Tan, No.1 Middle School of Hengdong County, Hunan Province, 42 1400

Nitrogen has five positive valences of+1, +2, +3, +4 and +5, and can form six oxides, such as N2O, NO, N2O3, NO2, N2O4 and N2O5. These oxides are gases except N2O5 at room temperature, in which N2O3 and N2O5 are acid anhydrides of nitrous acid and nitric acid respectively. Except NO2 is reddish brown gas, the others are colorless. From the view of chemical valence, all oxides of nitrogen are oxidizing, except N2O5 which is mainly reducing, and NO2 which is mainly oxidizing. Nitrogen oxides are all toxic air pollutants.

First, the important knowledge of NO and NO2

Two sides of 1 NO: ①NO is one of the main gases that cause air pollution; NO can also combine with hemoglobin, making hemoglobin lose its oxygen-carrying function, leading to human poisoning. ②NO has the functions of transmitting signals in human vascular system, dilating blood vessels, enhancing immunity and enhancing memory.

2.2 the formation and nature. Nitric oxide and nitrogen dioxide

① Nitrogen and oxygen combine directly under lightning conditions: N2+O2·2NO.

②NO is a colorless gas, insoluble in water, and easily combined with oxygen in the air at room temperature: 2NO+O2=2NO2, so NO cannot be collected by exhausting air.

(3) NO2 is a reddish-brown gas with pungent smell, which is not easy to be oxidized by oxygen, but easy to react with water and sodium hydroxide solution: 3NO2+H2O=2HNO3+NO, 2NO2+2NaOH=NaNO3+NaNO2+H2O. In this reaction, NO2 is both oxidant and reductant, and nitric acid can be produced and NO2 can be absorbed through two reactions.

(4) NO2 has strong oxidizability, which can oxidize reducing gases such as SO2, H2S and HI, making wet potassium iodide starch test paper blue.

⑤ Under normal circumstances, NO2 can self-combine: 2NO2 N2O4, so pure NO2 is not pure, and the volume of 1molNO2 collected under standard circumstances should be less than 22.4L.

⑥NO2 can react with H2O and O2: 4NO2+O2+2H2O = 4NO3.

NO can react with H2O and O2: 4NO+3O2+2H2O = 4NO3.

Second, the important test points of nitrogen oxides.

Nitrogen oxides in 1.STS knowledge

Example1:The harm of NO to the environment lies in: ① destroying the ozone layer; ② Oxidation of some metals at high temperature; ③ Causing acid rain; ④ Binding with human hemoglobin.

A.①③ B.③④ C.①②③ D.①③④

Analysis: The mechanism of NO destroying the ozone layer: O3+NO=O2+NO2, O3=O2+O, NO2+O=NO+O2.

Therefore, NO is a catalyst in the reaction of converting ozone into oxygen; Thunderstorms will produce a lot of NO2, which may form acid rain and corrode metals and buildings. NO can combine with hemoglobin, making hemoglobin lose its oxygen-carrying function. The correct answer is D.

2.2 Dissolution calculation. Nitrogen oxides and oxygen in water

Example 2: A 13mL test tube is filled with NO2 and O2, and inverted in a water tank with enough water. If the residual gas in the test tube is 1 ml after the reaction is completed, try to determine the volumes of NO2 and O2 in the original mixed gas.

Analysis: The last remaining gas may be O2 or NO. Let the volume of NO2 in the mixed gas be xmL.

If the residual gas is NO, the NO2 of reaction 3NO2+H2O=2HNO3+NO is 3mL, and the NO2 of reaction 4NO2+O2+2H2O=4HNO3 is (x-3) ml, so: (x-3) = 4 (13-x), and the solution is x = 65438+.

If the residual gas is O2, and the ratio of NO2 to O2 participating in the reaction is 4∶ 1, there is: x = 4 (13-x- 1), the solution is x=9.6, and O2 is 3.4mL at this time.

A: The volumes of NO2 and O2 in the original mixed gas are 1 1 mL, 2mL or 9.6 mL and 3.4mL respectively.

