxide. Write the equations of four possible reactions between these substances
16. Substances are given: potassium carbonate (solution), potassium bicarbonate (solution), carbon dioxide, magnesium chloride, magnesium. Write the equations of four possible reactions between these substances.
17. Substances are given: sodium nitrate, phosphorus, bromine, potassium hydroxide (solution). Write the equations of four possible reactions between these substances.
Tasks C2. (solutions and answers)
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3mg + 2NH3 \u003d MG3N2 + 3H2 4mg + 10hn03 \u003d 4mg (NO3) 2 + N2O + 5H2O NH3 + HNO3 \u003d NH4NO3 |
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4CA + 10HnO3 (concluded) \u003d 4ca (NO3) 2 + N2O + 5H2O 4CA + 10HNO3 (RSC) \u003d 4CA (NO3) 2 + NH4NO3 + 3H2O P + 5HNO3 \u003d H3PO4 + 5NO2 + H2O 3CA + 2P \u003d CA3P2 |
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Na2SO3 + 2kmnO4 + 2KOH \u003d NA2SO4 + 2K2MNO4 + H2O 3NA2SO3 + 2KmnO4 + H2O \u003d 3NA2SO4 + 2MNO2 + 2KOH Na2SO3 + H3PO4 \u003d NaH2PO4 + NaHSO3 3KOH + H3PO4 \u003d K3PO4 + 3H2O |
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Cu + 4hnO3 (concluded) \u003d Cu (NO3) 2 + 2NO2 + 2H2O 3CU + 8HNO3 (RSC) \u003d 3CU (NO3) 2 + 2NO + 4H2O CUS + 8HNO3 (conc) \u003d CUSO4 + 8NO2 + 4H2O 2CU + 2NO \u003d 2CUO + N2 |
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S + 6HNO3 (concluding) \u003d H2SO4 + 6NO2 + 2H2O S + 2H2SO4 (CON) \u003d 3SO2 + 2H2O H2S + 2HNO3 (concluded) \u003d S + 2NO2 + 2H2O H2S + 3H2SO4 (concludes) \u003d 4SO2 + 4H2O |
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2FeCl3 + 2NAI \u003d 2NACL + 2FECL2 + I2 FECL3 + 3CSOH \u003d FE (OH) 3 ↓ + 3CSCL H2SO4 + 2CSOH \u003d CS2SO4 + 2H2O Na2Cr2O7 + 2CSOH \u003d Na2Cro4 + CS2CRO4 + H2O Na2Cr2O7 + 6NAI + 7H2SO4 \u003d CR2 (SO4) 3 + 3i2 + 4NA2SO4 + 7H2O |
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2Al + 3Cl2 \u003d 2AlCl3 2ki + CL2 \u003d I2 + 2KCL 2ki + 2h2SO4 (concluding) \u003d i2 + k2SO4 + SO2 + 2H2O 2AL + 6H2SO4 (concaten) \u003d Al2 (SO4) 3 + 3SO2 + 6H2O |
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C + 2H2SO4 (CON) \u003d CO2 + 2SO2 + 2H2O 3C + 8H2SO4 + 2K2CR2O7 \u003d 3CO2 + 2CR2 (SO4) 3 + 2K2SO4 + 8H2O K2Cr2O7 + 2H2SO4 \u003d 2KHSO4 + 2CRO3 + H2O |
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NaOH + HCl \u003d NaCl + H2O NaHCO3 + HCl \u003d NaCl + CO2 + H2O NaHCO3 + NaOH \u003d Na2Co3 + H2O Si + 4Naoh \u003d Na4SiO4 + 2H2 |
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2AL (chalk. PI) + 6H2O \u003d 2AL (OH) 3 + 3H2 NaOH + HNO3 \u003d Nano3 + H2O 8Al + 30hnO3 \u003d 8Al (NO3) 3 + 3NH4NO3 + 9H2O 2AL + 2NAOH + 6H2O \u003d 2NA + 3H2 (Permissible na3) |
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Na2S + H2S \u003d 2NAHS 3NA2S + 2AlCl3 + 6H2O \u003d 3H2S + 2AL (OH) 3 + 6NACL Na2S + CL2 \u003d 2NACL + S H2S + CL2 \u003d 2HCl + S |
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Na2O + Fe2O3 \u003d 2NAFEO2 2HI + Na2O \u003d 2NAI + H2O Na2O + CO2 \u003d Na2Co3 Fe2O3 + 6Hi \u003d 2Fei2 + i2 + 3H2O |
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K3 + ALCL3 \u003d 2AL (OH) 3 + 3KCL K3 + 3H2S \u003d Al (OH) 3 + 3KHS + 3H2O H2S + 2RBOH \u003d RB2S + 2H2O ALCL3 + 3RBOH \u003d AL (OH) 3 + 3RBCL |
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K2CO3 + CO2 + H2O \u003d 2KHCO3 2K2CO3 + H2O + MgCl2 \u003d (MgOH) 2CO3 + CO2 + 4KCL 2khco3 + MgCl2 \u003d MgCO3 + 2KCL + CO2 + H2O CO2 + 2MG \u003d C + 2MGO |
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5nano3 + 2p \u003d 5NanO2 + P2O5 5Br2 + 2p \u003d 2PBr5 4p + 3KOH + 3H2O \u003d 3KH2PO4 + PH3 BR2 + 2KOH (HOT) \u003d KBRO + KBR + H2O 3BR2 + 6KOH (mountains) \u003d 5kbr + KBRO3 + 3H2O |
Tasks C3.
