This method is suitable for common chemical equations with little difficulty. For example, KClO3→KCl+O2↑ In this reaction formula, the number of oxygen atoms on the right is 2 and the number of oxygen atoms on the left is 3, so the minimum common multiple is 6, so the pre-coefficient of KClO3 should be matched with 2, and the pre-coefficient of O2 should be matched with 3, and the formula becomes: 2KClO3→KCl+3O2↑, because the number of potassium atoms and chlorine atoms on the left becomes 2, the pre-coefficient of KCl should be matched with 2, and the short-term should be changed into an equal sign.

2KClO3==2KCl+3O2↑

(2) parity balance method

This method is suitable for chemical equations in which an element appears on both sides for many times and the total number of atoms on both sides is odd and even, such as C2H2+O2→CO2+H2O. The equation starts from the oxygen atom that appears first and most times. There are two oxygen atoms in O2, and the total number of oxygen atoms should be even regardless of the pre-chemical coefficient Therefore, the coefficient of H2O on the right side should be matched with 2 (if other molecular coefficients have scores, it can be matched with 4), from which it can be inferred that C2H2 is the first 2, and the formula becomes: 2ch2 h 2+O2 = = CO2+2H2O, from which it can be seen that the former CO2 coefficient should be 4, and the last elemental O2 should be 5, and the condition can be expressed as follows:

2C2H2+5O2==4CO2+2H2O

(3) Balance of observation method

Sometimes a substance with complex chemical formula will appear in the equation. We can deduce the coefficients of other chemical formulas from this complex molecule, such as Fe+H2O-Fe3O4+H2. Fe3O4 has a complex chemical formula. Obviously, Fe in Fe3O4 comes from the element Fe and O comes from H2O, so Fe is pre-matched with 3 and H2O is pre-matched with 4. The formula is: 3Fe+4H2O = Fe3O4+H2.

3Fe+4H2O==Fe3O4+4H2↑

Note: In the chemical equation of this item, footprints are not bold.

[Edit this paragraph] Balanced poetry collection

This part of the poem includes six short poems. The first five poems introduce you to five balancing methods of chemical reaction equations, and the sixth poem tells you how to use these five methods flexibly and skillfully in the actual balancing process. If you can remember and understand these six poems, then you can proudly say, "There is not a chemical reaction equation in the world that I will not balance ..."

Simple equilibrium method of disambiguation reaction

First, three valence states are marked,

Subtract two from the third system.

If there are some divisors that need to be reduced,

If you look at it calmly, it will be complete.

Description:

1, disambiguation reaction is also called autoxidation-reduction reaction. In disambiguation, some atoms (or ions) of the same element are oxidized, while others are reduced. For example:

KCIO3 → KCIO4+KCI

Sulfur+potassium hydroxide → K2S+K2SO3+H2O

2. This poem introduces a simple balanced method of disambiguation reaction. Balancing in this way is simple and accurate, and the speed can be described as rapid!

Explanation:

1, first mark three valence states: it means a simple balancing method for disproportionation reaction. The first part is to clearly mark the valence states of elements in different substances in the reaction formula. For example:

S0+KOH → K2S-2+K2S+4O3+H2O

2. The third system of subtracting the two: it means that the change value (absolute value) of any two valences is the coefficient of the third party.

3. If there is a divisor, it needs to be reduced: it means that if the three coefficients obtained in the second step have common divisors, they need 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, and the variable value is ∣0-(-2)∣= 2, so the coefficient before K2SO3 is 2.

In S and K2SO3, S0→S+4, and the variable value is ∣0-4∣= 4, so the coefficient before K2S is 4.

In K2S and K2SO3, the variable value of S-2→S+4 is ∣(-2)-4∣= 6, so the coefficient before S is 6.

Because 2, 4 and 6 have a common divisor of 2, they are reduced to 1, 2 and 3, and the reduced coefficients are substituted into the reaction formula to obtain:

3S+KOH → 2K2S+K2SO3+H2O

4. Leisure observation will match: it means that after the conversion coefficient is substituted into the reaction formula, it can be balanced through leisure observation.