3. Inferring the nitrogen oxides in the problem

Example 3: After passing 3 volumes of NO2 gas through the following 3 devices respectively filled with ① saturated NaHCO3 solution ② concentrated H2SO4③Na2O2, the remaining gas is collected into a gas container by drainage method, and the gas in the gas container is (measured at the same temperature and pressure).

A. 1 volume NO B.2 volume NO2 and 0.5 volume O2

C.2 volume O2 D.0.25 volume O2

Analysis: NO2 reacts with saturated NaHCO3 solution: 3NO2+H2O=2HNO3+NO, NaHCO3+HNO3=NaNO3+H2O+CO2 to obtain 1 volume NO, 2 volumes of CO2 and a small amount of water vapor are absorbed by concentrated sulfuric acid and react with Na2O2: 2co2+Na2CO3+O2. 2NO+O2=2NO2, resulting in 1 volume NO2 and 0.5 volume O2. Finally, collecting and reacting by drainage method: 4NO2+O2+2H2O = 4NO3, and collecting 0.25 volume of O2 in a gas container. The correct answer is D.

Calculation of the reaction between nitric acid and metal by conservation method

The calculation of the reaction between nitric acid and metal is one of the hot topics in college entrance examination in recent years. Electron conservation method, atom conservation method and comprehensive conservation method are the main methods to solve this kind of problem. The following author analyzes this for reference.

I. Atomic conservation

Example 1: A certain amount of mixture of Fe and Fe2O3 is put into 25mL2mol/L nitric acid solution, and no solid remains after the reaction, resulting in 224mLNO gas (standard condition). Then, NaOH solution of 1mol/L is added to the reacted solution, and the volume of the added NaOH solution is at least ().

A.25 B. 30mL C. 40mL D.50ml.

Analysis: The reaction of Fe and Fe2O3 with nitric acid is complicated, and Fe(NO3)2 and Fe(NO3)3 may be generated, which cannot be solved by the conservation of iron. The final state analysis method can be used: after all the iron elements in the solution are precipitated, the solutes in the solution may be NaNO3 and NaOH. When the volume of NaOH solution is the least, only NaNO3 can be used, and the problem can be solved by using the conservation of sodium and nitrogen elements.

N(NaNO3)= n(Na+)=n(NaOH) (sodium conservation)

N (nano3) = n (NO3-) = n (HNO3)-n (NO) (nitrogen conservation)

So: n (NaOH) = n (HNO3)-n (NO)

v(AQ)=[0.025 L×2mol/L—0.224 L/22.4 L？ mol— 1]/ 1 L？ Molar-1= 0.04L = 40ml

Correct answer: C.

Second, the law of conservation of electrons

Example 2: A copper-silver alloy with a mass of 5 1.6g was dissolved in a sufficient amount of dilute HNO3, and 6.72L gas was collected under standard conditions. What is the mass of copper in the alloy?

Analysis: When dilute nitric acid reacts with copper and silver, NO:

3Cu+8HNO3==3Cu(NO3)2+2NO↑+4H2O

3Ag+4HNO3==3AgNO3+NO↑+2H2O

Suppose that Cu in the alloy is xg, Ag is (5. 16-x) g, and it is composed of Cu? —2e→Cu2+，Ag—e→Ag+； HNO3+3e→NO↑ It can be seen that 1molCu can lose 2mol electrons, 1molAg can lose 1mol electrons, and it takes 3mol electrons to release 1molNO, so according to the electron conservation:

2x/64+1x (5.16-x)/108 = 3× 6.72/22.4 solution: x =19.2g.

Third, the comprehensive application of conservation law.

Example 3: The mixture of 13.2mg and Cu was put into a sufficient amount of dilute nitric acid solution, and no solid remained after the reaction, resulting in 4.48LNO gas (standard condition). After the reaction, excessive KOH solution was added to the solution to precipitate all the metal ions, and the mass of the precipitate was ().

A.2 1.6g b . 23.4g c . 26.8g d . 3 1.9g

Analysis: The final precipitates are Mg(OH)2 and Cu(OH)2. If the mass of Mg and Cu is calculated by electron conservation method, the calculation amount is large, the process is complex and easy to make mistakes. Precipitation can be divided into two parts: metal cation and hydroxide ion, which are calculated by electron conservation and element conservation.

