"Integration of this Chapter" Elements and Material World PPT

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"Integration of this Chapter" Elements and Material World PPT

Part One: Breakthrough 1

Judgment that a large number of ions coexist

Typical example 1 A group of ions that can coexist in large amounts in both strongly acidic and strongly alkaline solutions is ()

A.Na+, Cu2+, Cl-, SO_4^2"-" B.K+, Ca2+, NO_3^"-", CO_3^2"-"

C.Mg2+, NH_4^+, SO_4^2"-", Na+ D.Na+, K+, Cl-, SO_4^2"-"

Answer:D

Summary of methods 1. Types of ions that cannot coexist in large amounts

(1) Metathesis reaction type

① Ions that generate poorly soluble or slightly soluble substances cannot coexist in large amounts.

② The ions that generate gases and volatile substances cannot coexist in large amounts.

③Generate difficult-to-ionize substances (water, weak acid, ammonia monohydrate, etc.).

(2) Oxidation-reduction type

Ions with strong oxidizing properties and ions with strong reducing properties will undergo redox reactions and cannot coexist in large amounts.

2. "Six major points of attention" when judging the coexistence of ions

(1) Pay attention to judge the coexistence of ions, "can", "cannot", "certainly", "possibly" and "must not".

(2) Pay attention to the acidity and alkalinity of the solution mentioned in the question. The presence of H+ should be considered for acidic solutions and the presence of OH- for alkaline solutions.

(3) Pay attention to the color of the solution indicated in the question. When the solution is colorless, colored ions cannot exist in large quantities.

(4) Pay attention to the correct understanding of "transparent solution" and do not think that "colored" is opaque. CuSO4 solution is a blue transparent solution.

(5) Note that oxidizing ions and reducing ions will undergo redox reactions and cannot coexist in large amounts. For example, ClO- is oxidizing in acidic, alkaline, and neutral solutions, and it interacts with reducing ions Fe2+, S2-, and HS- , I-, SO_3^2"-", HSO_3^"-", etc. cannot coexist in large quantities. Another example is the "NO_3^"-" +H+" combination, which has strong oxidizing properties and can undergo oxidation-reduction reactions with Fe2+, S2-, HS-, I-, SO_3^2"-", HSO_3^"-", etc. but cannot be used in large quantities. coexist.

Point training 1-1 In a colorless and transparent acidic solution, the ion group that can coexist in large amounts is ()

A.Mg2+, Ag+, NO_3^"-", Cl-

B.Cu2+, Na+, SO_4^2"-", Cl-

C.Ba2+, K+, NO_3^"-", Cl-

D.Zn2+, Na+, OH-, SO_4^2"-"

Answer:C

Analysis: Ag+ and Cl- in option A cannot coexist in large amounts, and the generated AgCl is insoluble; Cu2+ in option B appears blue in aqueous solution; OH- in option D cannot exist in large amounts in acidic solutions; each ion in option C can coexist and are colorless.

Point training 1-2 The following ions must be able to coexist in large amounts in the specified solution ()

A. In weakly acidic solution: Mg2+, SO_4^2"-", NH_4^+, Cl-

B. In a solution containing SO_4^2"-": K+, Ag+, CO_3^2"-", Br-

C. In the solution containing NO_3^"-": I-, SO_4^2"-", H+, Cl-

D. In strong acidic solution: SO_3^2"-", K+, Cl-, Na+

Answer:A

This chapter integrates PPT, the second part of the content: Breakthrough 2

Judgment and writing of ionic equations

Typical example 2 Ionic equation Ba2++SO_4^2"-" BaSO4↓ may represent ()

①Reaction between soluble barium salt solution and soluble sulfate solution

②Reaction between barium hydroxide solution and soluble sulfate solution

③Reaction between dilute sulfuric acid and soluble barium salt solution

④Reaction of barium hydroxide solution and dilute sulfuric acid

A.①② B.①②④

C.②③ D.①③

Answer:D

Analysis: In ②, if Ba(OH)2 solution reacts with CuSO4 or MgSO4 solution, in addition to BaSO4 precipitation, there will also be Cu(OH)2 or Mg(OH)2 precipitation. In ④, in addition to BaSO4 precipitation, water is also generated, so there should also be water in the ionic equation.

Method rules 1. Steps for writing ionic equations (taking the reaction of CaCO3 and hydrochloric acid as an example)

2. The general idea of ​​judging whether the ionic equation is correct or incorrect

[Micro Tips] (1) Write the ion symbol in the ionic equation for concentrated hydrochloric acid and concentrated nitric acid, and write the chemical formula for concentrated sulfuric acid.

(2) When writing ionic equations, the acid salts of weak acids such as HCO_3^"-", HS-, HSO_3^"-" cannot be written separately, while HSO_4^"-" should be written separately.

(3) The four common slightly soluble substances are Ag2SO4, CaSO4, MgCO3, and Ca(OH)2.

3. Write the ionic equation of the reaction between acid salt and base using the "Determining 1 method"

(1) A small amount of substance is defined as "1 mol". If a small amount of substance has two or more ions participating in the reaction, the ratio of the amount of substances participating in the reaction ions is consistent with the ratio of the composition of the substance.

