Basic principle: 1. The atomic number of each element before and after the reaction is equal, that is, the mass is conserved; 1. The total number of electrons obtained by oxidant is equal to the total number of electrons lost by reductant, that is, electron conservation; 1. The total reduction price of oxidant is equal to the total increase price of reductant. Mark the valence of valence-changing elements. 1. column change: lists the rising and falling values of the valence of elements before and after the reaction. 3. Find the total number: make the total number of rising and falling valence equal. 4. Matching coefficient: Balance the stoichiometry of other substances through observation, and then turn the single line into an equal sign. 5. Check the conservation: check whether the two sides of the equation are "mass conservation". Generally, there are the following rules: 1. If the valence of an element in the oxidant and reductant changes completely, then the equilibrium should start from the oxidant and reductant, that is, the reactants should be considered first. 1. If the valence of an element in the oxidant (or reductant) changes only partially, the equilibrium should start from the oxidation product and the reduction product, that is, the product should be considered first. 4. Autoredox reaction equation. Balance should start with the product. (reverse balance) 1. Many elements in the same reactant are valence-changing, so the substance can be regarded as a whole, that is, the algebraic sum of the increasing value and decreasing value of each valence-changing element in a molecule of the substance can be obtained. There are two methods you can try: ① Zero-valence Method Usage: Schilling can't determine that all valuable elements in a substance are zero by conventional methods, and then calculate the increase and decrease value of each element. And make the rise and fall of the valence of elements equal, and finally balance the stoichiometry of other substances through observation. Example 1: Try to balance Fe3C+HNO3-Fe (NO3) 3+NO2+CO2+H2O. Analysis: The specific valence of Fe and C in the complex Fe3C cannot be determined by the conventional valence analysis method. At this time, the valence of each element constituting the substance can be made zero. According to the valence fluctuation method, Fe3C → Fe(NO3)3 and CO2 increase 13, HNO3 → NO2 decrease 13 (except Fe and C, only N changes valence). It is easy to obtain Fe3C+22HNO3 = 3Fe (NO3) 3+13NO2+CO2+11H2O. ② Usage of average pricing method: When atoms of the same element in the same reactant appear twice and their valence states are different, they can be treated equally, that is, assuming their valence states are the same. According to the principle that the algebraic sum of valence in compounds is zero, the average price is marked. If diatomic molecules appear in the equation, the number of related atoms will be doubled. Example 2: Try to balance NH4NO3- HNO3+ N2+ H2O. Analysis: The average valence of N in NH4NO3 is+1. The valence relationships of elements are: NH4NO3 → HNO3:+ 1 →+5 liters, 4× 1 valence, NH4NO3 → N2:+ 1 → 0 decrease, 1×2 valence, which is easy to obtain. The use of the whole pricing method. When there are several compounds in a compound, it can be treated as a whole. The price of the whole compound is based on the principle that the algebraic sum of valence of elements in the compound is zero. Example 3: Trial balance analysis of S+Ca (OH) 2-CASX+Ca2O3+H2O: the average valence of N in NH4NO3 is+1(3 in-NH4 and +5 in NO3). The increasing and decreasing relationships of elemental valence are: NH4NO3 → HNO3:+1 → 5 liters, 4× 1 valence, NH4NO3 → N2:+ 1 → 0 decrease, 2×2 valence, 2 (x+/kloc-0 If the measurement number is a fraction, then multiply each measurement number by the same integer to change the fraction into an integer. This balance method of setting the measurement number of key chemical formula as 1 is called normalization method. Practice: Select the most complicated chemical formula in the chemical equation and set its coefficient as 1. Then infer in turn. Step 1: Let the coefficient of NH3 be 1 1nh3+O2-NO+H2O. Step 2: N atoms and H atoms in the reaction are transferred to NO and H2O respectively. Step 3: Derive O2 coefficient from the total number of oxygen atoms at the right end [edit this paragraph]. (5) Balance this part of the poem with a balanced collection of poems. The first five poems are for you. The sixth poem tells you how to use these five methods flexibly and skillfully in the actual balance process. If you can remember and understand these six short poems, then you can proudly say, "There is no chemical reaction equation in the world that I can't balance …" The simple balancing method of disproportionation reaction is to mark three valence states first and subtract the third system from the two. If there is reduction, it means: 1, disproportionation reaction is also called autooxidation-reduction reaction, some atoms (or ions) of the same element are oxidized and others are reduced. For example, kcio3 → kcio4+kcis+KOH → K2S+K2SO3+H2O2. This poem introduces disproportionation. Explanation: 1, first mark three valence states: The first part of the simple balancing method for disproportionation reaction is to clearly mark the valence states of elements in different substance molecules in the reaction formula. For example, S0+KOH → K2S-2+K2S+4O3+H2O2, the third system for subtracting the two: representing the change value (absolute value) of any two valences. That is, the coefficient of the third party. 