量变引起质变（Quantitative change causes qualitative change）

量变引起质变（Quantitative change causes qualitative change） 量变引起质变(Quantitative change causes qualitative change) Interactions of two substances, because on the one hand, the amount of the change and cause the qualitative change of the product, or as a result of drug to join a different order and lead to different results, or as the reaction progresses, the concentration of the acid changed, product is changed, the above examples are common in the middle school chemistry, frequently appear in the college entrance examination. In order to better grasp the characteristics of this kind of reaction, we will according to the double decomposition reaction of chemical reaction, REDOX reaction, complexation reaction, hydrolysis reaction, substitution reaction, addition reaction type 59 quantitative change causes qualitative classified into 11 types of chemical reaction, the points above are as follows: 1. REDOX - REDOX type 1. The reaction of sodium (small amount, excessive) with oxygen 2Na + O2 Na2O2 + 2Na = 2Na2O 2. The reaction of carbon (small amount, excessive) with oxygen C + O2 CO2 + C2CO 2C + O2 2CO Note: carbon monoxide is produced when oxygen is inadequate and can cause gas poisoning. 3. The reaction of carbon and oxygen (small amount, excessive) 2C + O2 2CO + O2 2CO2 C + O2 2CO2 4. The reaction of silicon dioxide and coke (small amount, excessive) SiO2 + 2C Si + 2CO arrow Si + C SiC SiO2 + 3C 2CO Note: when making the coarse silicon, control the amount of coke, otherwise the coarse silicon will not be obtained, and the silicon carbide will be obtained. 5. Reaction of red phosphorus and chlorine gas (small amount, excessive) 2 p + 3 Cl2 2 PCl3 PCl3 + Cl2 = PCl5 Note: the experiment saw the phenomenon: white smoke, the composition of smoke PCl5, the composition of fog is PCl3. Hydrogen sulphide reacts with oxygen (a small amount of excess) 2H2S + O2 2S + 2H2O + O2SO2 2H2S + 3O2 2 + 2H2O Note: H2S is toxic, and the combustion product SO2 is also toxic, which cannot be used to treat H2S exhaust. 7. Hydrogen sulfide (small amount, excessive) reaction with oxygen 2H2S + 3O2 2SO2 + 2H2O SO2 + 2H2S = 3S + 2H2O 2H2S + 2H2O The reaction of sulfur (small amount, excessive) with concentrated solution 3S + 6NaOH (thick) 2Na2S + Na2SO3 + 3H2O 9. Reaction of potassium sulfide solution with chlorine gas (small amount, excessive) K2S + KCL Cl2 = 2 + S left Cl2 S + 3 + 4 + 6 h2o = - H2SO4 HCL K2S + 4 Cl2 + 4 h2o + 6 = - H2SO4 HCL + 2 KCL 10. Reaction of potassium iodide solution with chlorine gas (small amount, excessive) 2 ki + KCL Cl2 = 2 + I2 Cl2 I2 + 5 + 6 h2o + 10 = 2 hio3 HCL Cl2 ki + 6 + 6 h2o = 2 hio3 + 10 HCL + 2 KCL 11. The reaction of iron (small amount, excessive) with dilute nitric acid Fe + 4HNO3 (sparse) = Fe (NO3) 3 + NO arrow + 2H2O 2Fe (NO3) 3 + Fe = 3Fe (NO3) 2 3Fe + 8HNO3 (sparse) = 3Fe (NO3) 2 + 2NO up Copper reacts with nitric acid (thicker and thinner) Cu + 4HNO3 (thick) = Cu (NO3) 2 + 2NO2 arrow + 2H2O 3Cu + 8HNO3 (sparse) = 3Cu (NO3) 2 + 2NO up 13. Zinc and sulfuric acid (strong dilute) reaction Zn + 2H2SO4 (thick) = ZnSO4 + SO2 arrow + 2H2O Zn + H2SO4 (thin) = ZnSO4 + H2 arrow Note: it is often used to calculate the properties of sulfuric acid. 14. The reaction of ferrous bromide solution with chlorine gas (small amount, excessive) 6 febr2 + 3 cl2 = 2 fecl3 2 febr3 + 3 + 4 febr3 fecl3 cl2 = 2 + 3 br2 febr2 2 + 3 cl2 fecl3 br2 + 2 = 2 Note: this reaction is a hot spot for the writing and quantitative determination of ion equations. 