(2) Based on the amount of ion material in a small amount of material, determine the amount of ion material in the excess material that actually participates in the reaction.

(3) Written according to the idea of ​​"neutralize first and then precipitate".

For example, the ionic equation for adding excess NaOH solution to Ca(HCO3)2 solution is:

For point training 2, add dilute sulfuric acid dropwise to the Ba(OH)2 solution. Please complete the following questions:

(1) Write the ionic equation of the reaction: _________.

(2) In the following three cases, the ionic equation that is the same as (1) is ________(fill in the letter).

A. Add Ba(OH)2 solution drop by drop to NaHSO4 solution until the solution becomes neutral.

B. Add Ba(OH)2 solution dropwise to the NaHSO4 solution until _________ just completely precipitates.

C. To the NaHSO4 solution, add Ba(OH)2 solution dropwise to excess.

This chapter integrates PPT, the third part: Breakthrough Three

A method to quickly and accurately memorize concepts related to redox reactions

Typical Example 3 Industrially, the highly efficient drinking water treatment agent sodium ferrate can be prepared by the following method

(Na2FeO4):Fe2O3+3Na2O2 2Na2FeO4+Na2O,

Which of the following statements is correct ()

A. Fe2O3 is the reducing agent in the reaction

B.Fe2O3 can be produced by burning iron wire in oxygen

C.3 mol Na2O2 reaction, 3 mol electrons are transferred

D. Fe in Na2FeO4 has a valence of +4, has strong oxidizing properties, and can disinfect and sterilize

Answer:A

Analysis: The valence of Fe element increases from +3 valence to +6 valence. Fe2O3 is used as reducing agent in the reaction, item A is correct; iron wire burns in oxygen to generate Fe3O4, item B is wrong; 3 mol Na2O2 reaction, 6 mol electrons are transferred , Item C is wrong; Fe in Na2FeO4 has a valence of +6, has strong oxidizing properties, and can be disinfected. Item D is wrong.

Knowledge induction

Clarify the relationship between concepts related to redox reactions

Method rules 1. Associative memory keywords

(1) Memorize the formula: "When it rises, it loses oxygen, and when it falls, it regains oxygen." The meaning is: increased valence → loss of electrons (electron pair deviation) → oxidation → reducing agent → reduction (ability to lose electrons). The valence decreases → electrons are obtained (electron pair bias) → reduced → oxidizing agent → has oxidizing properties (electron-obtaining ability).

(2) Memory formula: "The dosage properties are the same, and the others are opposite." The meaning is that the oxidizing agent has oxidizing properties and the reducing agent has reducing properties. "Other opposites" mean that the oxidant is reduced, a reduction reaction occurs, and a reduction product is obtained; the reducing agent is oxidized, an oxidation reaction occurs, and an oxidation product is obtained.

2. Learn the correct analysis methods

Valence changes are the premise and basis for analyzing all redox reaction problems. Correctly marking the valence of each element is the key and breakthrough for analyzing redox reactions. The specific method is to find the price and determine the type (whether it is a redox reaction). Point up and down, and determine the others (the valence increases, electrons are lost, and an oxidation reaction occurs; the valency decreases, electrons are gained, and a reduction reaction occurs).

This chapter integrates PPT, the fourth part: Breakthrough Four

Laws of electron conservation and their applications

Typical example: 41.52 g of a certain iron oxide (FexO) is dissolved in a sufficient amount of hydrochloric acid, and 112 mL of Cl2 under standard conditions is passed into the resulting solution, which completely oxidizes the iron element into Fe3+. x value is ()

A.0.80 B.0.85 C.0.90 D.0.93

Answer:A

Analysis: According to the conservation of electrons gained and lost, the number of electrons lost when FexO is oxidized to Fe3+ is equal to the number of electrons gained by Cl2. Under standard conditions, 112 mL Cl2 obtains 0.01 mol of electrons, then:

Method Law Conservation Law - Gained and lost electrons are equal

That is, in the redox reaction, the total number of increases in the valence of an element is equal to the total number of decreases. The essence is that the total number of electrons lost and the total number of electrons gained in the reaction are equal.

Application: Balancing of redox reaction equations and related calculations.

By using the law of conservation, you can put aside the cumbersome reaction process, you don't need to write chemical equations, and you don't need to investigate the intermediate reaction process. You just need to divide the material into the initial state and the final state, and think holistically from the two aspects of electron gain and electron loss, and you can quickly Get correct results.

Point-to-Point Training 4 Under certain conditions, _____ and fluorine gas can react

2RO_3^n"-" +F2+4OH- 2RO_4^2"-" +2F-+2H2O, so it can be seen that the valence of element R in RO_3^n"-" is ()

A.+4 B.+5

C.+6 D.+7

Answer:B

Analysis: Suppose the valence of R in _____ is +x valence, then the valence of R before and after the reaction changes from +x valence → +6 valence, and the valence of F changes from 0 valence → -1 valence. According to electron conservation, we can know that 2×( +6-x)=2×[0-(-1)], that is, x=5.

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