Qiu Zhijun, yancheng middle school City, Jiangsu Province 22400 1
In many chemistry test questions, mastering certain decomposition methods can often get twice the result with half the effort for students to learn chemistry well. Now I will talk about the application of several common conservation in chemical calculation in middle school based on my own teaching practice.
I. Conservation of mass
Conservation of quality is the focus of chemistry teaching in grade three. In any chemical reaction, the types and quantities of atoms have not changed before and after the reaction, so the total mass of atoms has not changed, so the total mass of reactants participating in the chemical reaction is equal to the total mass of products generated by the reaction.
Example 1: In the reaction formula X+2Y=R+2M, it is known that the molar mass ratio of R to M is 22∶9. When 1.6gX completely reacts with Y, and 4.4gR is generated, the mass ratio of Y to M in this reaction is ().
A 16:9 B 23:9 C 32:9 D 46:9
Analysis: The molar mass ratio of R and M is equal to the molecular formula quantity ratio of R and M. But the mass ratio of R and M is 1: 2, so the mass ratio of R and M is 22: 18, that is, 1 1: 9. So to generate 4.4gR, you must generate 3.6gM. According to the law of conservation of mass, the total mass of X and Y consumed should be equal to the total mass of R and M produced, that is, the mass of Y consumed is: m (r)+m (m)-m (x) = 4.4g+3.6g-1.6g = 6.4g .. So the mass ratio of Y and M in the reaction is 6.4g/3.6g =/. Therefore, this question should choose a.
Note: The law of conservation of mass can also be extended to: the total mass of the system before and after the reaction is equal to the total mass of the system after the reaction if there are reactants remaining or substances that do not participate in the reaction.
Example 2: Ag dilute sulfuric acid solution and Bg dilute barium chloride solution just react completely to generate Cg hydrochloric acid, and the mass fraction of the original dilute sulfuric acid solution is calculated.
Analysis: the reaction is H2SO4+bacl2 = baso4+2hcl, and the water in the solution does not participate in the reaction. The total mass of the system before the reaction is (a+b) g, then the total mass of BaSO4 generated after the reaction is (a+b-c) g. Because the amounts of BaSO4 and H2SO4 are equal, the mass fraction of solute in dilute Ag H2SO4 is 98 (a+b-c)/233a× 100%.
Second, charge conservation.
Charge conservation means that the total number of positive charges of cations in the whole system (or some substances in the system) is equal to the total number of negative charges of anions, and the whole system (or some substances in the system) is still electrically neutral.
Example 3: Dissolve the mixture of KCl and KBr 13.4g in water to make 500mL solution, then introduce excess Cl2, and evaporate the solution after full reaction to obtain solid 1 1. 175g. Then, the molar ratio of K+, Cl- and Br- in the stock solution is:
a 1:2:3 B 3:2: 1 C 1:3:2D 2:3: 1
Analysis: It is troublesome to solve this problem according to the general thinking method: it is necessary to find out the amount of KCl and KBr in the original mixture first, which should be combined with the total mass of KCl and KBr 13.4g and the final mass of solid KCl 1 1. 175g ... If this problem is solved by using the charge conservation of solute in the solution, It is not difficult to find that n (k+) = n (Cl-)+n (br-). Substituting the four options A, B, C and D into the above formula, we can see that only option B meets the meaning of the question, so the answer to this question is B.
Example 4: What is the relationship between the following ion concentrations c(Na+), c(H+), C (OH-), C (HCO3-) and C (CO32-) in Na2CO3 solution? Does this relationship still exist in NaHCO3? Does this relationship still exist in the mixed solution of NaHCO3 and Na2CO3?
Analysis: In Na2CO3 solution, only Na+ and H+ are cations, while CO32-,HCO3-and OH- are anions. According to the charge conservation in the whole solution system, n (Na+)+n (H+) = n (OH-)+n (HCO3-)+2n (CO32-), and the two sides are divided by the total volume of the solution, that is, C (Na+)+C (H+) = C (OH-)+. In the sodium bicarbonate solution and the mixed solution of sodium carbonate and sodium bicarbonate, only Na+ and H+ are cations, while only CO32-,HCO3- and OH- are anions. According to the charge conservation of the whole solution system, the relation C (Na+)+C (H+) = C (OH-)+C (HCO3-)+2C (CO32-) still exists.
Third, save materials.
Conservation of matter means that the ratio of the amount of matter between the atoms of two elements in a certain system is a constant value and does not change with the existing forms of these elements.
Example 5: In Na2CO3 solution, the following ion concentrations are c(Na+), c(H2CO3), C (HCO3-),
What does C (CO32-) have to do with it? Does this relationship still exist in NaHCO3? Equal amount of substance
Does this relationship still exist in the mixed solution of NaHCO3 and Na2CO3?
