Solutions. The tasks of the exam in chemistry with solutions: the relationship of various classes of inorganic substances was dissolved in concentrated nitrogen
Live Journal
Facebook.
Twitter.The chemical properties of most elements are their ability to dissolve in aqueous medium and acids. Studying the characteristics of copper is associated with a low-active action under normal conditions. A feature of its chemical processes is the formation of compounds with ammonia, mercury, nitric and low solubility of copper in water is not able to cause corrosion processes. Special is inherent in it chemical propertiesallowing to use a connection in different industries.
Description of the element
Copper is considered the oldest of metals, which learned how to produce people even before our era. This substance is obtained from natural sources in the form of ore. Copper is called element chemical table With the Latin name Cuprum, whose sequence number is 29. In periodic system It is located in the fourth period and refers to the first group.
Natural substance is pink-red severe metal With soft and forging structure. The temperature of its boiling and melting is more than 1000 ° C. It is considered a good conductor.
Chemical structure and properties
If you study the electronic formula of the copper atom, you can find that it has 4 levels. On the valence 4S orbital is only one electron. During chemical reactions from an atom, 1 to 3 negatively charged particles can be cleaving from 1 to 3, then copper compounds are obtained with a degree of oxidation +3, +2, +1. Its bivalent derivatives possess the greatest stability.
IN chemical reactions It acts as a low-effective metal. Under normal conditions, the solubility of copper in water is absent. In dry air, corrosion does not observe, but when heated, the metal surface is covered with black bivalent oxide. Copper chemical resistance is manifested in the action of anhydrous gases, carbon, row organic compounds, phenolic resins and alcohols. It is characterized by the complexation reaction with the separation of painted compounds. Copper has low similarities with alkaline group metals associated with the formation of a derivatives of a monovalent row.
What is solubility?
This is the process of forming homogeneous systems in the form of solutions when the interaction of one compound with other substances. Their components are individual molecules, atoms, ions and other particles. The degree of solubility is determined by the concentration of the substance, which was dissolved during the preparation of a saturated solution.
Unit of measurement most often are percentages, volumetric or weights. The solubility of copper in water, as well as other compounds of a solid species, is subject to changes in temperature conditions. This dependence is expressed with the help of curves. If the indicator is very small, the substance is considered insoluble.
Copper solubility in aquatic environment
Metal manifests corrosion resistance under action sea water. This proves its inertia under normal conditions. The solubility of copper in water (fresh) is practically not observed. But in a humid environment and under the action of carbon dioxide on a metal surface, the formation of a green film, which is the main carbonate:
Cu + Cu + O 2 + H 2 O + CO 2 → CU (OH) 2 · CUCO 2.
If we consider its monovalent compounds in the form of salts, then their insignificant dissolution is observed. Such substances are subject to rapid oxidation. As a result, copper compounds are bivalent. These salts have good solubility in the aquatic environment. Their complete dissociation on ions occurs.
Solubility in acids
The usual conditions of copper reactions with weak or dilute acids do not contribute to their interaction. The chemical metal process with alkalis is not observed. The solubility of copper in acids is possible if they are strong oxidizing agents. Only in this case the interaction proceeds.
Copper solubility in nitric acid
Such a reaction is possible due to the fact that the process of a strong reagent occurs. Nitric acid in a diluted and concentrated form shows oxidative properties with copper dissolution.
In the first embodiment, during the reaction, copper nitrate and nitrogen is obtained by a bivalent oxide in a ratio of 75% to 25%. The process with dilute acid nitrogen can be described by the following equation:
8HNO 3 + 3CU → 3CU (NO 3) 2 + NO + NO + 4H 2 O.
In the second case, there is copper nitrate and nitrogen oxides bivalent and quadricular, the ratio of which 1 to 1. In this process, 1 mole of metal and 3 praying acid nitrogen concentrated are involved. In the dissolution of copper, a strong heating of the solution occurs, as a result of which the thermal decomposition of the oxidant and the release of the additional volume of nitrogen oxides are observed:
4HNO 3 + Cu → Cu (NO 3) 2 + NO 2 + NO 2 + 2H 2 O.
The reaction is used in low-tonnage production associated with the processing of scrap or removal of coating from waste. However, this method of dissolution of copper has a number of disadvantages associated with the release of a large number of nitrogen oxides. For their capture or neutralization, special equipment is necessary. These processes are very expensive.
Copper dissolution is considered completed when occurs full termination Development of volatile nitrous oxides. The reaction temperature ranges from 60 to 70 ° C. The next step is the descent of the solution from its day there are small pieces of metal, which has not reacted. Water is added to the resulting liquid and filtering.
Solubility in sulfuric acid
In the usual state, this reaction does not proceed. The factor determining the dissolution of copper in sulfuric acid is its strong concentration. Diluted medium cannot oxidize metal. The dissolution of copper in the concentrated proceeds with the separation of sulfate.
The process is expressed by the following equation:
Cu + H 2 SO 4 + H 2 SO 4 → CUSO 4 + 2H 2 O + SO 2.
Properties of medium sulfate
A two-axis salt is also called sulk acid, designate it as: Cuso 4. It is a substance without a characteristic smell that does not show volatility. In anhydrous form, salt has no color, it is opaque, having high hygroscopicity. In copper (sulfate) solubility is good. Water molecules, connecting to salts, can form crystallohydrate compounds. An example serves as a blue pentahydrate. His formula: CUSO 4 · 5H 2 O.
Crystal hydrates are inherent in the transparent structure of the bluish shade, they show a bitter, metal taste. Their molecules are capable of lose linked water over time. In nature, there are in the form of minerals, which are attributed to halcantitis and butt.
Exposed to copper sulfate. Solubility is an exothermic reaction. In the process of hydration of salt, a significant amount of heat is distinguished.
Copper solubility in hardware
As a result of this process, pseudoplays from Fe and Cu are formed. For metallic iron and copper, limited mutual solubility is possible. The maximum values \u200b\u200bare observed at a temperature indicator 1099.85 ° C. The degree of solubility of copper in the solid form of iron is equal to 8.5%. These are small indicators. The dissolution of metallic iron in the solid form of copper is about 4.2%.
The decrease in temperature to room values \u200b\u200bmakes mutual processes minor. When moltening the metal copper, it is capable of wetting the iron in solid form. When obtaining Pseudo-locomotives Fe and Cu use special blanks. They are created by pressing or stove of iron powder located in a clean or alloyed form. Such blanks are impregnated with liquid copper, forming pseudoplans.
Dissolution in ammonia
The process often flows when NH 3 is passed into a gaseous form over a split metal. The result is the dissolution of copper in ammonia, the release of Cu 3 N. This compound is called alone nitride.
Salts are exposed to ammonium solution. The addition of such reagent to copper chloride leads to the fallout of the precipitate in the form of hydroxide:
CUCL 2 + NH 3 + NH 3 + 2H 2 O → 2NH 4 CL + CU (OH) 2 ↓.
Ammonia excess contributes to the formation of an integrated type compound having dark blue color:
Cu (OH) 2 ↓ + 4NH 3 → (OH) 2.
This process is used to determine bivalent copper ions.
Solubility in cast iron
In the structure of ductile perlite cast iron, in addition to the main components there is an additional element in the form of ordinary copper. It is it that increases the graphitization of carbon atoms, contributes to an increase in liquid process, strength and hardness of alloys. Metal positively affects the level of perlite in the final product. Copper solubility in cast iron is used to doping the source composition. The main goal of such a process is to obtain a carpet alloy. It will have elevated mechanical and corrosive properties, but reduced embrittlement.
If the copper content in the cast iron is about 1%, then the strength indicator for stretching is equal to 40%, and the fluidity increases to 50%. This significantly changes the characteristics of the alloy. Increasing the amount of metal alloying up to 2% leads to a change in strength to a value of 65%, and the yield indicator becomes 70%. With greater copper content in the composition of the cast iron, ferrous graphite is harder. The introduction into the structure of the alloying element does not change the technology of forming a viscous and soft alloy. The time that is discharged for annealing coincides with the duration of such a reaction at no copper impurity. It is about 10 hours.
The use of copper for the manufacture of cast iron with a high silicon concentration is not able to completely eliminate the so-called formation of the mixture during annealing. As a result, a product with low elasticity is obtained.
