高中化学讲义整理2（A high school chemistry handout）.doc

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1、高中化学讲义整理2（A high school chemistry handout）Summary of chemical oxygen equation in high school1. SO2 and water reaction2. SO2 + Ca (OH) 2 = CaSO3 + H2o3. SO2 and sufficient sodium hydroxide solution SO2 + 2NaOH = Na2SO3 + H2o4. SO2 SO2 into saturated baking soda solution SO2 + NaHCO3 = NaHSO3 + CO25.

2、Sulfur dioxide catalytic oxidation 2SO2 + O2 = 2SO3 (heating, catalyst)2. 2. 2. 2Concentrated sulfuric acid and copper reactionConcentrated sulfuric acid and carbon reaction C + 2H2SO4 = CO2High school chemistry magnesium, aluminum and iron chemical equation summary1. Calcium carbonate solution heat

3、 decomposition Ca (HCO3) 2 = CaCO32. Magnesium chloride and ammonia water MgCl2 + 2NH3 H2O = Mg (OH) 2 + 2NH4Cl3. Aluminum and hydroxide solution 2Al + 2NaOH + 2H2o = 2NaAlO2 + 3H2 upAluminum reaction 4Al + 3MnO2 = 3Mn + 2Al2O3 (high temperature)5. Alumina and hydrochloric acid Al2O3 + 6HCl = 2AlCl3

4、 + 3H2O6. Alumina and sodium hydroxide solution Al2O3 + 2NaOH = 2NaAlO2 + H2OAluminum hydroxide is divided into 2 al (OH) 3Aluminum hydroxide and hydrochloric acid Al (OH) 3 + 3 HCL = AlCl3 + 3 h2o9, aluminum hydroxide and sodium hydroxide solution Al (OH) 3 + NaOH = NaAlO2 + 2H2O10, aluminum chlori

5、de and excess ammonia. AlCl3 + 3NH3 + 3H2O = Al (OH) 3Aluminium chloride and an excess of sodium hydroxide AlCl3 + 4NaOH = NaAlO2 + 3NaCl + 2H2oIron and water vapor 3FeIron and sufficient amount of dilute hydrochloric acid Fe + 2HCl = FeCl2 + H2Fe + 4HNO3 = Fe (NO3) 3 + NO arrow + 2 h2o15. Iron (com

6、plete overdose) and a small amount of dilute nitric acidFe 3 + 8 hno3 = 3 fe (NO3) 2 + 2 no write + 4 h2oIron hydroxide is decomposed into two fe (OH) 3High chemical alkali metal chemical equation2Na + O2 = = Na2O2 (ignition) 2Na + S = = Na2S2Na + 2H2O = = 2NaOH + H2 + H2 + 2 + 2 co2 = = = 2Na2CO3 +

7、 O22Na2O2 + 2H2O = = 4NaOH + O2 = 2Li2O = 2Li2O (ignition)Na2CO3 + 2HCl = = 2NaCl + H2O + CO2NaHCO3 + HCl = NaCl + H2O + CO2 arrow2NaHCO3 = = Na2CO3 + H2O + CO2 arrow (heating)2Na + CuSO4 + 2H2SO4 = = Cu (OH) 2 + Na2SO4 + H2 up2NaHCO3 + Ca (OH) 2 = = Na2CO3 + CaCO3High school chemistry knowledge poi

8、nt summary: ion coexistence problem1. Due to the complex decomposition reaction, ions cannot coexist in large quantities.(1) there is gas. The acid roots of weak acids, such as CO32 -, SO32 -, S2 -, HCO3 -, HSO3 -, HS -, etc., cannot co-exist in large quantities.(2) precipitation generation. Such as

9、 Ba2 +, Ca2 +, Mg2 +, Ag + and so on cannot coexist with SO42 -, CO32 - and so on.The Mg2 +, Fe2 +, Ag +, Al3 +, Zn2 +, Cu2 +, Fe3 + cant coexist with OH - abundance; Pb2 + and Cl -, Fe2 + and S2 -, Ca2 + and PO43 -, Ag + and I - cannot coexist in large quantities.(3) a weak electrolyte is produced.