3. If there is a divisor, it needs to be reduced: it means that if there are common divisors in the three coefficients obtained in the second step, it needs to be reduced and then added to the reaction formula. According to the poetic requirements, the analysis is as follows: in S and K2S, S0 →S-2, the variable value is ∣0-(-2)∣= 2, so the variable value is ∣0-4∣= 4, so the coefficient before K2S is 4. In K2S and K2SO3, the valence change value is ∣(-2)-4∣= 6, so the coefficient before S is 6. Substituting the conversion coefficient into the reaction formula: 3s+KOH → 2k2s+k2so3+H2O2, and you will get a match after leisurely observation: it means that after substituting the conversion coefficient into the reaction formula, you can get a balance after leisurely observation. Observation shows that there is 6 k on the right, so 6 should be added before KOH, and 6 h after adding 6 on the left, so 3 should be added before H2O. So we get the equilibrium chemical reaction equation: 3s+6koh = 2k2s+K2SO3+3H2O. It's too late to say now, but it's too soon. As long as you master this method, you can complete the balance process in a few seconds in actual combat. Therefore, it is not an exaggeration to say "fast". The weak points in the simple balancing method of double hydrolysis reaction should be clearly remembered and coefficient charges should be added. Reaction formula. Balance is the conservation of mass. Description: Double hydrolysis reaction refers to the reaction of a strong acid and weak base salt with another strong base and weak acid salt, and the hydrolysis reaction goes to the end due to mutual promotion. For example, the reaction between AI3+(SO4)3 and Na2CO3. The characteristic of this method is that the coefficient can be written directly and the balance process can be completed instantly. Explanation: 1, remember who is weak: "NH4+ ion is a special case) The acid anion corresponding to weak acid (such as CO32- is the acid anion corresponding to weak acid H2CO3) is the object of addition coefficient (equilibrium). 2. What is the addition coefficient? It means adding a certain coefficient before the selected object, so that the charge number of metal cation (or NH4+) corresponding to weak base is equal to that of acid anion corresponding to weak acid. 3. Water is often added to the reaction formula. H2O。 For example, write the chemical equation of hydrolysis reaction when two solutions of AI2(SO4)3 and Na2CO3 are mixed. According to the poetic requirements, the analysis is as follows: (1) According to the principle of hydrolysis, first write the hydrolysis products: AI2 (SO4) 3+Na2CO3-AI (OH) 3 ↓+CO2 ↑+Na2SO4 ↑, because "whoever is weak will choose". CO32- has two negative charges. In order to make it "charge equal", a coefficient 3 must be added before CO32-, so we get: AI2 (SO4) 3+3Na2CO3-2ai (OH) 3 ↓+3CO2 ↑+3Na2S4 ↑, "Water is often added in the reaction formula". Because there are six H's in the product, "3H2O" should be added to the reactant. Thus, a balanced double hydrolysis reaction equation is obtained: AI2 (SO4) 3+3Na2CO3+3H2O = 2ai (OH) 3 ↓+3CO2 ↑+3Na2SO4. Odd numbers appear the most, and then they become even numbers. Observe the simplification of balance. Explanation: This poem introduces the steps of balancing chemical reaction equations by odd-numbered spouse method. The advantage of this method is that it can adapt to the balance of various chemical reaction equations, and it is simple, fast and can directly add coefficients. It is especially effective for balancing the chemical reaction equations of some organic compounds (especially hydrocarbons). However, this method is not suitable for balancing chemical reaction equations with complex reactants and products. In this case, it is often troublesome to use this method. Explanation: 1, the odd number appears, and then the odd number becomes an even number: these two sentences are the first step of the odd-numbered spouse method. " Oddest occurrence "refers to finding the element that appears the most times before and after the reaction, and then finding the item with odd number of atoms;" Then turn the odd number into an even number "means to multiply the found odd number by an even number (usually adding the smallest even number 2 in front of the molecule). 