Solubility in mercury
When mixing mercury with metals of other elements, amalgams are obtained. This process can be passed at room temperature, because in such conditions PB is a liquid. The solubility of copper in mercury passes only during heating. Metal must be pre-crushed. When wetting the liquid rutty of solid copper, the mutual penetration of a single substance to another or the diffundation process occurs. The solubility value is expressed as a percentage and is 7.4 * 10 -3. In the reaction process, a hard simple amalgam is obtained, similar to cement. If it is very warm, it softens. As a result, such a mixture is used for fixing products from porcelain. There are also complex amalgams with the optimal content of metals in it. For example, in the dental alloy there are elements of copper and zinc. Their number in percent is as 65: 27: 6: 2. Amalgam with such a composition is called silver. Each component of the alloy performs a specific function that allows you to get a high-quality seal.
Another example is an amalgamid alloy, in which there is a high copper content. It is also called a copper alloy. The amalgam is present from 10 to 30% Cu. The high content of copper prevents the reaction of tin with mercury, which does not allow to form a very weak and corrosive alloy phase. In addition, a decrease in the number in silver seals leads to cheaper. To prepare the amalgam, it is desirable to use an inert atmosphere or a protective fluid that forms a film. Metals included in the alloy are capable of quickly oxidized by air. The process of heating the amalgam dumplings in the presence of hydrogen leads to distillation of mercury, which allows separating the elementary copper. As you can see, this topic is easy to explore. Now you know how copper interacts not only with water, but also with acids and other elements.
1) Copper nitrate was calcined, the resulting solid precipitate was dissolved in sulfuric acid. Through the solution, hydrogen sulfide was missed, the resulting black precipitate was abused, and the solid residue was dissolved when heated in concentrated nitric acid.
2) Calcium phosphate was sparkled with coal and sand, then the above-mentioned substance was burned in an excess of oxygen, the combustion product was dissolved in an excess of caustic soda. A solution of barium chloride was adhered to the resulting solution. The resulting precipitate was treated with an excess of phosphoric acid.
| Show | |
|---|---|
Ca 3 (PO 4) 2 → P → P 2 O 5 → Na 3 PO 4 → Ba 3 (PO 4) 2 → BAHPO 4 or BA (H 2 PO 4) 2 CA 3 (PO 4) 2 + 5C + 3SIO 2 → 3Casio 3 + 2p + 5Co |
|
3) Copper was dissolved in concentrated nitric acid, the resulting gas was mixed with oxygen and dissolved in water. In the resulting solution, zinc oxide was dissolved, then a large excess solution of sodium hydroxide was added to the solution.
4) At dry sodium chloride, they were focused by concentrated sulfuric acid at low heating, the produced gas was missed into the barium hydroxide solution. A solution of potassium sulfate was adhered to the resulting solution. The resulting precipitate was sparkled with coal. The resulting substance was treated with hydrochloric acid.
5) Aluminum sulfide sandpaper was treated with hydrochloric acid. At the same time, gas was separated and a colorless solution was formed. A solution of ammonia was added to the resulting solution, and the gas was missed through a solution of lead nitrate. The precipitate obtained was treated with hydrogen peroxide solution.
| Show | |
|---|---|
Al (OH) 3 ← ALCL 3 ← AL 2 S 3 → H 2 S → PBS → PBSO 4 Al 2 S 3 + 6HCl → 3H 2 S + 2AlCl 3 |
|
6) Aluminum powder was mixed with sulfur powder, the mixture was heated, the resulting substance was treated with water, while gas was separated and the precipitate was formed to which the excess of potassium hydroxide was added to complete dissolution. This solution was evaporated and calcined. An excess solution of hydrochloric acid was added to the resulting solid substance.
7) Potassium iodide solution was treated with chlorine mortar. The resulting precipitate was treated with sodium sulfite solution. A solution of barium chloride was added to the resulting solution, and after separation of the sediment, the silver nitrate solution was added.
8) The gray-green chromium oxide (III) powder was splashing with an excess of a pitch, the resulting substance was dissolved in water, and a dark green solution was obtained. A hydrogen peroxide was added to the obtained alkaline solution. It turned out a solution of yellow, which, when adding sulfuric acid, acquires orange color. When the hydrogen sulfide is passed through the obtained acidified orange solution, it will beter and again becomes greine.
| Show | |
|---|---|
CR 2 O 3 → KCRO 2 → K → K 2 CRO 4 → K 2 CR 2 O 7 → CR 2 (SO 4) 3 CR 2 O 3 + 2KOH → 2KCRO 2 + H 2 O |
|
9) Aluminum was dissolved in a concentrated solution of potassium hydroxide. Through the resulting solution was passed carbon dioxide until the sediment is discontinued. The precipitate was filtered and calcined. The resulting solid residue was sparkled with sodium carbonate.
10) silicon was dissolved in a concentrated solution of potassium hydroxide. An excess of hydrochloric acid was added to the resulting solution. The lane solution is heated. The separated precipitate was filtered and calcined with calcium carbonate. Write the equations of the described reactions.
11) Copper (II) oxide is heated in carbon monoxide. The resulting substance was burned in the atmosphere of chlorine. The product of the reaction was dissolved in water. The resulting solution was divided into two parts. To one part, a solution of potassium iodide was added to the second - a solution of silver nitrate. And in that and in another case, the formation of precipitate was observed. Write the equations of the four reactions described.
12) Copper nitrate was calcined, the resulting solid was dissolved in dilute sulfuric acid. The solution of the resulting salt was subjected to electrolysis. The substance highlighted on the cathode was dissolved in concentrated nitric acid. The dissolution proceeded with the separation of brown gas. Write the equations of the four reactions described.
13) Iron burned in chlorine atmosphere. The resulting substance was treated with an excess solution of sodium hydroxide. A brown precipitate was formed, which was filtered and calcined. The residue after the calcination was dissolved in hydrochildoic acid. Write the equations of the four reactions described.
14) Metal aluminum powder was mixed with a solid iodine and added a few drops of water. A solution of sodium hydroxide was added to the salt obtained before the precipitate is dropped. The resulting precipitate was dissolved in hydrochloric acid. With the subsequent addition of the sodium carbonate solution, the precipitate was again observed. Write the equations of the four reactions described.
15) As a result of incomplete combustion of coal, gas was obtained, in the current of which heated iron (III) oxide. The resulting substance was dissolved in hot concentrated sulfuric acid. The resulting salt solution was subjected to electrolysis. Write the equations of the four reactions described.
16) Some zinc sulfide was divided into two parts. One of them was treated with nitric acid, and the other was subjected to firing in the air. In the interaction of the excreted gases, a simple substance was formed. This substance is heated with concentrated nitric acid, and brown gas was separated. Write the equations of the four reactions described.
17) Potassium chlorate was heard in the presence of a catalyst, while the colorless gas was separated. Iron Okle's Iron Burning in the atmosphere of this gas was obtained. It was dissolved in excess of hydrochloric acid. A solution containing sodium dichromate and hydrochloric acid was added to the solution obtained.
| Show | |
|---|---|
1) 2KCLO 3 → 2xl + 3o 2 2) ZFE + 2O 2 → Fe 3 O 4 3) Fe 3 O 4 + 8nsі → FECL 2 + 2FECL 3 + 4N 2 4) 6 FECL 2 + Na 2 Cr 2 O 7 + 14 NSI → 6 FECL 3 + 2 CRCL 3 + 2NACL + 7N 2 O 18) Iron burned in Chlorine. The resulting salt was added to the sodium carbonate solution, while the brown precipitate fell. This precipitate was filtered and calcined. The resulting substance was dissolved in hydrochloric acid. Write the equations of the four reactions described. 1) 2fe + 3Cl 2 → 2FeCl 3 2) 2FeCl 3 + 3NA 2 CO 3 → 2FE (OH) 3 + 6NACL + 3CO 2 3) 2Fe (OH) 3 Fe 2 O 3 + 3H 2 O 4) Fe 2 O 3 + 6HI → 2FEI 2 + I 2 + 3H 2 O |
|
19) Potassium iodide solution was treated with an excess of chlorine water, while the formation of the precipitate was observed first, and then its complete dissolution. The hydrogen-containing acid formed at the same time was isolated from the solution, dried and carefully heated. The resulting oxide reacted with carbon monoxide. Record the equations of the described reactions.
20) Chromium (III) sulfide powder was dissolved in sulfuric acid. At the same time, gas was separated and the painted solution was formed. An excess solution of ammonia was added to the resulting solution, and the gas was missed through lead nitrate. The black precipitate obtained was turned after treatment with hydrogen peroxide. Record the equations of the described reactions.
21) Aluminum powder heated with sulfur powder, the resulting substance was treated with water. The precipitate was treated with an excess of concentrated solution of potassium hydroxide to its complete dissolution. A solution of aluminum chloride solution was added to the resulting solution and the formation of a white precipitate was again observed. Record the equations of the described reactions.