It is observed that there is 6 K on the right, so 6 should be added before KOH, and 6 H on the left after adding 6, so 3 should be added before H2O, so the equilibrium chemical reaction equation is obtained:

3S+6KOH = 2K2S+K2SO3+3H2O

Note: it's too late to say it's too early. As long as this method is mastered, the balance process can be completed in only a few seconds in "actual combat". So it is not an exaggeration to say "fast".

Simple equilibrium method of double hydrolysis reaction

Remember who is weak,

How to add coefficient electricity?

Water is often added to the reaction formula,

Conservation of mass means balance.

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 AI2(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, whoever is weak will choose, remember clearly: "whoever is weak will choose" refers to choosing the metal ion corresponding to weak base in two kinds of salts (for example, AI3+ is the metal cation corresponding to weak base AI(OH)3; NH4+ ion is a special case) and the acid anion corresponding to weak acid (for example, CO32- is the acid anion corresponding to weak acid H2CO3) as the object of adding coefficient (balance).

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 radical anion corresponding to weak acid.

3. Water is often added to the reaction formula, and mass conservation means balance: it means that after adding appropriate coefficients in front of the two salts, N is often added to the reaction formula for mass conservation. 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, write the hydrolysis product first:

Ai2 (SO4) 3+Na2CO3-Aluminum hydroxide ↓+CO2↑+Na2SO4

(2) Because we want to "choose who is weak", we should choose AI3+ and CO32-.

(3), add coefficient charges, etc. Because AI3+ has three positive charges and AI2(SO4)3 has two AI3+, it has six positive charges. CO32- has two negative charges. To make it "charge, etc.", we must add a coefficient of 3 before CO32-, so we get:

AI2(SO4)3+3 na 2co 3——2al(OH)3↓+3co 2 ↑+ 3 na 2co 4

(4) "Water is often added to the reaction formula". Because there are six H's in the product, "3H2O" should be added to the reactant. In this way, a balanced double hydrolysis reaction equation is obtained:

AI2(SO4)3+3 na 2co 3+3H2O = 2AI(OH)3↓+3co 2 ↑+ 3 na 2 so 4

Odd spouse method

Most odd numbers appear,

Then turn odd numbers into even numbers.

Observe the principle of balance, Jane,

Two, four, no, six.

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 equation of the combustion of some organic compounds (especially hydrocarbons). However, this method is not suitable for the balance of complex chemical reaction equations of reactants and products. In this case, it is often troublesome to use this method.

Explanation:

1, the most 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. "Finding odd numbers at most" means finding the elements that appear most frequently before and after the reaction in the reaction formula, and then finding the items with odd numbers of atoms on this basis; "Changing an odd number into an even number" means multiplying the found odd number by an even number (usually adding the smallest even number 2 before the molecule).

2. Observe the principle of balance. If two or four can't do it, you can find six: it means that you can observe the balance after changing the odd number into the even number. If the balance is uneven, try the larger even number 4 in turn. If not, use 6 again. ...

Example 1: Please balance the reaction formula:

Copper+nitric acid (concentrated)-copper nitrate+nitrogen dioxide =+H2O

According to the poetic requirements, the analysis is as follows:

In this reaction formula, before and after the reaction, Cu appeared twice, H appeared twice, N appeared three times and O appeared four times. Obviously, oxygen is the most frequent element before and after the reaction, and the number of H2O in the product is 1, which is an odd number, so we should add a factor of 2 before H2O to make the odd number become an even number:

Cu+HNO3 (concentrated) -Cu (NO3) 2+NO2 =+2H2O

After adding 2 in front of H2O, there are four H's on the right, so we should add 4 in front of HNO3, 4 on the left and 3 N on the right, so we should add 2 in front of NO2, so that we can get a balanced chemical reaction equation:

Cu+4hNO3 (concentrated) = Cu (NO3) 2+2no2 ↑+2h2o

Example 2: Please balance the reaction formula:

C2H6 +O2 —— CO2 +H2O

Analysis: It is observed that oxygen is the element with the highest frequency before and after, so the coefficient 2 is added before H2O, and then it is uneven after observation, and then it is changed to 4, but it still doesn't work, and then it is changed to 6. Observe the trim as follows:

2C2H6+7O2 = 4CO2+6H2O

Redox reaction cross equilibrium method

Price increases and price decreases add up,

The total price change is probably the back fork.