Write the reaction equations with which the following transformations can be carried out:
t0, Sakk. + CH3CL, ALCL3 + SL2, UV + con water., T0
1. Ethin → X1 → Toluene → X2 → X3 → C6H5-CH2-SHOSN
H2SO4 scan. H2SO4 conc. T0 + BR2 + KOH aqueous., T0
2. Potassium → Potassium ethylate → X1 → CH2 \u003d CH2 → X2 → X3
H2O 12000 T0, Cat. + CH3CL, ALCL3 + CL2, UV
3. Aluminum carbide → X1 → X2 → Benzene → X3 → X4
H2O + H2O + KMNO4 + H2SO4 CACO3 T0
4. CAC2 → Ethin → Ethanal → X1 → X2 → x3
CH3CL, ALCL3 + KMNO4 + H2SO4 + CH3N, H2SO4
5. Methane → X1 → Benzene → X2 → Benzoic Acid → X3
BR2, light + con (alcohol.) HBR Na
6. CH3-CH2-CH (CH3) -CH3 → X1 → X2 → X1 → X3 → CO2
NamnO4 + NaOH Electrolysis CL2, Light Koh, H2O H2SO4, T0
7. CH3CHO → X1 → C2N6 → X2 → X3 → (C2N5) 2O
H2O, HG2 + + KMNO4 + H2SO4 + NaOH + CH3I + H2O, H +
8. C2N2 → X1 → CH3COs → X2 → X3 → Acetic acid
O2 + H2, Cat. + Na + HCl + KMNO4 + H2SO4
9. SN4 → NNO → H.1 → H.2 → H.1 → H.3
C, T + C2H5CL, ALCL3 BR2, Hν KOH (alcohol.) KMNO4, H2O
10. C2N2.→ H.1 → S.6 N.5 FROM2 N.5 → H.2 → H.3 → H.4
AG (NH3) 2] OH CL2, Hν NaOH (alcohol.) + CH3OH, H2SO4 polymerization
11. CH3-CH2-SNO → x1 → x2 → x3 → x4 → x5
H2SO4, 2000C Cat., T + OH + HCl + KMNO4 + H2O
12. Ethanol → X1 → X2 → AG2C2 → X2 → X3
CT., T + CL2, FECL3, T + HONO2, H2SO4 + KMNO4 + H2SO4
13. C2N2.→ H.1 → H.2 → S.6 N.5 SN3 → SN3 -FROM6 N.4- NO2 → X3.
electrolysis + CL2, Hν + NaOH + H2O H2SO4 (CON), T \u003c1400
14. CH3Cone → X1 → C2N6 → X2 → X3 → x4
H2, Ni, T + HBR, H2SO4 + KOH (alcohol.) + O2, T, PD2 + + OH
15. CH3CHO → X1 → X2 → Ethylene → CH3Cly → x3
Tasks C3. (solutions and answers)
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1) 3C2H2 → C6H6 2) C6H6 + CH3CL → C6H5CH3 + HCl 3) C6H5CH3 + CL2 → C6H5CH2CL + HCl 4) C6H5CH2CL + KOH → C6H5CH2OH + KCL 5) C6H5CH2OH + HCOOH → C6H5CH2OOCH + H2O |
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1) 2K + 2C2H5OH → 2C2H5OK + H2 2) C2H5OK + H2SO4 → C2H5OH + K2SO4 3) C2H5OH → C2H4 + H2O 4) CH2 \u003d CH2 + BR2 → CH2BR-CH2BR 5) CH2BR-CH2BR + 2KOH → CH2OH-CH2OH + 2KBR |
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1) AL4C3 + 12H2O → 3CH4 + 4AL (OH) 3 2) 2ch4 → C2H2 + 3H2 3) 3C2H2 → C6H6 4) C6H6 + CH3CL → C6H5CH3 + HCl 5) C6H5CH3 + CL2 → C6H5CH2CL + HCl |
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1) CAC2 + 2H2O → CA (OH) 2 + C2H2 2) C2H2 + H2O → CH3CHO 3) 5ch3CHO + 2KMNO4 + 3H2SO4 → 5Ch3COOH + K2SO4 + 2MNSO4 + 3H2O 4) 2Ch3COOH + Caco3 → (CH3COO) 2CA + H2O + CO2 5) (CH3COO) 2CA → Caco3 + CH3-CO-CH3 |