22) Potassium nitrate was heated with powdered lead until the reaction stops. The mixture of products was treated with water, and then the resulting solution was filtered. The filtrate was acidified with sulfuric acid and processed potassium iodide. The above-mentioned substance was heated with concentrated nitric acid. In the atmosphere of the resulting brown gas burned red phosphorus. Record the equations of the described reactions.
23) Copper was dissolved in dilute nitric acid. An excess ammonia solution was added to the resulting solution, observing the formation of the precipitate first, and then its complete dissolution to the formation of a dark blue solution. The resulting solution was treated with sulfuric acid until the characteristic blue painting of copper salts appeared. Record the equations of the described reactions.
| Show | |
|---|---|
1) 3CU + 8HNO 3 → 3CU (NO 3) 2 + 2NO + 4H 2 O 2) CU (NO 3) 2 + 2NH 3 H 2 O → Cu (OH) 2 + 2NH 4 NO 3 3) Cu (OH) 2 + 4NH 3 H 2 O → (OH) 2 + 4H 2 O 4) (OH) 2 + 3H 2 SO 4 → CUSO 4 +2 (NH 4) 2 SO 4 + 2H 2 O |
|
24) Magnesium was dissolved in dilute nitric acid, and the gas is not observed. The resulting solution was treated with an excess solution of potassium hydroxide during heating. Allocated gas burned in oxygen. Record the equations of the described reactions.
25) A mixture of powders of potassium nitrite and ammonium chloride was dissolved in water and the solution is gently heated. The gas has rewarded with magnesium. The reaction product was introduced to an excess solution of hydrochloric acid, while the gas is not observed. The resulting magnesium salt in solution was treated with sodium carbonate. Record the equations of the described reactions.
26) Aluminum oxide spiled with sodium hydroxide. The product of the reaction was introduced into ammonium chloride solution. The separated gas with a sharp smell is absorbed by sulfuric acid. The average salt formed at the same time. Record the equations of the described reactions.
27) Chlorine reacted with a hot potassium hydroxide solution. When cooling the solution, the crystals of the bertolet salt fell. The obtained crystals were introduced into a solution of hydrochloric acid. The simplicity of the substance reacted with metal iron. The product of the reaction was heated with a new iron jetty. Record the equations of the described reactions.
28) Copper was dissolved in concentrated nitric acid. An excess ammonia solution was added to the resulting solution, observing the formation of a precipitate first, and then its complete dissolution. The resulting solution was treated with an excess of hydrochloric acid. Record the equations of the described reactions.
29) iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess solution of sodium hydroxide. The resulting brown precipitate was filtered and calcined. The resulting substance was sparkled with iron. Write the equations of four described reactions.
30) As a result of incomplete combustion of coal, gas was obtained, in the current of which heated iron oxide (III). The resulting substance was dissolved in hot concentrated sulfuric acid. The resulting salt solution was treated with an excess solution of potassium sulfide.
31) Some zinc sulfide was divided into two parts. One of them was treated with hydrochloric acid, and another coupling on the air. In the interaction of the excreted gases, a simple substance was formed. This substance was heated with concentrated nitric acid, and brown gas was separated.
32) sulfur sprayed with iron. The reaction product was treated with hydrochloric acid. Allocated with gas burned in an excess of oxygen. The combustion products are absorbed with an aqueous solution of iron sulfate (III).
Like all D-elements, painted brightly.
Just like in copper is observed failing electrons - with s-orbitals on D-orbital
Electronic building of an atom:
Accordingly, there are 2 characteristic degrees of copper oxidation: +2 and +1.
Simple substance:metal golden pink color. 
Copper oxides:CU2O copper oxide (I) \\ copper oxide 1 - red-orange
Cuo copper oxide (II) \\ copper oxide 2 is black.
Other CU (I) compounds, except oxide, unstable.
Compounds of CU (II) - firstly, stable, secondly, blue or greenish color.
Why green copper coins? Copper in the presence of water interacts with carbon dioxide air, forms Cuco3 - a green substance.
Another painted copper compound is copper sulphide (II) - black precipitate.
Copper, in contrast to other elements, stands in after hydrogen, therefore does not emit it from the acids:
- from hot Sulfuric acid: CU + 2H2SO4 \u003d CUSO4 + SO2 + 2H2O
- from Cold Sulfuric acid: Cu + H2SO4 \u003d Cuo + SO2 + H2O
- with concentrated:
Cu + 4HnO3 \u003d Cu (NO3) 2 + 4NO2 + 4H2O - with diluted nitric acid:
3CU + 8HNO3 \u003d 3 CU (NO3) 2 + 2NO +4 H2O
Example of the task of the USE C2 Option 1:
The copper nitrate was calcined, the resulting solid precipitate was dissolved in sulfuric acid. Through the solution, hydrogen sulfide was missed, the resulting black precipitate was subjected to a firing, and the solid residue was dissolved when heated in nitric acid.
2SU (NO3) 2 → 2CUO ↓ +4 NO2 + O2
Solid precipitate - copper oxide (II).
Cuo + H2S → CUS ↓ + H2O
Copper sulphide (II) - black precipitate.
"Clamps subjected" means, an oxygen interaction occurred. Do not confuse with the "calcination". Fur - heat, naturally, at high temperatures.
2SUS + 3O2 \u003d 2CUO + 2SO2
The solid residue is Cuo - if the copper sulphide reacted completely, CuO + CUS - if partially.
Cuo + 2hnO3 \u003d Cu (NO3) 2 + H2O
CUS + 2HNO3 \u003d CU (NO3) 2 + H2S
another reaction is also possible:
CUS + 8HNO3 \u003d CU (NO3) 2 + SO2 + 6NO2 + 4H2O
Example of the task of the USE C2 Option 2:
Copper was dissolved in concentrated nitric acid, the resulting gas was mixed with oxygen and dissolved in water. In the resulting solution, zinc oxide was dissolved, then a large excess solution of sodium hydroxide was added to the solution.
As a result of the reaction with nitric acid, Cu (NO3) 2, NO2 and O2 is formed.
NO2 was mixed with oxygen - it means oxidized: 2nO2 + 5O2 \u003d 2N2O5. Mixed with water: N2O5 + H2O \u003d 2HNO3.
Zno + 2hnO3 \u003d Zn (NO3) 2 + 2H2O
Zn (NO 3) 2 + 4NAOH \u003d Na 2 + 2Nano 3
Task number 1
Sodium heated in the atmosphere of hydrogen. When the water and the formation of a transparent solution was observed when the water was observed to the resulting water. Through this solution, brown gas was missed, which was obtained as a result of copper interaction with a concentrated nitric acid solution. Write the equations of four described reactions.
1) When sodium heated in a hydrogen atmosphere (T \u003d 250-400 ° C), sodium hydride is formed):
2NA + H 2 \u003d 2NAH
2) when the water is added to the sodium hydride, NaOH is formed, and hydrogen is distinguished:
NAH + H 2 O \u003d NaOH + H 2
3) when the interaction of copper with a concentrated solution of nitric acid is distinguished by brown gas - NO 2:
CU + 4HNO 3 (conc.) \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
4) when the bogged gas is passed through the alkali solution, the disproportionation reaction - nitrogen N +4 is oxidized at the same time and is restored to n +5 and n +3:
2NAOH + 2NO 2 \u003d Nano 3 + Nano 2 + H 2 O
(Reaction of Disproportionation 2N +4 → N +5 + N +3).
Task number 2.
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 the four reactions described.
Formula of iron scale - Fe 3 O 4.
In the interaction of iron scale with a concentrated nitric acid, iron nitrate is formed and nitrogen oxide NO 2 is distinguished:
Fe 3 O 4 + 10hno 3 (conc.) → 3Fe (NO 3) 3 + NO 2 + 5H 2 O
When the iron nitrate interacts with sodium hydroxide, the sediment is highlighted - iron hydroxide (III):
Fe (NO 3) 3 + 3NAOH → Fe (OH) 3 ↓ + 3Nano 3
Fe (OH) 3 - amphoteric hydroxidenot soluble in water, decomposes when heated on iron (III) oxide and water:
2Fe (OH) 3 → Fe 2 O 3 + 3H 2 O
When smelting iron oxide (III) with iron, iron (II) oxide is formed:
Fe 2 O 3 + Fe → 3Feo
Task number 3.
Sodium burned in the air. The resulting substance was treated with a chloride in heating. The obtained simplicity of yellow-green, when heated, reacted with chromium oxide (III) in the presence of potassium hydroxide. When processing a solution of one of the resulting salts, the barium chloride fell a yellow precipitate. Write the equations of the four reactions described.
