"End of Chapter Review Lesson" Iron Metal Materials PPT

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"End of Chapter Review Lesson" Iron Metal Materials PPT

Special topic summary and practice

Application of conservation method in chemical calculations

Conservation relationship characteristics

Conservation of mass Macro: conservation of mass of matter before and after reaction

Microscopically: conservation of the number of atoms of an element before and after the reaction

Conservation of electrons. In redox reactions, electrons are gained and lost.

Conservation of charge: In an electrolyte solution, the total number of positive charges carried by cations is equal to the total number of negative charges carried by anions.

In the ionic equation, the total number of charges carried by the reactants is equal to the total number of charges carried by the products, and the electrical properties are the same

Thematic sparring

1. In the mixed solution of sodium sulfate and aluminum sulfate, the concentration of Al3+ is 0.2 mol·L-1, SO2-4 is 0.4 mol·L-1, and the concentration of Na+ in the solution is ()

A. 0.1 mol·L－1 B. 0.2 mol·L－1

C． 0.3 mol·L－1 D. 0.4 mol·L－1

2. A mol of carbon monoxide and b mol of oxygen are introduced into a closed container with a volume of V L. After the full reaction is ignited, the ratio of the number of carbon atoms to oxygen atoms in the container is ()

A. a∶b B. a:2b

C． a∶(a＋2b) D. a∶[2(a+b)]

Average method - the "master of simplification" for mixture calculations

1. Basis: If XA>XB, then XA>X>XB, X represents the average relative atomic (molecule) mass, average concentration, average content, average production, average consumption, etc.

2. Application: If X is known, the range of XA and XB can be determined; if XA and XB are known, the range of X can be determined.

The key to solving the problem is to determine the range through the average value. The average value of many problems needs to be determined first according to the conditions before making a judgment. In fact, it is an extension of the extreme value method.

Thematic sparring

3． A total of 15 g of the two metal mixtures is put into a sufficient amount of hydrochloric acid. After full reaction, 11.2 L H2 (standard conditions) is obtained. The composition of the original mixture must not be ()

A. Mg and Ag B. Zn and Cu C. Al and Zn D. Al and Cu

B [This problem can be solved by the average molar electron mass (that is, the mass of metal required to provide 1 mol of electrons) method. During the reaction, H+ is reduced to form H2. According to the meaning of the question, 15 g of metal mixture can provide 1 mol e-, and its average molar electron mass is 15 g·mol-1. The molar electron masses of the metals Mg, Zn, and Al in the options are 12 g·mol-1, 32.5 g·mol-1, and 9 g·mol-1 respectively. The molar electron masses of Ag and Cu that cannot react with hydrochloric acid can be seen Make infinity. According to the mathematical average principle, the molar electron mass of one metal in the original mixture is greater than 15 g·mol-1, and the molar electron mass of the other metal is less than 15 g·mol-1. From this, it can be judged that the original mixture cannot be Zn and Cu. ]

4. After 95 g of MgCl2 powder containing a certain chloride impurity is dissolved in water and reacted with a sufficient amount of AgNO3 solution, the resulting AgCl is measured to be 300 g, then the impurity in the MgCl2 may be ()

A. NaCl B. AlCl3 C. KCl D. CaCl2

B　[The masses of each substance required to provide 1 mol Cl-(that is, the average mole Cl-mass) are respectively

Chemical formula of the substance MgCl2 NaCl AlCl3 KCl CaCl2

Average molar Cl-mass 47.5 58.5 44.5 74.5 55.5

The average value = 95 × 143.5300 ≈ 45.4, and the only one less than 45.4 is AlCl3. ]

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Update Time: 2024-10-01

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