15. Reaction of iodized ferrous solution with chlorine gas (small amount, excessive) FeI2 + Cl2 = 2 FeCl2 FeCl2 + I2 + Cl2 = 2 fecl3 FeI2 2 + 3 Cl2 fecl3 I2 + 2 = 2 16. The reaction of iodized ferrous solution with bromine (small amount, excessive) FeI2 + Br2 = 2 FeBr2 FeBr2 + I2 + Br2 = 2 febr3 FeI2 2 + 3 Br2 febr3 I2 + 2 = 2 2. Oxidation-reduction - complex reaction type The reaction of potassium sulfide and iodine (small, overdose) K2S + I2 = 2 left KI KI + S + I2 = KI3 18. Sulfur elemental (a small amount, excessive) reaction with concentrated sodium hydroxide solution 3S + 6NaOH (thick) 2Na2S + Na2SO3 + 3H2O Na2S + (x - 1) S = Na2Sx 3. Oxidation-reduction - type of complex decomposition 19. Reaction of chlorine water and sodium sulfite (small amount, excessive) solution Na2SO3 + Cl2 + H2O = 2 nacl + - H2SO4 - H2SO4 + Na2SO3 = Na2SO4 + H2O + SO2 write 20. Reaction of bromine and sodium sulfite (small amount, excessive) solution Na2SO3 + Br2 nabr + + H2O = 2 - H2SO4 - H2SO4 + Na2SO3 = Na2SO4 + H2O + SO2 write 21. The reaction of iodine water and sodium sulfite (small amount, excessive) solution Na2SO3 nai + + I2 + H2O = 2 - H2SO4 - H2SO4 + Na2SO3 = Na2SO4 + H2O + SO2 write 22. The reaction of sodium sulfide (small amount, excessive) solution with iron chloride solution Nacl fecl3 Na2S + 2 = 2 + 2 FeCl2 + S left Na2S + nacl FeCl2 = 2 + FeS left 23, the reaction of iron hydroxide and hydrogen iodate (small amount, excessive) 2 Fe (OH) 3 = 2 + 2 hi Fe (OH) 2 + I2 + 2 h2o Fe (OH) 2 + 2 hi = FeI2 + 2 h2o 4. Coagulation - - complex decomposition response type The reaction of iron hydroxide colloids and hydrochloric acid (small amount, excessive) Fe (OH) 3 (colloid) + 3 drops HCl, reddish-brown precipitate Fe (OH) 3 + 3HCl = FeCl3 + 3 h2o 25. Reaction of silica gel with sodium hydroxide (small amount, excessive) solution Silica gel + 3 drops NaOH, white precipitate SiO2 ? nH2O + 2NaOH = Na2SiO3 + (n + 1) H2O 5. Complex decomposition - the type of complex decomposition reaction 26. In aluminum chloride solution, drop into sodium hydroxide (a small amount of excess) solution AlCl3 NaOH + 3 = Al (OH) 3 left + 3 nacl Al (OH) 3 + 2 h2o + NaOH = NaAlO2 AlCl3 naoh + 4 = NaAlO2 + 3 nacl + 2 h2o Note: the method of sodium hydroxide in aluminum chloride solution is not used to prepare aluminum hydroxide by means of controlling the dosage of sodium hydroxide. 27. Add hydrochloric acid (small amount, excessive) to sodium dialuminate solution NaAlO2 + H2O + HCl = Al (OH) 3 + NaCl Al (OH) 3 + 3 HCl = AlCl3 + 3 H2O NaAlO2 HCL + 4 = AlCl3 + NaCl + 2 h2o Note: the method of adding hydrochloric acid to sodium dialuminate solution is not prepared by using this method to prepare aluminum hydroxide. 