1) Sodium peroxide is formed when burning sodium in air:
2NA + O 2 → Na 2 O 2
2) when the sodium peroxide interacts with a chloride chloride during heating, GAZ CL 2 is released:
Na 2 O 2 + 4HCl → 2NACL + CL 2 + 2H 2 O
3) In an alkaline medium, chlorine reacts when heated with amphoteric chromium oxide in the formation of chromate and potassium chloride:
CR 2 O 3 + 3CL 2 + 10KOH → 2K 2 CRO 4 + 6KCL + 5H 2 O
2Cr +3 -6E → 2Cr +6 | . 3 - oxidation
Cl 2 + 2E → 2CL - | . 1 - Restoration
4) The sediment of yellow (Bacro 4) is formed when the interaction of potassium chromat and barium chloride:
K 2 CRO 4 + BACL 2 → BACRO 4 ↓ + 2KCL
Task number 4.
Zinc was fully soluble in a concentrated solution of potassium hydroxide. The resulting transparent solution was evaporated, and then calcined. The solid residue was dissolved in the required amount of hydrochloric acid. Ammonium sulfide was added to the resulting transparent solution and the formation of white sediment was observed. Write the equations of four described reactions.
1) zinc reacts with potassium hydroxide with potassium tetrahydroxide formation (Al and BE behave similarly:
2) Potassium tetrahydroxycinat after calcination loses water and turns into potassium cincat:
3) cycling of potassium when interacting with hydrochloric acid forms zinc chloride, potassium chloride and water:
4) zinc chloride as a result of interaction with ammonium sulphide turns into an insoluble zinc sulfide - precipitate white color:
Task number 5.
Iodium hydrogen acid neutralized potassium bicarbonate. The resulting salt reacted with a solution containing potassium dichromate and sulfuric acid. When interacting formed simple substance With aluminum got salt. This salt was dissolved in water and mixed with a solution of potassium sulfide, as a result of which the precipitate was formed and gas was separated. Write the equations of the four reactions described.
1) iodium hydrogen acid is neutralized with an acidic salt of weak coalic acid, resulting in carbon dioxide and NaCl is formed:
Hi + KHCO 3 → Ki + CO 2 + H 2 O
2) Potassium iodide enters a redox reaction with potassium dichromate in an acidic medium, while CR +6 is reduced to CR +3, I - oxidizes to molecular i 2, which falls into a precipitate:
6ki + K 2 CR 2 O 7 + 7H 2 SO 4 → CR 2 (SO 4) 3 + 4K 2 SO 4 + 3i 2 ↓ + 7h 2 o
2Cr +6 + 6e → 2Cr +3 │ 1
2i - -2E → I 2 │ 3
3) In the interaction of molecular iodine with aluminum, aluminum iodide is formed:
2Al + 3i 2 → 2Ali 3
4) When the aluminum iodide interacts with the potassium sulfide solution, Al (OH) 3 is precipitated and H 2 S is released. The formation of Al 2 S 3 does not occur due to the total hydrolysis of salt in aqueous solution:
2ALI 3 + 3K 2 S + 6H 2 O → 2AL (OH) 3 ↓ + 6ki + 3H 2 S
Task number 6.
Aluminum carbide was completely dissolved in bromomic acid. A solution of potassium sulfite was added to the resulting solution, while the formation of white precipitate and the separation of colorless gas was observed. The gas was absorbed by a solution of potassium dichromate in the presence of sulfuric acid. The chromium salt was isolated and added to the solution of barium nitrate, sediment was observed. Write the equations of the four reactions described.
1) When dissolving aluminum carbide in bromomic acid, salt is formed - aluminum bromide, and methane is distinguished:
Al 4 C 3 + 12HBR → 4Albr 3 + 3CH 4
2) when the aluminum bromide interacts with the potassium sulfite solution, Al (OH) 3 is precipitated and sulfur - SO 2 is released:
2Albr 3 + 3K 2 SO 3 + 3H 2 O → 2AL (OH) 3 ↓ + 6kbr + 3SO 2
3) skipping sulfur gas through an acidified solution of potassium dichromate, while CR +6 is restored to CR +3, S +4 oxidizes to S +6:
3SO 2 + k 2 Cr 2 O 7 + H 2 SO 4 → CR 2 (SO 4) 3 + K 2 SO 4 + H 2 O
2Cr +6 + 6e → 2Cr +3 │ 1
S +4 -2E → S +6 │ 3
4) with the interaction of chromium sulfate (III) with a nitrate solution of the barium, chromium nitrate (III) is formed, and white barium sulfate falls into the sediment:
CR 2 (SO 4) 3 + 3BA (NO 3) 2 → 3Baso 4 ↓ + 2Cr (NO 3) 3
Task number 7.
Sodium hydroxide solution added aluminum powder. Through the solution of the resulting substance, excess carbon dioxide was missed. The precipitate was separated and calcined. The resulting product was splashing with sodium carbonate. Write the equations of four described reactions.
1) aluminum, as well as beryllium and zinc, is able to react both with an aqueous solutions of alkalis and with anhydrous alkalis when fusing. When treatment with an aqueous solution of sodium hydroxide, sodium tetrahydroxyaluminate and hydrogen are formed:
2) When carbon dioxide passes through an aqueous solution of sodium tetrahydrokoxaluminate, crystalline aluminum hydroxide falls. Since, by a condition through a solution, an excess of carbon dioxide is passed, a carbonate is formed, and sodium bicarbonate:
Na + CO 2 → AL (OH) 3 ↓ + NaHCO 3
3) Aluminum hydroxide is an insoluble metal hydroxide, therefore, when heated is decomposed on the corresponding metal oxide and water:
4) Aluminum oxide, which is an amphoteric oxide, when fusing with carbonates, displaces carbon dioxide with the formation of aluminates (not to be confused with tetrahydroxyalulums!):
Task number 8.
Aluminum reacted with sodium hydroxide solution. The separated gas was missed over the heated powder of copper oxide (II). The resulting substance was dissolved when heated in concentrated sulfuric acid. The resulting salt was isolated and added to a solution of potassium iodide. Write the equations of the four reactions described.
1) Aluminum (also beryllium and zinc) reacts with both aqueous solutions of alkalis and anhydrous alkalis when fusing. When treatment with an aqueous solution of sodium hydroxide, sodium tetrahydroxyaluminate and hydrogen are formed:
2NAOH + 2AL + 6H 2 O → 2NA + 3H 2
2) when hydrogen passing over the heated powder of copper oxide (II) Cu +2 is restored to Cu 0: The color of the powder changes from black (Cuo) to red (CU):
3) Copper dissolves in concentrated sulfuric acid to form copper sulfate (II). In addition, the sulfur dioxide is distinguished:
4) when the copper sulfate is added to the potassium iodide solution flows the redox reaction: Cu +2 is restored to Cu +1, i - oxidizes to i 2 (the molecular iodine falls into a precipitate):
CUSO 4 + 4KI → 2CUI + 2K 2 SO 4 + i 2 ↓
Task number 9.
Sodium chloride solution conducted electrolysis. The iron (III) chloride was added to the resulting solution. The precipitate was filtered and calcined. The solid residue was dissolved in hydrochloric acid. Write the equations of the four reactions described.
1) Electrolysis Sodium chloride solution:
Cathode: 2H 2 O + 2E → H 2 + 2OH -
ADD: 2CL - - 2E → CL 2
Thus, gas-shaped H 2 and CL 2 is released from the sodium chloride solution as a result of its electrolysis, and Na + ions remain in the solution. IN general The equation is written as follows:
2H 2 O + 2NACL → H 2 + 2NAOH + CL 2
2) when alkali (III) alkali (III) is added to the alkali solution, an exchange reaction occurs, as a result of which Fe (OH) 3 falls into the precipitate:
3NAOH + FECL 3 → Fe (OH) 3 ↓ + 3NACL
3) When calcinating iron hydroxide (III), iron (III) oxide and water are formed:
4) When dissolving iron (III) oxide in iodomic acid, FEI 2 is formed, while I 2 falls precipitated:
Fe 2 O 3 + 6HI → 2FEI 2 + i 2 ↓ + 3H 2 O
2fe +3 + 2e → 2fe +2 │1
2i - - 2E → I 2 │1
Task number 10.
Potassium chlorate was heard in the presence of a catalyst, while the colorless gas was separated. Iron Okle's Iron Burning in the atmosphere of this gas was obtained. It was dissolved in excess of hydrochloric acid. A solution containing sodium dichromate and hydrochloric acid was added to the solution obtained.