Does not involve redox,

Balance. Don't forget.

Among redox molecules,

Start from the right. Don't be afraid.

The back of the fork is surprisingly flat,

Cross again after the sudden change.

Description: 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 complex redox reactions, it is more convenient to balance in this way.

Explanation:

1, price increases and price decreases are added separately: this sentence means to introduce the first step of the cross-balance method, that is, to indicate the price states of price increase elements and price decrease elements first, and then add the price increases and price decreases respectively, so as to obtain the total number of price changes in the price states of price increase elements and the total number of price changes in price decrease elements.

For example, use the cross-balance method to balance the following equations:

FeS2+O2 —— SO2+Fe2O3

According to the poetic requirements, first show the valence of price-increasing elements and price-decreasing elements, and then get:

Fe+2s 2- 1+O20——S+4o 2-2+Fe2+3o 3-2

According to the poetic requirements, calculate the total price changes of price increase elements and price decrease elements. The valence increment of Fe2+→Fe3+ is 1, and that of S- 1→S+4 is 5. Because there are two S in FeS2, the total price increase of S is 5×2= 10, so the total price change of the price increase elements (Fe and S) is 65438. The valence reduction number of O0→O-2 is 2. Because O2 contains two O's, the total valence of the price reduction element O is 2×2=4. Therefore, the following formula is obtained:

1 1 4

FeS2 + O2 —— SO2 + Fe2O3

2. The total valence about the back fork: it means that after obtaining the total valence of the price-increasing elements and the total valence of the price-decreasing elements, if they have common divisors, they need to be reduced and then crossed (if they are 6 and 9, they need to be reduced to 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, so the following formula is obtained:

1 1 4

4 Fe S2+ 1 1o 2——SO2+fe2o 3

Observe the balance between left and right to get the answer:

4 Fe S2+ 1 1o 2 = 8so 2+2 fe2o 3

3. Redox is not involved. Don't forget when balancing: it means that if some reactants only partially participate in the redox reaction, and some do not, then the cross coefficient should be added to the number of molecules that the substance does not participate in the redox reaction, which is the pre-molecular coefficient of the substance.

For example, use the cross-balance method to balance the following equations:

Magnesium+nitric acid-magnesium nitrate+ammonium nitrate +H2O

According to the poetic requirements, the analysis is as follows:

The total valence of Mg is 2, the total valence of N is 8, and it is 1 and 4 after reduction. So there is no doubt that the pre-Mg coefficient is 4 and the pre-HNO3 coefficient seems to be 1. However, nine molecules of HNO3 in the product did not participate in the reaction, so the pre-HNO3 coefficient was not 1+9 =. Therefore, the following equilibrium reaction equation can be obtained:

4 mg+10 nitric acid = 4 mg (nitric acid) 2+ ammonium nitrate +3H2O

4. Within the redox molecule, don't be afraid to start from the right: it means that if the redox reaction is within the molecule, it should start from the product.

For example, use the cross-balance method to balance the following equations:

NH4NO3 —— N2+O2+H2O

According to the poetic analysis is as follows:

At first glance, this is a typical intramolecular redox reaction, starting with the product. N0→N-3 valence number -3 is N0→N+5 valence number is 5, so the total valence of N should be ∣5+(-3) ∣ = 2, and the total valence of O0→O-2 is 1. Observe the decoration:

2NH4NO3 = 2N2+O2+4H2O

5. Parity before and after crossing, and then crossing after parity: it means that if an atomic number before and after the reaction has parity after the cross coefficient, it is necessary to turn the odd number (multiplied by 2) into an even number.

For example, use the cross-balance method to balance the following equations:

FeS+kmno 4+h2so 4——k2so 4+mnso 4+Fe2(SO4)3+H2O+S↓

According to the poetic requirements, the analysis is as follows:

The total price increase of Fe and S is 3 (odd number), and the total price reduction of Mn is 5, so the cross coefficients are 3 and 5. However, there are 2 Fe (even number) in Fe2(SO4)3 and 2 K (even number) in K2SO4, so 3 and 5 should be multiplied by 2 to become even numbers 6 and 10 respectively, that is, 6 and 10. It can be concluded that:

10 FeS+6 kmno 4+24h2so 4 = 3k2so 4+6 mnso 4+5fe 2(SO4)3+24H2O+ 10S↓

Interpretation: The cross-balance method seems to be "complicated" when explaining, but in the actual balance process, it is only completed by the instantaneous thinking of the brain, so as long as you really understand this poem, you will achieve the effect of instantaneous completion in the actual balance.