28. Carbon dioxide (small amount, excessive) in sodium aluminate solution 2 naalo2 + 3 H2O + CO2 = 2 al (OH) 3 left + Na2CO3 Na2CO3 nahco3 + H2O + CO2 = 2 NaAlO2 + 2 h2o + CO2 = Al (OH) 3 left + NaHCO3 Note: this method can be used to prepare aluminum hydroxide. Add barium hydroxide (a small amount of overdose) to the solution of alum 2 kal (SO4) 2 + 3 ba (OH) 2 = 2 + 2 al (OH) 3 left + K2SO4 baso4 left 2 al (OH) 3 left + K2SO4 + Ba (OH) 2 = BaSO4 left kalo2 + 2 + 4 h2o KAl (SO4) 2 + 2 ba (OH) 2 = 2 baso4 left + KAlO2 + 2 h2o Note: this reaction is more examined in the calculation of ion equation writing and combining images. 30. Clarification of carbon dioxide (small amount, excessive) in lime water Ca (OH) 2 + CO2 = CaCO3 + H2O, CaCO3 + H2O + CO2 = Ca (HCO3) 2 Ca (OH) 2 + 2 CO2 = Ca (HCO3) 2 1. Add dry ice (small amount, excessive) in aqueous solution of bleaching powder Ca (ClO) 2 + H2O + CO2 = CaCO3 + 2HClO CaCO3 + H2O + CO2 = Ca (HCO3) 2 Ca (ClO) 2 + 2 h2o + 2 = co2 Ca (HCO3) 2 + 2 hclo Note 1: the phenomenon of calcium hypochlorite solution is bubbled into carbon dioxide and clarify the phenomenon of limewater zhongtong into carbon dioxide, but for the product of the light, can appear different results, hypochlorous acid decomposition of light repeatedly out oxygen itself into hydrochloric acid at the same time. Hydrochloric acid can dissolve calcium carbonate, calcium carbonate, and emit carbon dioxide. Note 2: the illumination of the product after the solution of calcium hypochlorite to the carbon dioxide is the hot spot of the exam. 32. Drop in the solution of sodium carbonate into hydrochloric acid (small amount, excessive) Na2CO3 + HCl = NaHCO3 + NaCl NaHCO3 + HCl = NaCl + H2O + co 2 + 2 H2O + CO2 Note: the mixture of sodium bicarbonate and sodium bicarbonate drops into hydrochloric acid, first the sodium bicarbonate reacts with sodium bicarbonate, then the sodium bicarbonate reacts with hydrochloric acid. 33. Drop the sodium bicarbonate solution in the solution of barium hydroxide (small amount, excessive) Ba (OH) 2 + NaHCO3 = BaCO3 + NaOH + H2O Ba (OH) 2 + 2 = nahco3 BaCO3 left + Na2CO3 + 2 h2o 34. Sodium hydroxide solution in sodium bicarbonate solution (small amount, excessive) 2NaHCO3 + Ba (OH) 2 = BaCO3 + Na2CO3 + 2H2O Na2CO3 + Ba (OH) 2 = BaCO3 + 2 naoh NaHCO3 + Ba (OH) 2 = = BaCO3 left + NaOH + H2O In a barium hydroxide solution, a solution of ammonium bicarbonate (small amount, excess) Ba (OH) 2 + NH4HCO3 = BaCO3 + NH3, H2O + H2O Ba (OH) 2 + 2 nh4hco3 = (NH4) 2 co3 + BaCO3 left + 2 h2o 36. A barium hydroxide solution in a solution of ammonium bicarbonate (small amount, excessive) 2nh4 hco3 + Ba (OH) 2 = (NH4) 2CO3 + BaCO3 + 2 H2O (NH4) 2CO3 + Ba (OH) 2 = BaCO3 + 2 NH3 ? H2O NH4HCO3 + Ba (OH) 2 = BaCO3 37. Calcium hydroxide solution drops in sodium hydroxide solution (small amount, excessive) Ca (HCO3) 2 + NaOH = CaCO3 + NaHCO3 + H2O NaHCO3 + NaOH Ca (HCO3) 2 + 2 naoh = CaCO3 left + Na2CO3 + 2 h2o 38, sodium hydroxide solution (small amount, excessive) 2NaOH + Ca (HCO3) 2 = CaCO3 + Na2CO3 + 2H2O Na2CO3 + Ca (HCO3) 2 = CaCO3 + 2NaHCO3 NaOH + Ca (HCO3) 2 = CaCO3 NaHCO3 left + + H2O 39. Pour in the magnesium carbonate solution to clarify the lime water (small amount, excessive) Mg (HCO3) 2 + Ca (OH) 2 = MgCO3 + CaCO3 + 2 h2o MgCO3 + Ca (OH) 2 = Mg (OH) 2 + CaCO3 (HCO3) 2 + 2 Mg ca (OH) 2 + 2 = Mg (OH) 2 left caco3 left + 2 h2o 40. The solution of sodium hydroxide in phosphate solution (small amount, excessive) H3PO4 + NaOH = NaH2PO4 + H2O + NaOH NaH2PO4 = Na2HPO4 + H2O + NaOH Na2HPO4 = Na3PO4 + H2O H3PO4 naoh + 2 = Na2HPO4 + 2 h2o + 3 naoh H3PO4 = Na3PO4 + 3 h2o 1. Water droplets in sodium hydroxide solution (small amount, excessive) 3NaOH + H3PO4 + 3H2O 2Na3PO4 + H3PO4 + H3PO4 + H3PO4 + H3PO4 = 2NaH2PO4 2NaOH + H3PO4 = na2h2o NaOH + H3PO4 = NaH2PO4 + H2O 2. Drop in the phosphate solution to clarify the lime water (small amount, excessive) 2 h3po4 + Ca (OH) 2 = Ca 2 + 2 h2o (H2PO4) Ca (H2PO4) Ca (OH) 2 + 2 = 2 cahpo4 left + 2 h2o 2CaHPO4 + Ca (OH) 2 = Ca3 (PO4) 2 + 2H2O H3PO4 + Ca (OH) 2 = CaHPO4 2 h3po4 + 3 ca (OH) 2 = Ca3 (PO4) 2 left + 6 h2o Note: the water in the phosphate solution is used to clarify the lime water. When < 0.5, there is no precipitation, the product is Ca (H2PO4) 2 When < < 1, there is precipitation, the product is Ca (H2PO4) 2 and CaHPO4 When = 1, there is precipitation, and the product is CaHPO4 When < < 1.5, there is precipitation, and the product is CaHPO4 And Ca3 (PO4) 2 When it's 1.5, it's precipitated and the product is Ca3 (PO4) 2 Clarifying the water droplets into the phosphate solution (small, excessive) 3Ca (OH) 2 + 2H3PO4 = Ca3 (PO4) 2 + 6H2O Ca3 (PO4) 2 + H3PO4 = 3CaHPO4 CaHPO4 + H3PO4 = Ca (H2PO4) 2 Ca (OH) 2 + H3PO4 = CaHPO4 Ca (OH) 2 + 2 h3po4 = Ca 2 + 2 h2o (H2PO4) Note: to clarify that a small amount of phosphoric acid can be deposited in the water of lime water, and the precipitate will occur as the droplets of phosphoric acid drop into the water. The method can be used to identify phosphoric acid and clarify the lime water. 4. Sodium hydroxide solution in hydrogen sulfate solution (small amount, excessive) H2S + NaOH = NaHS + H2O NaHS + NaOH = Na2S + H2O + H2O + 2H2O 45. Sodium hydroxide solution into hydrogen sulphide gas (small amount, excessive) 2NaOH + H2S = Na2S + 2H2O NaOH + H2S = NaHS Clarify the amount of sulfur dioxide (small, excess) in the lime-water. Ca (OH) 2 + SO2 = CaSO3 + H2O CaSO3 + H2O + SO2 = Ca (HSO3) 2 Ca (OH) 2 + 2SO2 = = Ca (HSO3) 2 Note: both carbon dioxide and sulfur dioxide can make the lime water cloudy and the gas is too much to be clarified. Therefore, it is not possible to say that it is carbon dioxide that makes the limewater cloudy. 