1) When the potassium chlorate is heated in the presence of a catalyst (MNO 2, FE 2 O 3, Cuo, etc.), potassium chloride is formed and oxygen is released:
2) When burning iron in an oxygen atmosphere, iron scale is formed, which FE 3 O 4 formula (Iron Okalina is a mixed oxide Fe 2 O 3 and Feo):
3) When the iron scale is dissolved in an excess of hydrochloric acid, a mixture of iron (II) chlorides and (III) is formed:
4) In the presence of a strong oxidizing agent - sodium dichromate Fe +2 is oxidized to Fe +3:
6FeCl 2 + Na 2 Cr 2 O 7 + 14hCl → 6FeCl 3 + 2CrCl 3 + 2NACL + 7H 2 O
Fe +2 - 1e → Fe +3 │6
2Cr +6 + 6e → 2Cr +3 │1
Task number 11.
Ammonia was missed through bromomic acid. A solution of silver nitrate was added to the resulting solution. The precipitate was separated and heated with zinc powder. The metal formed during the reaction was carried out by a concentrated solution of sulfuric acid, while gas was separated with a sharp odor. Write the equations of the four reactions described.
1) When ammonia passes through bromomic acid, ammonium bromide (neutralization reaction) is formed:
NH 3 + HBr → NH 4 BR
2) When merging the solutions of ammonium bromide and silver nitrate proceeds the exchange of the exchange between the two salts, as a result of which a light yellow sediment falls - silver bromide:
NH 4 BR + AGNO 3 → AGBR ↓ + NH 4 NO 3
3) When the silver bromide is heated with zinc powder, the reaction of substitution proceeds - silver is highlighted:
2AGBR + Zn → 2Ag + ZnBr 2
4) Under action on a metal concentrated sulfuric acid, silver sulfate is formed and gas with an unpleasant odor - sulfur dioxide is released:
2AG + 2H 2 SO 4 (conc.) → AG 2 SO 4 + SO 2 + 2H 2 O
2AG 0 - 2E → 2Ag + │1
S +6 + 2e → s +4 │1
Task number 12.
9c278s
Chromium oxide (VI) reacted with potassium hydroxide. The resulting substance was treated with sulfuric acid, an orange-colored salt was isolated from the resulting solution. This salt was treated with bromomic acid. The resulting simple substance entered the reaction with hydrogen sulfide. Write the equations of the four reactions described.
1) Cro 3 chromium oxide (VI) is acidic oxide, therefore, it interacts with alkali with salt formation - potassium chromat:
CRO 3 + 2KOH → K 2 CRO 4 + H 2 O
2) Potassium chromat in an acidic medium is converted without changing the degree of chromium oxidation into Bichromat K 2 CR 2 O 7 - orange saline:
2K 2 CRO 4 + H 2 SO 4 → K 2 Cr 2 O 7 + K 2 SO 4 + H 2 O
3) in the processing of potassium bichromate with bromometroic acid CR +6 is reduced to CR +3, while the molecular bromine is released:
K 2 CR 2 O 7 + 14HBR → 2CrBr 3 + 2kbr + 3Br 2 + 7H 2 O
2Cr +6 + 6e → 2Cr +3 │1
2BR - - 2E → BR 2 │3
4) Bromine as a stronger oxidizer displaces sulfur from its hydrogen compound:
Br 2 + H 2 S → 2HBR + S ↓
Task number 13.
Magnesium powder heated in a nitrogen atmosphere. In the interaction of the resulting substance with water, gas was separated. The gas was missed through an aqueous solution of chromium sulfate (III), as a result of which a gray sediment was formed. The precipitate was separated and processed when heated by a solution containing hydrogen peroxide and potassium hydroxide. Write the equations of the four reactions described.
1) In the heating of magnesium powder in the nitrogen atmosphere, magnesium nitride is formed:
2) Magnesium nitride is completely hydrolyzed to form magnesium and ammonia hydroxide:
Mg 3 N 2 + 6H 2 O → 3MG (OH) 2 ↓ + 2NH 3
3) Ammonia has basic properties due to the presence of a nitrogen atom of a vapor electronic pair at the nitrogen atom and an exchange reaction with chromium sulfate (III), as a result of which the gray sediment is distinguished - CR (OH) 3:
6NH 3. H 2 O + CR 2 (SO 4) 3 → 2CR (OH) 3 ↓ + 3 (NH 4) 2 SO 4
4) hydrogen peroxide in an alkaline medium oxidizes CR +3 to CR +6, as a result of which potassium chromat is formed:
2CR (OH) 3 + 3H 2 O 2 + 4KOH → 2K 2 CRO 4 + 8H 2 O
CR +3 -3E → CR +6 │2
2O - + 2E → 2O -2 │3
Task number 14.
In the interaction of aluminum oxide with nitric acid, salt was formed. Salt was dried and cried. The solid residue formed during the calcination was subjected to electrolysis in the molten cryolite. The metal obtained under the electrolysis is heated with a concentrated solution containing potassium nitrate and potassium hydroxide, while gas with a sharp smell. Write the equations of the four reactions described.
1) in the interaction of amphoteric Al 2 O 3 with nitric acid, salt is formed - aluminum nitrate (exchange reaction):
Al 2 O 3 + 6HNO 3 → 2AL (NO 3) 3 + 3H 2 O
2) When calcining aluminum nitrate is formed aluminum oxide, as well as nitrogen dioxide and oxygen (aluminum belongs to a metal group (in a row of activity from alkaline-land to Cu inclusive), whose nitrates are decomposed to metal oxides, NO 2 and O 2):
3) Metal aluminum is formed with Al 2 O 3 electrolysis in the molten cryolate Na 2 ALF 6 at 960-970 o C.
Al 2 O 3 electrolysis circuit:
The dissociation of aluminum oxide proceeds in the melt:
Al 2 O 3 → AL 3+ + ALO 3 3-
K (-): Al 3+ + 3e → Al 0
A (+): 4alo 3 3- - 12E → 2AL 2 O 3 + 3O 2
Total equation Process:
Liquid aluminum is assembled at the bottom of the electrolyzer.
4) When the alumina is treated with a concentrated alkaline solution containing potassium nitrate, ammonia is distinguished, and the tetrahydroxyaluminate of potassium (alkaline medium) is formed:
8Al + 5KOH + 3KNO 3 + 18H 2 O → 3NH 3 + 8K
Al 0 - 3e → Al +3 │8
N +5 + 8e → N -3 │3
Task number 15.
8AA8S
A certain amount of iron sulfide (II) was divided into two parts. One of them was treated with hydrochloric acid, and the other was subjected to firing in the air. In the interaction of the excreted gases, a simple substance was formed. The resulting substance was heated with concentrated nitric acid, and brown gas was separated. Write the equations of the four reactions described.
1) When processing iron (II) sulfide, hydrochloride is formed by iron (II) chloride and hydrogen sulfide (exchange reaction) is released:
FES + 2HCL → FECL 2 + H 2 S
2) when firing iron (II) sulfide (II), iron is oxidized to the degree of oxidation +3 (FE 2 O 3 is formed) and sulfur dioxide is released:
3) When the interaction of two sulfur-containing compounds SO 2 and H 2 S flows the redox reaction (comparideration), as a result of which sulfur stands out:
2H 2 S + SO 2 → 3S ↓ + 2H 2 O
S -2 - 2E → S 0 │2
S +4 + 4e → s 0 │1
4) when sulfur heating with concentrated nitric acid is formed sulfuric acid and nitrogen dioxide (redox reaction):
S + 6HNO 3 (conc.) → H 2 SO 4 + 6NO 2 + 2H 2 O
S 0 - 6E → S +6 │1
N +5 + E → N +4 │6
Task number 16.
The gas obtained during the treatment of calcium nitride with water was missed above the chopped copper oxide (II) powder. The solid obtained was dissolved in concentrated nitric acid, the solution was evaporated, and the resulting solid residue was calcined. Make the equation of the four reactions described.
1) calcium nitride reacts with water, forming alkali and ammonia:
Ca 3 N 2 + 6H 2 O → 3CA (OH) 2 + 2NH 3
2) Passing ammonia over a hot copper (II) oxide powder, copper in oxide is restored to the metal, and nitrogen is released (hydrogen, coal, carbon monoxide, etc. are also used as reducing agents.
Cu +2 + 2e → Cu 0 │3
2N -3 - 6E → N 2 0 │1
3) Copper, located in a number of metal activities after hydrogen, interacts with concentrated nitric acid to form copper nitrate and nitrogen dioxide:
CU + 4HNO 3 (conc.) → Cu (NO 3) 2 + 2NO 2 + 2H 2 O
Cu 0 - 2e → Cu +2 │1
N +5 + E → N +4 │2
4) When calculating copper nitrate forms copper oxide, as well as nitrogen dioxide and oxygen (copper belongs to a metal group (in a row of activity from alkaline-land to Cu inclusive), whose nitrates are decomposed to oxide metals, NO 2 and O 2):
Task number 17.