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1) 2ch4 → C2H2 + 3H2 2) 3C2H2 → C6H6 3) C6H6 + CH3CL → C6H5CH3 + HCl 4) 5C6H5CH3 + 6KMNO4 + 9H2SO4 → 5CH3COOH + 3K2SO4 + 6MNSO4 + 14H2O 5) CH3COOH + CH3OH → CH3COOCH3 + H2O |
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1) CH3-CH2-CH (CH3) -CH3 + BR2 → CH3-CH2-CBR (CH3) -CH3 + HBR 2) CH3-CH2-CBR (CH3) -CH3 + KOH (alcohol.) → CH3-CH \u003d C (CH3) -CH3 + H2O + KBR 3) CH3-CH \u003d C (CH3) -CH3 + HBr → CH3-CH2-CBR (CH3) -CH3 4) 2ch3-CH2-CBR (CH3) -CH3 + 2NA → CH3-CH2-C (CH3) 2-C (CH3) 2-CH2-CH3 + 2NABR 5) 2C10H22 + 31O2 → 20CO2 + 22H2O |
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1) CH3CHO + 2NAMNO4 + 3NAOH → CH3COONA + 2NA2MNO4 + 2H2O 2) 2ch3Coona → C2H6 + 2NAHCO3 + H2 3) C2H6 + CL2 → C2H5CL + HCl 4) C2H5CL + KOH → C2H5OH + NaCl |
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1) C2H2 + H2O → CH3COH 2) 5ch3CHO + 2KMNO4 + 3H2SO4 → 5CH3COOH + K2SO4 + 2MNSO4 + 3H2O 3) CH3COOH + NaOH → CH3COONA + H2O 4) CH3COONA + CH3I → CH3COOCH3 + NAI 5) CH3COOCH3 + H2O → CH3COOH + CH3OH |
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1) CH4 + O2 → HCHO + H2O 2) HCHO + H2 → CH3OH 3) 2ch3oh + 2NA → 2Ch3ONA + H2 4) CH3ONA + HCL → 2Ch3OH + NaCl 5) 5ch3oh + 6kmnO4 + 9H2SO4 → 5CO2 + 6mnSO4 + 3K2SO4 + 19H2O |
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1) 3C2H2 → C6H6 2) C6H6 + C2H5CL → C6H5C2H5 + HCl 3) C6H5C2H5 + BR2 → C6H5-CHBR-CH3 + HBR 4) C6H5-CHBR-CH3 + KOH (alcohol) → C6H5-CH \u003d CH2 + KBR + H2O 5) 3C6H5-CH \u003d CH2 + 2KMNO4 + 4H2O → 3C6H5-CH (OH) -CH2OH + 2MNO2 + 2KOH |
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1) CH3-CH2-CHO + 2OH → CH3-CH2-Chooh + 2Ag + 4NH3 + H2O 2) CH3-CH2-Chooh + CL2 → CH3-CHCL-COOH + HCl 3) CH3-CHCL-COOH + NaOH (alcohol) → CH2 \u003d CH-COOH + NaCl + H2O 4) CH2 \u003d CH-COOH + CH3OH → CH2 \u003d CH-COOCH3 + H2O 5) NCH2 \u003d CH-COOCH3 → (-CH2-CH-) N |
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1) C2H5OH → CH2 \u003d CH2 + H2O 2) CH2 \u003d CH2 → C2H2 + H2 3) C2H2 + 2OH → C2AG2 ↓ + 4NH3 + 2H2O 4) C2AG2 + 2HCl → C2H2 + 2AGCL 5) 3C2H2 + 8KMNO4 → 3K2C2O4 + 2KOH + 8MNO2 + 2H2O |
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1) 3C2H2 → C6H6 2) C6H6 + CL2 → C6H5CL + HCl 3) C6H5CL + CH3CL + 2NA → C6H5-CH3 + 2NACL 4) C6H5-CH3 + HO-NO2 → CH3-C6H4-NO2 + H2O 5) 5ch3-C6H4-NO2 + 6KMNO4 + 9H2SO4 → 5Hooc-C6H4-NO2 + 6MNSO4 + 3K2SO4 + 14H2O |