1. Drop in the solution of sulfuric acid to clarify the lime water (small amount, excessive) 2H2SO3 + Ca (OH) 2 = Ca (HSO3) 2 + 2H2O Ca (HSO3) 2 + Ca (OH) 2 = 2 caso3 + 2 h2o 48. Sodium hydroxide solution (small amount, excessive) in the solution of sodium hydroxide. Ba (OH) 2 + NaHSO4 = BaSO4 Ba (OH) 2 + 2 nahso4 = BaSO4 left + Na2SO4 + H2O 49. Sodium hydroxide solution in sodium hydroxide solution (small amount, excessive) 2NaHSO4 + Ba (OH) 2 = BaSO4 + Na2SO4 + H2O Na2SO4 + Ba (OH) 2 = BaSO4 + 2NaOH NaHSO4 + Ba (OH) 2 = BaSO4 left + NaOH + H2O Note: the test of this equation is the hot spot of the college entrance exam. The sequence of drops and solutions is different, resulting in different amounts and different products. 50 sodium sulfite solution drips into hydrochloric acid (small amount, excessive) Na2SO3 + HCl = NaHSO3 + NaCl NaHSO3 + HCl = NaCl + H2O + SO2 write Nacl Na2SO3 HCL + 2 = 2 + H2O + SO2 write Vi. Complex decomposition - double hydrolytic type 51. The solution of sodium hydroxide by drop into aluminum chloride solution (small amount, excessive) 4 naoh + AlCl3 = 3 + 3 NaAlO2 NaAlO2 + 2 h2o + nacl AlCl3 + 6 h2o + 3 = 4 al (OH) 3 left nacl 3 naoh + AlCl3 = Al (OH) 3 left + 3 nacl Note: in sodium hydroxide solution, a small amount of aluminum chloride solution will not produce precipitation, and the solution of sodium hydroxide can be used to prepare aluminum hydroxide precipitation. 52. The solution of hydrochloric acid is dilate into sodium aluminate solution (small amount, excessive). 4HCl + NaAlO2 = NaCl + AlCl3 + 2H2O AlCl3 + 3NaAlO2 + 6H2O = 4Al (OH) 3 + 3NaCl HCl + NaAlO2 + H2O = Al (OH) 3 left + NaCl Note: a small amount of sodium aluminate in hydrochloric acid does not produce precipitation, and the method of sodium hydrochloric acid drop in sodium hydrochloric acid can also prepare aluminum hydroxide precipitation. Vii. Complex decomposition - complex type 3. In silver nitrate solution, the water is introduced into ammonia (small amount, excessive). AgNO3 + NH3 ? H2O = AgOH + NH4NO3 AgOH + 2NH3 ? H2O = [Ag (NH3) AgNO3 + 3 NH3 ? H2O = [Ag (NH3) 2] OH + NH4NO3 + 2 H2O Note: when the preparation of silver ammonia solution, excessive ammonia solution dissolved precipitate. In addition, the solution of silver ammonia should be used now, and the time is long. It will be dangerous to convert to AgN3. If there is residual, neutralize it with nitric acid and eliminate the danger. 54. The solution of copper sulphate drops into ammonia (small amount, excessive) CuSO4 + 2NH3 ? H2O = Cu (OH) 2 + (NH4) 2SO4 Cu (OH) 2 + 4NH3 H2O = [Cu (NH3) 4] (OH) 2 + 4 H2O CuSO4 + 6 NH3 ? H2O = [Cu (NH3) 4 (OH) 2 + (NH4) 2 so4 + 4 H2O Replace the type 55, methane and chlorine gas (a small amount, too much) to replace the reaction CH4 + Cl2, CH3Cl + HCl + Cl2, CH2Cl2 + HCl 9. Addition - - addition type The reaction of acetylene to bromine (a small amount of excess) HC ? CH + Br2 - CHBr = CHBr CHBr = CHBr + Br2 - CHBr2 - CHBr2 HC ? CH + 2 br2 - CHBr2 - CHBr2 10. Hydrolysis - complexation 57. The reaction of boron trifluoride (small amount, excessive) and water 3 h2o + BF3 = H3BO3 + 3 HF HF + BF3 = H [BF4] 59. The reaction of silicon tetrafluoride (small amount, excessive) and water 3 h2o + SiF4 = H2SiO3 + 4 hf hf + SiF4 = 2 H2 [SiF6] 11. Decompose - decarboxylation 59. Reaction of acetic acid with sodium hydroxide (small amount, excessive) CH3COOH + NaOH = CH3COONa + H2O CH3COONa + NaOHNa2CO3 + CH4