Silicon burned in the chlorine atmosphere. The resulting chloride was treated with water. The sediment distinguished at the same time. Then sparkled with calcium and coal phosphate. Make the equation of the four reactions described.
1) The reaction of the interaction of silicon and chlorine occurs at a temperature of 340-420 o C in the argon current to form silicon chloride (IV):
2) silicon chloride (IV) is completely hydrolyzed, at the same time hydrochloric acid is formed, and silicic acid falls precipitated:
SICL 4 + 3H 2 O → H 2 SiO 3 ↓ + 4HCl
3) When calcining, silicic acid decomposes to silicon oxide (IV) and water:
4) When fusing silicon dioxide with coal and phosphate, a redox reaction occurs, as a result of which calcium silicate, phosphorus are formed, and carbon monoxide is highlighted:
C 0 - 2E → C +2 │10
4p +5 + 20e → P 4 0 │1
Task number 18.
Note! Such a format of tasks is outdated, but nevertheless, the tasks of this species deserve attention, since in fact they need to write down the same equations that are found in the Cyah ege of the new format.
Substances are given: iron, iron scale, diluted hydrochloric and concentrated nitric acid. Write the equations of four possible reactions between all the proposed substances, without repeating the pairs of reagents.
1) hydrochloric acid reacts with iron, oxidizing it to the degree of oxidation +2, while hydrogen is released (reaction of substitution):
Fe + 2HCl → FECL 2 + H 2
2) Concentrated nitric acid passivates iron (i.e., a solid protective oxide film is formed on its surface), but under the influence high temperatures Iron is oxidized by concentrated nitric acid to the degree of oxidation +3:
3) The iron scale formula is FE 3 O 4 (a mixture of iron oxides Feo and Fe 2 O 3). Fe 3 O 4 enters the exchange reaction with hydrochloric acid, while a mixture of two chlorides of iron (II) and (III) are formed:
Fe 3 O 4 + 8HCl → 2FeCl 3 + FECL 2 + 4H 2 O
4) In addition, the iron scale comes into a redox reaction with concentrated nitric acid, with Fe +2 contained in it oxidizes to Fe +3:
Fe 3 O 4 + 10hno 3 (conc.) → 3Fe (NO 3) 3 + NO 2 + 5H 2 O
5) Iron Okalina and iron when they are sintering to react to the Corporation (the oxidizing agent and the reducing agent is the same chemical element):
Task №19.
Substances are given: phosphorus, chlorine, aqueous solutions of sulfuric acid and potassium hydroxide. Write the equations of four possible reactions between all the proposed substances, without repeating the pairs of reagents.
1) Chlorine - poisonous gas with high chemical activity, is particularly vigorously reacting with red phosphorus. In the atmosphere of chlorine phosphorus self-turn and burns a weak greenish flame. Depending on the ratio of the reacting substances, phosphorus (III) chloride or phosphorus chloride (V) may be obtained:
2p (Krasn.) + 3Cl 2 → 2PCL 3
2P (Krasn.) + 5Cl 2 → 2PCL 5
CL 2 + 2KOH → KCL + KCLO + H 2 O
If chlorine is passed through a hot concentrated alkali solution, the molecular chlorine is disproportionated on CL +5 and CL -1, as a result of which chlorate and chloride are formed, respectively:
3) as a result of interaction aqueous solutions Alkali and sulfuric acid are formed by an acidic or average sulfuric acid salt (depending on the concentration of reagents):
KOH + H 2 SO 4 → KHSO 4 + H 2 O
2KOH + H 2 SO 4 → K 2 SO 4 + 2H 2 O (neutralization reaction)
4) Strong oxidants such as sulfuric acid turn phosphorus into phosphoric acid:
2p + 5h 2 SO 4 → 2H 3 PO 4 + 5SO 2 + 2H 2 O
Task number 20.
Substances are given: nitrogen oxide (IV), copper, potassium hydroxide solution and concentrated sulfuric acid. Write the equations of four possible reactions between all the proposed substances, without repeating the pairs of reagents.
1) Copper, located in a number of metals' activity, the right of hydrogen is capable of oxidizing with strong oxidizing acids (H 2 SO 4 (conc.), HNO 3, etc.):
Cu + 2H 2 SO 4 (conc.) → CUSO 4 + SO 2 + 2H 2 O
2) As a result of the interaction of the KOH solution with concentrated sulfuric acid, acidic salt - potassium hydrosulfate are formed:
KOH + H 2 SO 4 (conc.) → KHSO 4 + H 2 O
3) by passing the brown gas NO 2 N +4 disproportionates on n +5 and n +3, as a result of which nitrate and potassium nitrite are formed, respectively:
2NO 2 + 2KOH → KNO 3 + KNO 2 + H 2 O
4) when the brown gas pass through the concentrated solution of sulfuric acid N +4 is oxidized to N +5 and sulfur dioxide is distinguished:
2NO 2 + H 2 SO 4 (conc.) → 2HNO 3 + SO 2
Task number 27.
Substances are given: chlorine, sodium hydrosulfide, potassium hydroxide (solution), iron. Write the equations of four possible reactions between all the proposed substances, without repeating the pairs of reagents.
1) Chlorine, being a strong oxidizing agent, reacts with iron, oxidizing it to Fe +3:
2fe + 3Cl 2 → 2FeCl 3
2) when chlorine passes through a cold concentrated alkali solution, chloride and hypochlorite (molecular chlorine is disproportionate on CL +1 and CL -1):
2KOH + CL 2 → KCL + KCLO + H 2 O
If chlorine is passed through a hot concentrated alkali solution, the molecular chlorine is disproportionate on CL +5 and CL -1, as a result of which chlorate and chloride are formed, respectively:
3CL 2 + 6KOH → 5KCl + KCLO 3 + 3H 2 O
3) chlorine, which has stronger oxidative properties, is able to oxidize the sulfur-part in the acid salt:
Cl 2 + NaHS → NaCl + HCl + S ↓
4) sour salt - sodium hydrosulfide in an alkaline medium turns into sulfide:
2NAHS + 2KOH → K 2 S + Na 2 S + 2H 2 O
CUCL 2 + 4NH 3 \u003d CL 2
Na 2 + 4HCl \u003d 2NACL + CUCL 2 + 4H 2 O
2Cl + K 2 S \u003d Cu 2 S + 2KCl + 4NH 3
When mixing solutions, hydrolysis occurs and in the cation of a weak base, and by anion of weak acid:
2Cuso 4 + Na 2 SO 3 + 2H 2 O \u003d Cu 2 O + Na 2 SO 4 + 2H 2 SO 4
2Cuso 4 + 2NA 2 CO 3 + H 2 O \u003d (CuOH) 2 CO 3 ↓ + 2NA 2 SO 4 + CO 2
Copper and copper compounds.
1) through a solution of copper chloride (II) with graphite electrodes passed permanent electricity. The electrolysis product ranged on the cathode was dissolved in concentrated nitric acid. At the same time, the gas was collected and missed through a solution of sodium hydroxide. The electrolysis produced on the anode on the anode was missed through a hot sodium hydroxide solution. Write the equations of the described reactions.
2) The substance obtained at the cathode with the electrolysis of the melt of copper chloride (II) reacts with gray. The resulting product was treated with concentrated nitric acid, and the gas distinguished was passed through a hydroxide solution of barium. Write the equations of the described reactions.
3) Unknown salt is colorless and staining the flame in yellow. With the easiest heating of this salt with a concentrated sulfuric acid, the liquid is distilled off in which copper dissolves; The last transformation is accompanied by the separation of brown gas and the formation of copper salt. With the thermal decay of both salts, one of the decomposition products is oxygen. Write the equations of the described reactions.
4) in the interaction of the solution of salt A with an alkali, a studentist-insoluble in water was obtained in water, which was dissolved in a colorless liquid B to form a blue solution. The solid product remaining after the careful evaporation of the solution, was calcined; At the same time, two gas was separated, one of which was brown, and the second is part of atmospheric air, and a black solid remains, which dissolves in liquid B to form a substance A. Write the equations of the described reactions.
5) The copper chips was dissolved in dilute nitric acid, and the solution was neutralized by caviar potassium. The highlighted substance of the blue color was separated, calcined (the color of the substance has changed to black), mixed with coke and reinforced. Write the equations of the described reactions.