2. Observe the truth of balance. If two or four can't do it, ask for six: it means to observe the balance after changing the odd number into the even number. If the match is uneven, try the larger even number 4 in turn, if not, use 6 again. Example: 1: Please balance the reaction formula: Cu+HNO3 (concentrated)-Cu (NO3) 2+NO2 =+H2O. According to poetic requirements, in this reaction formula, Cu appears twice, H appears twice, N appears three times and O appears four times before and after the reaction. Obviously, oxygen is the most frequent element, so a factor of 2 should be added before H2O to make the odd number even: Cu+HNO3 (concentrated) -Cu (NO3) 2+NO2 ↑+2H2O. After H2O is preceded by 2, there are four H's on the right, so HNO3 is preceded by 4, left by 4 N's and right by 3 N's, so NO2 is preceded by 2. Thus, a balanced chemical reaction equation was obtained: Cu+4HNO3 (concentrated) = Cu (NO3) 2+2NO2 ↑+2H2O Example 2: Please balance the reaction equation: C2H6+O2-CO2+H2O Analysis: It was observed that oxygen was the most frequently occurring element before and after, so a coefficient was added before H2O, and it was found to be uneven after observation, and then it was changed to 4, but it was still not possible. 6. The observation balance is as follows: 2c2h6+7O2 = 4co2+6H2O redox reaction cross balance method, the price increase and price decrease are added, and the total price change is about the back fork. There is no redox involved, so don't forget it on the balance. Within redox molecules, don't be afraid to start from the right. The back fork is surprisingly even. Note: This poem introduces the steps of balancing the redox reaction equation by cross balancing method and the problems that should be paid attention to when applying this method. For more complicated redox reactions, this method is more convenient for balancing. Note: 1, the price increase and price decrease are added together: this sentence means the first step of introducing the cross-balance method, that is, first indicate the price state of the price increase element and the price decrease element, and then get the total number of price state changes of the rising element and the total number of price state changes of the falling element in this way. For example, please use the cross balancing method to balance the following reaction formula: FeS2+O2-SO2+Fe2O3. According to the requirements of poetics, the valence of rising elements and falling elements should be marked first. So we get: Fe+2S2-1+O20-S+4O2-2+Fe2+3O3-2. According to the poetic requirements, we can calculate the total amount of price changes of price-increasing elements and price-decreasing elements. The price increments of Fe2+→Fe3+ are 1 and S- 1 → S respectively, so the total price increase of S is 5×2= 10, so the total price change of price increase elements (Fe and S) is1+10. The valence reduction number of O0→O-2 is 2. Because O2 contains two O's, the total valence change number of O's is 2×2=4. Then the following formula is obtained:114fe2+O2-SO2+Fe2O3, and the total valence is about the back fork: the total valence representing the valence of the price-increasing elements and the valence of the price-decreasing elements. You need to reduce and then cross (if it is 6 and 9, the reduction is 2 and 3). The implication is that if it is a prime number, it can be crossed directly. In this example, 1 1 and 4 are prime numbers, so they can be directly crossed. Then the following formula is obtained:1144fes2+11O2-SO2+Fe2O3, and the answer can be obtained by observing the balance: 4fes2+1kloc-0/O2 = 8so2+2fe2o3, which does not involve oxidation. For example, please use the cross-balance method to balance the following reaction formula: Mg+HNO3-Mg (NO3) 2+NH4NO3+H2O. According to the requirements of poetics, the total valence of mg is 2, the total valence of N is 8, and it is 1 and 4 after reduction, so the pre-mg coefficient is 4, but HNO3 is beyond doubt. However, it was observed that nine molecules of HNO3 in the product did not participate in the reaction, so the coefficient before HNO3 was not 1, but 1+9= 10. Therefore, the following equilibrium reaction equation can be obtained: 4 mg+10hno3 = 4 mg (NO3) 2+NH4NO3+3H2O4. Don't be afraid to start from the right: it means that if there is an intramolecular redox reaction, it should be cross-balanced from the product. For example, please use the cross-balance method to balance the following reaction formula: NH4NO3-N2+O2+H2O. Poetry analysis, at a glance, this is a typical intramolecular redox reaction, starting with the product. The valence reduction number of N0→N-3 is -3, and the valence increment of N0→N+5 is 5, so the total valence of n should be ∣ 5+(-3) ∣ = 2, and the total valence of o0 → o-2 is 4. After simplifying the crossover, the equilibrium is observed as follows: 2nh4no3 = 2n2+O2+4ho5, and the crossover is odd. You need to change the odd number (multiplied by 2) into an even number. For example, please use the cross-balance method to balance the following reaction formula: FeS+KMnO4+H2SO4-K2SO4+MnSO4+Fe2 (SO4) 3+H2O+S ↓ According to the poetic requirements, the analysis is as follows: the total valence of Fe and S increases to 3 (odd number), and the total valence of Mn decreases to 5. But there are two Fe's (even numbers) in Fe2(SO4)3 and two K's (even numbers) in K2SO4, so 