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1) CH3COOH + NaOH → CH3Sona + H2O 2) 2ch3Sona + 2H2O → C2H6 + 2NAHCO3 + H2 3) C2H6 + CL2 → C2H5CL + HCl 4) C2H5CL + NaOH → C2H5OH + NaCl H2SO4, T.<140° 5) 2C2H5OH → C2H5-O-C2H5 + H2O |
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1) CH3CHO + H2 → C2H5OH 2) C2H5OH + HBR → C2H5BR + H2O 3) C2H5BR + KOH (alcohol) → C2H4 + KBR + H2O 4) 2C2H4 + O2 → 2ch3cho 5) CH3CHO + 2OH → CH3COOH + 2AG ↓ + 4NH3 + H2O |
Tasks C4 (solutions and answers)
1. 3 Cu + 8.Hno3 \u003d 3.Cu (NO3) 2 + 2NO + 4.H2.O (1)
m (p-ra HnO3) \u003d 115.3 g
m (hnO3) \u003d 115.3. 0.3 \u003d 34.59 g
n (hnO3) \u003d 34.59 g / 63 \u003d 0.55 mol
n (Cu) \u003d 6.4 g / 64 g / mol \u003d 0.1 mol
HNO3 - in excess
n (hnO3 Izr.) \u003d 0.1. 8/3 \u003d 0.27
n (hno3 is over.) \u003d 0.55 mol - 0.27 mol \u003d 0.28 mol
Cu (NO3) 2 + 2NAOH \u003d Cu (OH) 2 + 2NanO3 (2)
n (Cu (NO3) 2) \u003d 0.1 mol (by equation 1)
n (NaOH) \u003d 0.2 mol (by equation 2)
HNO3 + NaOH \u003d Nano3 + H2O (with excess HNO3) (3)
n (NaOH) \u003d 0.28 mol
n (NaOH common.) \u003d 0.2 + 0.28 \u003d 0.48 mol
m (NaOH) \u003d 0.48 mol. 40 g / mol \u003d 19.2 g
Ω (NaOH) \u003d m (p.) / m (p-ra) \u003d 0.096or 9,6%
2. AGNO3 + NaCl \u003d AGCL↓ + Nano3
m (NaCl) \u003d 1170 0.005 \u003d 5.85 g
n (NaCl) \u003d 5.85 g / 58.5 g / mol \u003d 0.1 mol
m (AGNO3) \u003d 1275 0.002 \u003d 2.55 g
n (agno3) \u003d 2.55 g / 170 g / mol \u003d 0,015 mol
AGNO3 - in a shortcoming
n (nano3) \u003d 0,015 mol
m (Nano3) \u003d 0.015 mol. 85 g / mol \u003d 1.28 g
n (AgCl) \u003d 0,015 mol
m (AgCl) \u003d 0.015 mol. 143.5 g / mol \u003d 2.15 g
m (p-ra) \u003d 1275 g + 1170 g - 2.15 g \u003d 2442.85
Ω (nano3) \u003d 1.28 g / 2442.85 g \u003d 0,00052 or 0,052%
3. NH4Cl + NaOH \u003d NaCl + NH3 + H2O
m (NH4Cl) \u003d 107 g. 0.2 \u003d 21.4 g
n (NH4Cl) \u003d 21.4 g / 53.5 g / mol \u003d 0.4 mol
m (NaOH) \u003d 150 g. 0.18 \u003d 27 g
n (NaOH) \u003d 27 g / 40 g / mol \u003d 0,675 mol
NH4CL - in a shortcoming
n (NH3) \u003d 0.4 mol
m (NH3) \u003d 0.4 mol. 17 g / mol \u003d 6.8 g
n (NaCl) \u003d 0.4 mol
m (NaCl) \u003d 0.4 mol. 58.5 g / mol \u003d 23.4 g
m (p-ra) \u003d 107 g + 150 g - 6.8 g \u003d 250.2 g
Ω (NaCl) \u003d 23.4 g / 250.2 g \u003d 0.0935 or 9.35%