6) A copper chips was added to the mercury nitrate solution (II). After the reaction is completed, the solution was filtered, and the filtrate dropped dropwise to a solution containing the caustic soda and ammonium hydroxide. In this case, a short-term formation of a precipitate was observed, which was dissolved with the formation of a bright blue solution. When an excess solution of sulfuric acid was added to the resulting solution, a change in color occurred. Write the equations of the described reactions.
7) Copper (I) oxide was treated with concentrated nitric acid, the solution was carefully evaporated and the solid residue was calcined. Gaseous reaction products were missed through a large amount of water and a magnesium chips was added to the resulting solution, the gas used in medicine was separated. Write the equations of the described reactions.
8) The solid forming when the malachite is heated, heated in the atmosphere of hydrogen. The reaction product was treated with concentrated sulfuric acid, introduced into a solution of sodium chloride, containing copper sawdust, as a result, a precipitate was formed. Write the equations of the described reactions.
9) Salt obtained by dissolving copper in dilute nitric acid was subjected to electrolysis using graphite electrodes. The substance elected on the anode was introduced into interaction with sodium, and the resulting reaction product was placed in a carbon dioxide vessel. Write the equations of the described reactions.
10) The solid product of the thermal decomposition of the malachite was dissolved when heated in concentrated nitric acid. The solution was carefully evaporated, and the solid residue was calcined by obtaining a substance of black, which heated in an excess of ammonia (gas). Write the equations of the described reactions.
11) A solution of diluted sulfuric acid was added to the black powder substance and heated. A solution of caustic natra was poured into the resulting solution of blue, until the sediment is discontinued. The precipitate was filtered and heated. The reaction product was heated in a hydrogen atmosphere, resulting in a red substance. Write the equations of the described reactions.
12) The unknown substance of red is heated in chlorine, and the product of the reaction was dissolved in water. The resulting solution was added alkali, the precipitated blue color was filtered and calcined. When the product is heated, which has a black color, the source of red substance was obtained with coke. Write the equations of the described reactions.
13) The solution obtained by reacting copper with concentrated nitric acid was evaporated and the precipitate was calcined. Gaseous products are completely absorbed by water, and hydrogen missed over the solid residue. Write the equations of the described reactions.
14) Black powder, which was formed when burning a metal of red in an excess of air, was dissolved in 10%-ofsert acid. The resulting solution was added alkali, and the precipitated blue color was separated and dissolved in an excess of ammonia solution. Write the equations of the described reactions.
15) The substance of the black color was obtained by calcining the precipitate, which is formed by the interaction of sodium hydroxide and copper sulfate (II). When this substance is heated with coal, a metal is obtained, which dissolves in concentrated sulfuric acid. Write the equations of the described reactions.
16) Metal copper was treated with iodine heating. The resulting product was dissolved in concentrated sulfuric acid during heating. The resulting solution was treated with potassium hydroxide solution. The precipitated precipitate was hampered. Write the equations of the described reactions.
17) The immochloride of copper (II) was added an excess solution of soda. The precipitate was hampered, and the resulting product was heated in a hydrogen atmosphere. The resulting powder was dissolved in dilute nitric acid. Write the equations of the described reactions.
18) Copper was dissolved in dilute nitric acid. An excess ammonia solution was added to the resulting solution, observing the formation of the precipitate first, and then its complete dissolution to the formation of a dark blue solution. The resulting solution was treated with sulfuric acid until the characteristic blue painting of copper salts appears. Write the equations of the described reactions.
19) Copper was dissolved in concentrated nitric acid. An excess ammonia solution was added to the resulting solution, observing the formation of the precipitate first, and then its complete dissolution to the formation of a dark blue solution. The resulting solution was treated with an excess of hydrochloric acid. Write the equations of the described reactions.
20) The gas obtained by the interaction of iron sawdust with a solution of hydrochloric acid was missed over the heated oxide of copper (II) to the complete reduction of the metal. The resulting metal was dissolved in concentrated nitric acid. The resulting solution was subjected to electrolysis with inert electrodes. Write the equations of the described reactions.
21) iodine was placed in a tube with concentrated hot nitric acid. The separated gas was missed through water in the presence of oxygen. The resulting solution was added copper hydroxide (II). The resulting solution was evaporated and the dry solid residue was calcined. Write the equations of the described reactions.
22) Orange copper oxide was placed in concentrated sulfuric acid and heated. An excess solution of potassium hydroxide was poured to the resulting blue solution. The bleached blue precipitate was filtered, dried and cried. Solid black substance In the glass tube, heated and missed ammonia over him. Write the equations of the described reactions.
23) The oxide of copper (II) was treated with a solution of sulfuric acid. In the electrolysis of the resulting solution on an inert anode, gas is released. The gas was mixed with nitrogen oxide (IV) and absorbed with water. Magnesium was added to the dilute solution of the resulting acid, as a result of which two salts were formed in the solution, and the release of the gaseous product did not occur. Write the equations of the described reactions.
24) copper (II) oxide is heated in carbon monoxide. The resulting substance was burned in the atmosphere of chlorine. The product of the reaction was dissolved in water. The resulting solution was divided into two parts. To one part, a solution of potassium iodide was added to the second - a solution of silver nitrate. And in that, and in another case, the formation of a precipitate was observed. Write the equations of the described reactions.
25) copper (II) nitrate was calcined, the resulting solid was dissolved in dilute sulfuric acid. The solution of the resulting salt was subjected to electrolysis. The substance highlighted on the cathode was dissolved in concentrated nitric acid. The dissolution proceeds with the separation of brown gas. Write the equations of the described reactions.
26) Shanchaic acid heated with a small amount of concentrated sulfuric acid. The separated gas was missed through a solution of calcium hydroxide. In which the sediment fell. Part of the gas was not absorbed, it was missed over the solid substance of the black color, obtained by calcining the copper nitrate (II). As a result, a dark red solid was formed. Write the equations of the described reactions.
27) Concentrated sulfuric acid reacted with copper. The released at gas was completely absorbed by the excess solution of potassium hydroxide. The copper oxidation product was mixed with the calculated amount of sodium hydroxide until the precipitate is stopped. The latter was dissolved in excess of hydrochloric acid. Write the equations of the described reactions.
Copper. Copper compounds.