10、 Such as OH -, CH3COO -, PO43 -, HPO42 -, H2PO4 -, F -, ClO -, AlO2 -, SiO32 -, CN -, C17H35COO -, etc. Some acid weak acid roots such as HCO3 -, HPO42 -, HS -, H2PO4 -, HSO3 - cant coexist with OH - much; NH4 + and OH - cant do a lot of coexistence.(4) there are some ions that can be hydrolyzed, an

11、d there are conditions in the solution. Such as AlO2 -, S2 -, CO32 -, C6H5O - must be present in solution under alkaline conditions; Such as Fe3 +, Al3 +, etc. must be present in solution under acidic conditions. These two ions cant be in the same solution at the same time, which is the bihydrolysis

12、 reaction between ions. Such as 3 alo2 al3 + 3 + 4 + 6 h2o = al (OH) 3 left, etc.2. Due to oxidation-reduction reactions, ions cannot coexist in large quantities.(1) ions with strong reducing properties cannot coexist in large quantities with highly oxidized ions. If S2 -, HS -, SO32 -, I - and Fe3

13、+ cant coexist in large quantities.(2) in an acidic or alkaline medium, the REDOX reaction cannot co-exist in large quantities. Such as MnO4 -, Cr2O7 -, NO3 -, ClO - and S2 -, HS -, SO32 -, HSO3 -, I -, Fe2 +, etc. SO32 - and S2 - can coexist in alkaline conditions, but under acidic conditions, 2S2

14、- + SO32 - + 6H + = 3S + 3H2O reactions cannot be combined. H + and S2O32 - cant co-exist in large Numbers.A hydrolyzed cation can not co-exist in the aqueous solution in large quantities (bihydrolysis).Examples: Al3 + and HCO3 -, CO32 -, HS -, S2 -, AlO2 -, ClO -, etc. Fe3 + and CO32 -, HCO3 -, AlO

15、2 -, ClO - cant coexist in large Numbers.4. The ions in the solution can not co-exist in large quantities.Such as Fe2 +, Fe3 + and SCN - cannot coexist in large Numbers; Fe3 + does not co-exist in large quantities.5. The additional conditions given in the question should be noted in the examination.

16、Acid solution (H +), alkaline solution (OH -), the solution of gas after adding aluminum powder, H + or OH - = 1 x 10-10 mol/L solution of water and electricity.Colored ion MnO4 -, Fe3 +, Fe2 +, Cu2 +, Fe (SCN) 2 +. The MnO4 -, NO3 - is highly oxidized under acidic conditions.S2O32 - REDOX reaction

17、occurs under acidic conditions: S2O32 - + 2H + = SNote that they require a lot of coexistence or not a lot of coexistence.6. The following points should be paid special attention to:(1) note the effect of the acidity of the solution on the REDOX reaction between ions. For example, Fe2 + and NO3 - ca

18、n coexist, but in strong acid conditions (i.e., Fe2 +, NO3 -, H + encounter) cant coexist; MnO4 - and Cl - cant coexist in strong acid conditions; S2 - with SO32 - can coexist with sodium and potassium, but not in acidic conditions.(2) acid salt containing hydrogen weak acid root ions cannot coexist

19、 with strong base (OH -), strong acid (H +).If HCO3 - + OH - = CO32 - + H2O (HCO3 - is ionized further);HCO3 - + H + = write CO2 + H2OHigh school chemistry knowledge point summary: oxidation sex, reductive force weak judgement(1) according to the valence of the elementThe elements in the substance h

20、ave the highest price, which is only oxidizing; The element has the lowest price, and the element is only reductive; The element has an intermediate valence, which is both oxidizing and reductive. For the same element, the higher the valence state, the stronger the oxidation. The lower the valence s

21、tate, the stronger the reducibility.(2) according to the REDOX reaction equationIn the same oxidation-reduction reaction, oxidizing: oxidant oxidation productReductive: reductive product of The stronger the oxidative properties of oxidants, the weaker the reduction products of the corresponding redu

22、ction products; The more reductive the reductant, the weaker the oxidation of the corresponding oxidation products.(3) according to the ease of reactionNote: the strong and weak oxidation of the REDOX is only related to the difficulty of the electrons gain and loss, but not the number of electrons i

23、n the electron. The stronger the electronic power, the stronger the oxidation. The stronger the loss of electronic power, the stronger the reducibility.No REDOX reaction occurs between adjacent valence states of the same element.Common oxidants:The active non-metal, such as Cl2, Br2, O2, etc.The ele

24、ments (such as Mn, etc.) are in high valence oxide, such as MnO2, KMnO4, etc(S, N, etc.) the oxygen-containing acid in high valence, such as strong H2SO4, HNO3, etcThe salt, such as KMnO4, KClO3, FeCl3 and K2Cr2O7, are in high valence when the elements (such as Mn, Cl, Fe, etc.) are in high valenceThe peroxide, such as Na2O2, H2O2, etc