NH3 + H3PO4 \u003d NH4H2PO4
n (H3PO4) \u003d N (NH3) \u003d 0.4 mol
m (H3PO4) \u003d 0.4 mol. 98 g / mol \u003d 39.2 g
m (p-ra H3PO4) \u003d 39.2 g / 0.6 \u003d 65.3 g
4. CAH2 + 2HCl \u003d CaCl2 + 2H2
n (H2) \u003d 11.2 l / 22.4 l / mol \u003d 0.5 mol
m (H2) \u003d 0.5 mol. 2 g / mol \u003d 1 g
n (Cah2) \u003d 0.25 mol
m (CAH2) \u003d 0.25 mol. 42g / mol \u003d 10.5 g
m (HCl in p-re) \u003d 200 g. 0.15 \u003d 30 g
n (HCl cutting.) \u003d 0.5 mol
m (HCl cutag.) \u003d 0.5 mol. 36.5 g / mol \u003d 18.25 g
m (HCl Ost.) \u003d 30 g - 18.25 g \u003d 11.25 g
m (p-ra) \u003d 200 g + 10.5 g - 1 g \u003d 209.5 g
Ω (HCl) \u003d 11.75 g / 209.5 g \u003d 0,056 or 5,6%
5. Lioh + HNO3 \u003d LINO3 + H2O
m (p-ra lioh) \u003d 125 ml. 1.05 g / ml \u003d 131.25 g
m (Lioh) \u003d 131.25 g. 0.05 \u003d 6.57 g
n (lioh) \u003d 6.57 g / 24 g mol \u003d 0.27 mol
m (p-ra HnO3) \u003d 100 ml. 1.03 g / ml \u003d 103 g
m (hno3) \u003d 103 g. 0.05 \u003d 5.15 g
n (hno3) \u003d 5.15 g / 63 g / mol \u003d 0.082 mol
HNO3 - in the disadvantage, therefore, the environment is alkaline
n (LINO3) \u003d 0,082 mole
m (LINO3) \u003d 0.082 mol. 69 g / mol \u003d 5.66 g
m (p-ra) \u003d 131.25 g + 103 g \u003d 234.25 g
Ω (LINO3) \u003d 5.66 g / 234.25 g \u003d 0,024 or 2,4%
6. 3Cl2 + 6Naoh (mountains) \u003d 5NAcl + Naclo3 + 3H2O
m (p-ra NaOH) \u003d 228.58 ml. 1.05 g / ml \u003d 240 g
m (NaOH) \u003d 240 g. 0.05 \u003d 12 g
n (NaOH) \u003d 12 g / 40 g / mol \u003d 0.3 mol
n (CL2) \u003d 0.15 molm (CL2) \u003d 0.15 mol. 71 g / mol \u003d 10.65 g
n (NaCl) \u003d 0.25 molm (NaCl) \u003d 0.25 mol. 58.5 g / mol \u003d 14,625 g
n (NaClO3) \u003d 0.05 molm (NaClO3) \u003d 0.05 mol. 106.5 g / mol \u003d 5,325 g
m (p-ra) \u003d 240g + 10.65 g \u003d 250.65 g
Ω (NaCl) \u003d 14,625 g / 250.65 g \u003d 0,0583 or 5,83%
Ω (Naclo3) \u003d 5.325 g / 250.65 g \u003d 0,0212 or 2,12%
7. P2O5 + 3H2O \u003d 2H3PO4
m (H3PO4 in p-re) \u003d 60 g. 0.082 \u003d 4.92 g
n (H3PO4 in p-re) \u003d 4.92 g / 98 g / mol \u003d 0.05 mol
n (p2O5) \u003d 1.42 g / 142 g, mol \u003d 0.01 mol
n (H3PO4 image.) \u003d 0.02 mol
n (KOH) \u003d 3.92 g / 56 g / mol \u003d 0.07 mol
n (H3PO4 commonly.) \u003d 0.02 mol + 0.05 mol \u003d 0.07 mol
n (H3PO4): N (KOH) \u003d 1: 1, therefore
H3PO4 + KOH \u003d KH2PO4 + H2O
n (KH2PO4) \u003d 0.07 mol
8. CAC2 + 2H2O \u003d CA (OH) 2 + C2H2
n (C2H2) \u003d 4,48 l / 22.4 l / mol \u003d 0.2 mol