① According to the conservation of elements, the mass of metal cations = the mass of the mixture of Mg and Cu =13.2g. ..

② The number of electrons to be obtained when nitric acid is converted into NO: 4.48L/22.4L? Mol-1× 3 = 0.6 mol.

The number of electrons obtained by Mg and Cu is also 0.6mol, and the formed cation has a positive charge of 0.6mol. Combined with 0.6mol of hydroxide ion, the mass of 0.6mol of hydroxide ion is 0.6 mol× 17 g? Mol-1=10.2g. 。

So the precipitation quality =13.2g+10.2g = 23.4g.

Correct answer: B.

Ammonia knowledge storage

I. Diversity of components

There are two reversible changes in the process of ammonia dissolving in water:

NH3+H2O NH3？ H2O NH4 ++ Oh-

Therefore, ammonia water is a complex mixed system, with H2O, NH3, NH3? H2O； The existing ions are NH4++, OH- and a small amount of H+. And there is a relationship between the quantity and concentration of substances:

c(H2O)>c( NH3？ H2O)>c(NH3)>c(OH-)& gt； c(NH4+)& gt； Hydrogen ions.

Second, the diversity of chemical properties.

Ammonia water has many characteristics because it contains many components.

1. Instability: NH3? When heated, H2O easily decomposes and generates NH3 and H2O.

Therefore, the laboratory can adopt the method of heating concentrated ammonia water to prepare NH3, or mix concentrated ammonia water with solid alkali (or quicklime) at room temperature to quickly prepare NH3. Because of the instability of ammonia, ammonia should be sealed in a brown or dark bottle and placed in a cold and dark place.

2. Weak alkalinity: the nature of OH- expression. With C(OH-)> C(H+) in ammonia water, ammonia water is weakly alkaline, which has the common characteristics of alkali.

① It can make colorless phenolphthalein red, purple litmus test solution blue and wet red litmus test paper blue. The laboratory can use red litmus paper to test the presence of ammonia.

② It can react with acid to generate ammonium salt. When concentrated ammonia meets volatile acids (such as concentrated hydrochloric acid and nitric acid), it will produce white smoke: NH3+HCl==NH4Cl NH3+HNO3=NH4NO3. The laboratory can use this method to test the presence of ammonia and concentrated ammonia water with concentrated hydrochloric acid or hydrogen chloride and concentrated hydrochloric acid with concentrated ammonia water.

③ It can react with acid oxides. In industry, the weak alkalinity of ammonia water is used to absorb sulfuric acid industrial tail gas to prevent environmental pollution: SO2+2 NH3? H2O ==(NH4)2SO3 +H2O .

Because ammonia water is alkaline, it can't be dried with acidic desiccants such as concentrated sulfuric acid and P2O5, and alkaline lime is commonly used.

3. Sedimentation: the nature of OH-. OH- in ammonia water can react with various metal ions to form insoluble hydroxide, and ammonia water is a good precipitant. Such as: Al3++3NH3? H2O ==Al(OH)3↓+ 3NH4+

Because the generated Al(OH)3 precipitate is insoluble in excessive ammonia water, aluminum salt solution and ammonia water are commonly used to prepare pure Al(OH)3 in the laboratory.

4. Reducibility: NH3, NH3? The nature of H2O, in which nitrogen has the lowest price, so ammonia shows weak reducibility and can be oxidized by strong oxidant.

Such as ammonia and chlorine: 3Cl2+8NH3? H2O ==6NH4Cl + N2 + H2O

5. Oxidation: the nature of H2O and ionized H+, in which hydrogen has the highest valence state, so ammonia water is weak in oxidation and can react with strong reducing agent. Such as ammonia and sodium: 2na+2h2o = 2 NaOH+H2 =

6. Complexity: Characteristics of NH3. NH3 in ammonia water can form complex ions with metal ions of various transition elements. For example, prepare silver ammonia solution: Ag++2NH3 ==[Ag(NH3)2]+

Because ammonia can react with calcium chloride, ammonia can be absorbed by calcium chloride in large quantities, so it is not suitable to dry ammonia with anhydrous calcium chloride.