Universal balance method

English letters represent numbers,

Mass conservation and electric conservation equations.

One term is a solution equation,

If there is a score, remove the denominator.

Description: This poem introduces the steps of universal balance method. The advantage of this method is worthy of the name-omnipotent! It can be used to balance any chemical reaction equation and ionic equation. If you master this method skillfully, then you can proudly say, "There is no chemical reaction equation in the world that I can't balance." ; The disadvantage of this method is that the equilibrium speed is affected by this law for chemical equations with more reactants and products. 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, use the general balancing method to balance the following equations:

Copper+nitric acid (concentrated)-copper nitrate+nitrogen dioxide =+H2O

According to the requirements of the poem, the coefficient before each molecule is expressed in English letters, so the following reaction equation is obtained:

Answer? 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 multivariate linear equations according to the law of mass conservation and the law of charge conservation (if not ion equations, only according to the law of mass conservation).

List the following equations according to the requirements of poetry:

A = C

B = 2E

B = 2C + D

3B = 6C + 2D + E

3. One is the equation: the third step to express this method is to make an unknown number in the equation set "1" and then solve the equation set.

According to the requirements of poetry, we set B = 1 and substitute it into the equations to get the following equations:

A = C

1 = 2E

1 = 2C + D

3 = 6C + 2D + E

Solution: A= 1/4, C= 1/4, D= 1/2, E= 1/2.

Substitute the values of a, b, c, d and e into the reaction equation ① to obtain:

1/4cu+HNO3 (concentrated)-1/4cu (NO3) 2+1/2no2 =+1/2h2o. ...

Note: In the actual balancing process, which one should be "1" depends on the concrete analysis of specific problems and the simplicity of solving equations. Generally, the coefficient of the term with complex molecular formula is "1".

4. If there is a fraction, the denominator is removed: the fourth step of this method is to substitute the solution of the equation obtained by solving the equation in the third part into the chemical reaction equation. If there is a fraction, the two sides of the chemical reaction equation should be multiplied by the least common multiple of each denominator. So that the denominator is removed and the fraction becomes an integer.

According to the poetic requirements, both sides of Equation ② are multiplied by 4 to obtain:

Cu+4hNO3 (concentrated) = Cu (NO3) 2+2no2 ↑+2h2o

Balanced decision song

Quick observation type,

Firstly, disproportionation hydrolysis is used.

Whether the parity can be crossed again,

Four skills are poor and omnipotent.

Description: This poem expounds how to correctly use these five balance methods introduced by the author in actual balance.

Explanation:

1, quick observation type: it means that after seeing the test questions, the first step is to observe what kind of reaction it belongs to.

2. disproportionation and hydrolysis first: refers to the simple equilibrium method of disproportionation reaction first and the simple equilibrium method of double hydrolysis reaction first.

3. Can parity cross again? This means that it is neither disproportionation nor double hydrolysis. Then look at the amount of reactants and products. If there are few, use the parity method; if there are many, use the cross-balance method.

4, the four methods are poor and omnipotent: it means that in case of a situation, that is, the first four methods can not be solved, then come up with the last trick-"universal balance method."

In order to help students master the above five balance methods, we provide the following exercises:

(What happens when FeCI3 and Na2S are mixed? Write the reaction equation and balance it.

Tip: Use "Double Hydrolysis Simple Equilibrium Method".

(2) Balance the following reaction formula:

Potassium chlorate-Potassium chlorate +KCI

C2H2+O2 —— CO2+H2O

Zinc+nitric acid-zinc nitrate+ammonium nitrate +H2O

H2S+ Nitric Acid-Sulfur+Nitrogen +H2O

Tip: Use various methods and compare which method is simpler for a specific reaction.