Analysis: In Na2CO3 solution, the atoms of elements C and Na come from the solute Na2CO3, so n (Na) = 2n (C). Na element only exists in the form of Na+ in the system, so n (Na) = n (Na+); Element c exists in the system in three forms: HCO3-, CO32- and H2CO3, so n (c) = n (H2CO3)+n (HCO3-)+n (CO32-), that is, n (Na+) = 2 [n (H2CO3)+n (HCO3-)+. Divide the two sides by the total volume of the solution, that is, c (Na+) = 2 [c (H2CO3)+c (HCO3-)+c (CO32-)]. In the sodium bicarbonate solution, the atoms of elements C and Na both come from the solute sodium bicarbonate, so n (na) = n (c). Na only exists in the form of Na+, and C also exists in the form of HCO3-, CO32-and H2CO3. Similarly, C (Na+) = C (H2CO3)+C (HCO3-)+
2n (Na) = 3n (c) in the mixed solution of NaHCO3 and Na2CO3 such as bicarbonate. Na element still exists in the form of Na+, and C element still exists in the form of HCO3-, CO32- and H2CO3. Similarly, 2c (Na+) = 3 [c (H2CO3)+c (HCO3-)+c (CO32-)].
It can be seen that in Na2CO3 solution, NaHCO3 solution and the mixed solution of NaHCO3 and Na2CO3, the charge conservation relations of their whole systems are the same, but their material conservation relations are different.
Fourth, the conservation of gain and loss electrons
The conservation of gain and loss electrons means that in the redox reaction, the total number of electrons gained by oxidant is equal to the total number of electrons lost by reductant, that is, something gains electrons in the reaction, and something loses electrons.
Can SbF5 burn in air? Why?
Analysis: The burning of a substance in air generally refers to the process that the substance is oxidized by oxygen in the air, that is, oxygen can get electrons from the substance. Because in SbF5, the element Sb is already the highest price, and it can no longer lose electrons. Although the price of element F is the lowest, it is weak in reducibility and cannot lose electrons. So oxygen can't get electrons from SbF5, so SbF5 can't burn in the air.
Example 7: After the simple substance M fully reacts with excess hot concentrated nitric acid, the molar ratio of simple substance to nitric acid participating in the reaction is measured as 1: 4. What is the valence of this element M in the product?
Analysis: The simple substance that can react with hot concentrated nitric acid may be metal simple substance or nonmetal simple substance C, P, S, etc. If it is a simple metal, the reaction products are nitrate, NO2 and H2O；; ; If it is a nonmetallic simple substance, the reaction products are hydrates, NO2 and H2O corresponding to the highest valence oxides of nonmetallic elements.
(1) If it is nonmetallic simple substance M, it can be known from M∽4HNO3∽4NO2 that the total number of electrons of N element in HNO3 is 4e-, and the total number of electrons lost by nonmetallic simple substance M is 4e-, so m should be +4 in the product.
(2) If the metal is a simple substance M, let M represent the valence of +X in the product, which is determined by the following formula.
The total number of electrons lost in M∽4 hno3∽M(NO3)x ∽( 4-x)NO2:M is Xe-, and the total number of electrons of N element in HNO 3 is (4-x) E-. According to the conservation of gain and loss electrons: x = 4-x, the solution is x = 2, so m should appear in the product.
Verb (abbreviation for verb) element conservation
Conservation of elements means that the atomic number of some elements remains unchanged before and after the reaction without considering other substances participating in the reaction process.
Example 8: A KOH solid was exposed to air, and it was found that it contained 7.65% water, K2CO34.32 and the rest KOH through analysis. If the a g sample is put into bmL 1mol/L hydrochloric acid to make it fully react, the residual acid is just neutralized with 25.52mLcmol/L KOH solution, and the obtained solution is evaporated to obtain the expression of solid mass (unit: g).
A can only contain b, and b should contain a, c, d, a, b and C.
Analysis: The reaction process is: K2CO3 and KOH in the sample react with the added hydrochloric acid to generate KCl, and the excess hydrochloric acid is neutralized with KOH to generate KCl. Therefore, there is only one solute in the final solution, and the mass of the solid is also the mass of KCl. In the final KCl, K+ comes from KOH in the sample and K2CO3 in the added reagent, so if the mass of KCl is calculated by the conservation of potassium, there must be two letters A and C in the expression. Cl- only comes from hydrochloric acid, so if the mass of KCl is calculated by the conservation of chlorine, there is only one letter B in the expression. Therefore, this question should choose a.
Example 9:Cl- reacts with Ag+ to generate AgCl, and 10% of AgCl generated each time is photolyzed to generate Ag and Cl2, and Cl2 is disproportionated into HClO3 and HCl in water. The Cl- thus generated and the remaining Ag+ generate AgCl precipitate, and so on until the end. The existing solution contains 1. 1molNaCl. Add enough AgNO3 _ 3 _ 3 solution to it, and find out how many g insoluble matter can be generated finally.
Analysis: The key to this problem is the change of the number of particles in the solution before and after the reaction. The reactions involved in the inscription are: Cl-+Ag+= AgCl ↓, 2AgCl = 2AG+Cl2, 3Cl2+3H2O = HCLO3+5HCl. Therefore, Cl- and Ag+ must be consumed in the reaction, and there must be Ag, AgCl and HClO3 in the product, and the molar ratio of Ag to AgCl is 1: 9. According to the conservation of electrons and the conservation of Cl and Ag in the reaction, it is known that: 55cl-∽ 60ag+∽ 6ag ∽ 54agcl ∽ HCLO3. Therefore, the total mass of insoluble Ag and AgCl can be obtained from the amount of Cl- in NaCl: 0.12mol×108g/mol+1.08mol×143.5g/mol =167.94g. ..