Third, the particularity of ammonia water

1. density: the density of ammonia water is less than that of water, and the greater the mass fraction of solute, the smaller the density.

2. Electrolyte: Ammonia can conduct electricity, but the electrolyte is NH3? H2O, not ammonia.

3. Quantity concentration of substance: When calculating the quantity concentration of ammonia, the solute is NH3, not NH3? H2O, but the amount of solute is the total amount of ammonia that begins to dissolve.

4. ionic equation: ① When ammonia water is the reactant, use the chemical formula NH3? H2O .

(2) When ammonia water is a product, it is generated by the reaction between concentrated solutions or under heating conditions, and is expressed by "NH3↑+H2O"; For the reaction between unheated dilute solutions, "NH3? H2O "means.

Iv. Analysis on the problem of ammonia water

Example 1. (National College Entrance Examination in 2000) A student's extracurricular activity group made the following experiments with the device shown on the right:

(1) Inject a red solution into the test tube, heat the test tube, and the color of the solution will gradually become lighter, and then turn red again after cooling, so the original solution may be a solution; The reason why the solution becomes lighter from red when heated is that.

(2) inject a colorless solution into the test tube, heat the test tube, the solution turns red, and after cooling, it returns to colorless, so this solution may be a solution; The reason why the solution turns from colorless to red when heated is.

Analysis: This experimental device is a closed system. The color of the stock solution changes when it is heated, which is caused by the volatilization of some substances in the gas. After cooling, the volatile gas is dissolved in the solution to obtain the stock solution. Phenolphthalein and magenta are the most common substances in middle schools that can make the solution change between red and colorless, so it is not difficult to analyze that NH3 is the basic gas that causes the discoloration of magenta solution, and SO2 is the reason for the discoloration of phenolphthalein solution.

Answer:

(1) NH3 in dilute ammonia water and phenolphthalein dilute ammonia water escapes, so the color of the solution becomes lighter.

⑵ The magenta SO2 gas dissolved with SO2 escapes, and the magenta solution returns to red.

Example 2. If ω 1 and ω2 respectively indicate that the concentration is one mole? L- 1 and b mol? L- 1 mass fraction of ammonia, and 2a = b, then the following inference is correct (the density of ammonia is less than that of pure water) ().

a 2ω 1 =ω2 B 2ω2 =ω 1 Cω2 > 2ω 1Dω 1 & lt； ω2 & lt； 2ω 1

Analysis: according to the conversion relationship between solute mass fraction and substance mass concentration, combined with the particularity of ammonia water, the higher the concentration, the lower the density, and the correct answer is C.

Example 3. The white solid can be composed of one or more of (NH4)2SO4, AgNO3, BaCl2 and NaOH. A small amount of this white solid is dissolved in water to obtain a clear solution, which is divided into two equal parts, A and B. In part A, the phenolphthalein test solution showed red color, while in part B, when KI solution was added, yellow precipitate was precipitated, but it was insoluble when dilute nitric acid was added. According to the above experimental phenomena, point out what substances exist and write the relevant ionic equation.

Analysis: The obvious breakthrough given by the title: ① The phenolphthalein test solution turns red, indicating that there are alkaline or alkaline substances in the sample, and there must be NaOH；; ② Adding KI produces yellow precipitate insoluble in dilute nitric acid, which indicates that there is AgNO3 in the sample. Are these the only two substances? Obviously, we should pay attention to another feature in the stem: dissolving into a clear solution, because if only AgNO3 and NaOH are dissolved in water, AgOH will precipitate, and then Ag2O will decompose, which is contradictory to the clear solution. Then why is there no precipitation? By comparing the possible substances and associating with the preparation of silver ammonia solution, it can be judged that there is (NH4)2SO4 in the solid, which reacts with NaOH to generate NH3? H2O combines with Ag+ to form silver ammonia solution. But (NH4)2SO4 can't contain BaCl2, otherwise BaSO4 will precipitate when the sample is dissolved.

A: There are (NH4)2SO4, AgNO3 and NaOH in the sample.

The ionic reaction is NH4++OH-= NH3? H2O

ag++ 2 NH3 =[Ag(NH3)2]+[Ag(NH3)2]++ I—= = AgI↓+2 NH3