3 and 5 should be multiplied by 2 to become even numbers 6 and 10 respectively. That is, 6 and 10 are the coefficients that should actually cross. It is concluded that10fes+6mno4+24h2so4 = 3k2so4+6mnso4+5fe2 (SO4) 3+24h2o+10s↓ shows that the cross-balance method seems to be "more complicated" in explanation. Therefore, as long as you really understand this poem, you will achieve immediate results in the actual balance. The English letters of the cosmic balance method represent the numbers and equations of mass conservation and electric conservation. An item is an equation. If there is a fraction, remove the denominator. Description: This poem introduces the steps of the cosmic balance method. The advantage of this method is worthy of the name-omnipotent! It can be used to balance any chemical reaction equation with ionic equation. If you master this method skillfully, you can proudly say, "There is not a chemical reaction equation in the world that I can't balance. The disadvantage of this method is that for chemical equations with many reactants and products, the equilibrium speed is affected by this method, but it is not absolute, because its speed depends on your ability to solve multiple linear equations. If you have a good grasp of the skills of understanding equations, then the speed of balancing chemical equations with universal balancing method is ideal. Explanation: 1, English letters indicate numbers: "Numbers" refer to molecular coefficients that need to be balanced. The first step of universal balancing method is to express the coefficients before each molecular formula in English letters. For example, please use the general equilibrium method to balance the following reactions: Cu+HNO3 (concentration) -Cu (NO3). Cu+B？ Nitric acid (concentrated) -C? Copper nitrate 2+D? NO2↑+E？ H2O...① 2. Mass-electricity conservation equation: The second step of this method is to list multiple linear equations according to the laws of mass conservation and charge conservation (if it is not an ion equation, according to the poetic requirements, the following equations are listed: A = CB = 2EB = 2C+D3 = 6C+2D+E3, and one item is to solve the equation: it means that the third step of this method is in the equation "/kloc-" According to the poetic requirements, we make b = 65438. Substitute the following equation: a = c1= 2e1= 2c+D3 = 6c+2d+e, and get the solution: A = 1/4, C = 1/4, D = 1/2. We get: 1/4cu+HNO3 (concentrated) -65438 Generally speaking, the coefficient of the term with complex molecular formula is "1". 4. If there is a score, the denominator is removed: the fourth step of this method is to substitute the solution of the equation set obtained by solving the equation set in the third part into the chemical reaction equation. If there is a fraction, it must be multiplied by the least common multiple of each denominator on both sides of the chemical reaction equation, thus removing each denominator. Turn fractions into integers. According to the poetic requirements, two sides of Equation ② are multiplied by 4 to obtain: Cu+4HNO3 (concentrated) = Cu (NO3) 2+2NO2 ↑+2H2O equilibrium decision song. Quickly observe and set the type, first use disproportionation and hydrolysis. Can parity cross again? Description: This poem expounds how to correctly use the five balance methods introduced by the author in actual balance. Explanation: 1 Quickly observe and determine the type: it means that after seeing the test questions, the first step is to observe what type of response it belongs to. 2. Hydrolysis by disproportionation first: it means that if it is a disproportionation reaction, the simple equilibrium method of disproportionation reaction is used first. If it is a double hydrolysis reaction, first use the simple equilibrium method of double hydrolysis reaction. 3. Can I cross even and odd? This means that it is neither disproportionation nor double hydrolysis. Then look at the amount of reactants and products. If it is less, use the parity method. If there are many, use the cross-balance method. 4. Four methods are all-around: It means that if there is an emergency, the first four methods can't be solved. Then come up with the last trick-"universal balance method." In order to facilitate students to master the above five balance methods, the following exercises are provided: (What happens when FeCI3 and Na2S are mixed? Write the reaction equation and balance it. Tip: Use simple double hydrolysis equilibrium method. (2) Balance the following reaction equations: kcio3-kcio4+kci2h2+O2-CO2+H2O Zn+nitric acid -Zn(NO3)2+nh4no 3+H2O H2S+ nitric acid -S+NO+. And compare which method is more convenient for a specific reaction. (6) Let the' n' equilibrium be set, and the number of substances containing the most elements in the reaction formula is n, and then deduce them one by one, so that the equilibrium can be achieved. For example, if the square formula KCIO3-KCIO4+kci sets the coefficient of KCIO4 as n, then the coefficient of KCIO3 is 4N/3 and the coefficient of KCl is N/3. Finally, multiply left and right.