1. CUCL 2 CU + SL 2
on the cathode on the anode
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
6NAOH (mountains) + 3Cl 2 \u003d NaClo 3 + 5NACL + 3H 2 O
2. CUCL 2 CU + SL 2
on the cathode on the anode
CUS + 8HNO 3 (conc. Mountains.) \u003d Cuso 4 + 8NO 2 + 4H 2 O
or cus + 10hno 3 (conc.) \u003d Cu (NO 3) 2 + H 2 SO 4 + 8NO 2 + 4H 2 O
4NO 2 + 2BA (OH) 2 \u003d Ba (NO 3) 2 + Ba (NO 2) 2 + 2H 2 O
3. Nano 3 (TV.) + H 2 SO 4 (conc.) \u003d HNO 3 + NaHSO 4
CU + 4HNO 3 (conc.) \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
2Nano 3 2Nano 2 + O 2
4. Cu (NO 3) 2 + 2NAOH \u003d Cu (OH) 2 ↓ + 2Nano 3
Cu (OH) 2 + 2HNO 3 \u003d Cu (NO 3) 2 + 2H 2 O
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
Cuo + 2hno 3 \u003d Cu (NO 3) 2 + H 2 O
5. 3CU + 8HNO 3 (RSC) \u003d 3CU (NO 3) 2 + 2NO + 4H 2 O
Cu (NO 3) 2 + 2Koh \u003d Cu (OH) 2 ↓ + 2Kno 3
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
Cuo + C CU + CO
6. Hg (NO 3) 2 + Cu \u003d Cu (NO 3) 2 + HG
Cu (NO 3) 2 + 2NAOH \u003d Cu (OH) 2 ↓ + 2Nano 3
(OH) 2 + 5H 2 SO 4 \u003d Cuso 4 + 4NH 4 HSO 4 + 2H 2 O
7. Cu 2 O + 6HNO 3 (conc.) \u003d 2CU (NO 3) 2 + 2NO 2 + 3H 2 O
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
4NO 2 + O 2 + 2H 2 O \u003d 4HNO 3
10hno 3 + 4mg \u003d 4mg (NO 3) 2 + N 2 O + 5H 2 O
8. (CuOH) 2 CO 3 2CUO + CO 2 + H 2 O
Cuo + H 2 Cu + H 2 O
CUSO 4 + CU + 2NACL \u003d 2CUCL ↓ + Na 2 SO 4
9. 3CU + 8HNO 3 (RSC) \u003d 3CU (NO 3) 2 + 2NO + 4H 2 O
on the cathode on the anode
2NA + O 2 \u003d Na 2 O 2
2NA 2 O 2 + CO 2 \u003d 2NA 2 CO 3 + O 2
10. (CuOH) 2 CO 3 2CUO + CO 2 + H 2 O
Cuo + 2hno 3 Cu (NO 3) 2 + H 2 O
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
11. Cuo + H 2 SO 4 CUSO 4 + H 2 O
CUSO 4 + 2NAOH \u003d CU (OH) 2 + Na 2 SO 4
Cu (OH) 2 Cuo + H 2 O
Cuo + H 2 Cu + H 2 O
12. CU + CL 2 CUCL 2
CUCL 2 + 2NAOH \u003d CU (OH) 2 ↓ + 2NACL
Cu (OH) 2 Cuo + H 2 O
Cuo + C CU + CO
13. Cu + 4hno 3 (conc.) \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
4NO 2 + O 2 + 2H 2 O \u003d 4HNO 3
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
Cuo + H 2 Cu + H 2 O
14. 2CU + O 2 \u003d 2CUO
CUSO 4 + NaOH \u003d Cu (OH) 2 ↓ + Na 2 SO 4
Cu (OH) 2 + 4 (NH 3 · H 2 O) \u003d (OH) 2 + 4H 2 O
15. CUSO 4 + 2NAOH \u003d CU (OH) 2 + Na 2 SO 4
Cu (OH) 2 Cuo + H 2 O
Cuo + C CU + CO
Cu + 2H 2 SO 4 (conc.) \u003d Cuso 4 + SO 2 + 2H 2 O
16) 2CU + i 2 \u003d 2CUI
2CUI + 4H 2 SO 4 2cuso 4 + i 2 + 2SO 2 + 4H 2 O
Cu (OH) 2 Cuo + H 2 O
17) 2CUCl 2 + 2NA 2 CO 3 + H 2 O \u003d (CuOH) 2 CO 3 + CO 2 + 4NACL
(CuOH) 2 CO 3 2CUO + CO 2 + H 2 O
Cuo + H 2 Cu + H 2 O
3CU + 8HNO 3 (RSC) \u003d 3CU (NO 3) 2 + 2NO + 4H 2 O
18) 3CU + 8HNO 3 (RSC) \u003d 3CU (NO 3) 2 + 2NO + 4H 2 O
(OH) 2 + 3H 2 SO 4 \u003d Cuso 4 + 2 (NH 4) 2 SO 4 + 2H 2 O
19) CU + 4HNO 3 (conc.) \u003d Cu (NO 3) 2 + 2NO + 2H 2 O
Cu (NO 3) 2 + 2NH 3 · H 2 O \u003d Cu (OH) 2 ↓ + 2NH 4 NO 3
Cu (OH) 2 + 4NH 3 · H 2 O \u003d (OH) 2 + 4H 2 O
(OH) 2 + 6HCl \u003d CUCL 2 + 4NH 4 CL + 2H 2 O
20) Fe + 2HCl \u003d FECL 2 + H 2
Cuo + H 2 \u003d Cu + H 2 O
CU + 4HNO 3 (conc.) \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
2CU (NO 3) 2 + 2H 2 O 2CU + O 2 + 4HNO 3
21) i 2 + 10hno 3 \u003d 2HIO 3 + 10NO 2 + 4H 2 O
4NO 2 + 2H 2 O + O 2 \u003d 4HNO 3
Cu (OH) 2 + 2HNO 3 Cu (NO 3) 2 + 2H 2 O
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
22) Cu 2 O + 3H 2 SO 4 \u003d 2Cuso 4 + SO 2 + 3H 2 O
CUSO 4 + 2KOH \u003d CU (OH) 2 + K 2 SO 4
Cu (OH) 2 Cuo + H 2 O
3CUO + 2NH 3 3CU + N 2 + 3H 2 O
23) Cuo + H 2 SO 4 \u003d Cuso 4 + H 2 O
4NO 2 + O 2 + 2H 2 O \u003d 4HNO 3
10hno 3 + 4mg \u003d 4mg (NO 3) 2 + NH 4 NO 3 + 3H 2 O
24) CUO + CO CU + CO 2
Cu + Cl 2 \u003d CUCL 2
2cucl 2 + 2ki \u003d 2cucl ↓ + i 2 + 2KCL
CUCL 2 + 2AGNO 3 \u003d 2AGCL ↓ + CU (NO 3) 2
25) 2CU (NO 3) 2 2CUO + 4NO 2 + O 2
Cuo + H 2 SO 4 \u003d Cuso 4 + H 2 O
2CUSO 4 + 2H 2 O 2CU + O 2 + 2H 2 SO 4
CU + 4HNO 3 (conc.) \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
26) H 2 C 2 O 4 CO + CO 2 + H 2 O
CO 2 + CA (OH) 2 \u003d Caco 3 + H 2 O
2CU (NO 3) 2 2CUO + 4NO 2 + O 2
CUO + CO CU + CO 2
27) Cu + 2H 2 SO 4 (conc.) \u003d Cuso 4 + SO 2 + 2H 2 O
SO 2 + 2KOH \u003d K 2 SO 3 + H 2 O
CUSO 4 + 2NAOH \u003d CU (OH) 2 + Na 2 SO 4
Cu (OH) 2 + 2HCl CUCL 2 + 2H 2 O
Manganese. Connections manganese.
I. MARGANAN.
On the air, the manganese is covered with an oxide film that protects it even when heated from further oxidation, but in a small-cereble state (powder) it is oxidized quite easily. Manganese interacts with gray, halogens, nitrogen, phosphorus, carbon, silicon, boron, forming compounds with a degree of +2:
3mn + 2p \u003d Mn 3 P 2
3mn + N 2 \u003d Mn 3 n 2
Mn + Cl 2 \u003d MnCl 2
2mn + Si \u003d Mn 2 Si
When interacting with oxygen, the manganese forms manganese oxide (IV):
Mn + O 2 \u003d MNO 2
4mn + 3O 2 \u003d 2mn 2 O 3
2mn + O 2 \u003d 2mno
When heated, the manganese interacts with water:
Mn + 2H 2 O (pairs) Mn (OH) 2 + H 2
In the electrochemical row of stresses, the manganese is located to hydrogen, so it is easily dissolved in acids, forming a soldier of manganese (II):
Mn + H 2 SO 4 \u003d MNSO 4 + H 2
Mn + 2HCl \u003d MnCl 2 + H 2
With concentrated sulfuric acid, the manganese reacts when heated:
Mn + 2H 2 SO 4 (conc.) MNSO 4 + SO 2 + 2H 2 O
With nitric acid under normal conditions:
Mn + 4hno 3 (conc.) \u003d Mn (NO 3) 2 + 2NO 2 + 2H 2 O
3mn + 8hno 3 (spz ..) \u003d 3mn (NO 3) 2 + 2NO + 4H 2 O
Alkali solutions for manganese practically do not act, but it reacts with alkaline oxidizing melts, forming manganats (VI)
Mn + KCLO 3 + 2KOH K 2 MNO 4 + KCL + H 2 O
Manganese can restore oxides of many metals.
3mn + Fe 2 O 3 \u003d 3mno + 2fe
5mn + Nb 2 O 5 \u003d 5mno + 2NB
II. Manganese compounds (II, IV, VII)
1) Oxides.
The manganese forms a number of oxides, the acid-basic properties of which depend on the degree of oxidation of manganese.
MN. +2 O Mn. +4 O 2 Mn 2 +7 O 7.
basic amphoteric acid
Manganese Oxide (II)
Oxide of manganese (II) is obtained by restoring other oxides of manganese hydrogen or carbon oxide (II):
MNO 2 + H 2 MNO + H 2 O
MNO 2 + CO MNO + CO 2
The main properties of manganese oxide (II) are manifested in their interaction with acids and acid oxide:
MNO + 2HCl \u003d MnCl 2 + H 2 O
Mno + SiO 2 \u003d MNSIO 3
Mno + N 2 O 5 \u003d Mn (NO 3) 2
Mno + H 2 \u003d Mn + H 2 O
3mno + 2Al \u003d 2mn + Al 2 O 3
2mno + O 2 \u003d 2mno 2
3mno + 2KCLO 3 + 6KOH \u003d 3K 2 MNO 4 + 2KCL + 3H 2 O
- Wipers and waves, laws and formulas
- How the brain deceives our senses
- Secrets of the reaction of metal sodium with water
- Nominal parts of speech, their general features
- I do not die the soul in the cherished lyre
- During or during the right?
- Franco Agamenone live score, schedule and results - Tennis Agamemenon, King of the leader of Ahetsev in the Trojan War