Mathematical Matching Methods Matching Perfect Equalization Method How to Balance Chemical Equations

Let me take another example, you are not representative (quadratic coefficient is 1). 2x2-4x+5 The first step is to isolate the constant term (if it is a quadratic equation, move the constant term to the other side of the equal sign); That is: (2x2-4x)+5 in the second step, let the quadratic term coefficient be 1; That is, in the third step of 2(x2-2x)+5, the square of half of the coefficient of the primary term is allocated (plus) in parentheses, and in order to ensure that the value of the formula remains unchanged, the product of the coefficient outside the parentheses and half of the square of the coefficient of the primary term is subtracted from the parentheses; That is, the square of 2 x2-2x+(2 2) -2*(2 2) + the square of 5 in the fourth step, write the part in parentheses as a binomial perfect square.

That is: 2 (x-1) squared + 3 It should be said that the method steps are unique (or the expression is similar), you understand and remember the steps I talked about above, and the same problem will be solved. Give it a try.

According to the coefficient before the first formula of x, the constant is formulated, and the upper formula can be formulated as x2-4x+4-2, and the previous one can be perfectly squared.

The coefficient of the primary term is divided by 2, which is the constant term that needs to be added in the square, and then subtracted after addition, and merged with the original constant term. The constant in the square is the coefficient of the primary term divided by 2 (with a sign).

Steps:1).Find the most complex chemical formula in the equation and assign its stoichiometric number to "1";

2).The number of atoms of each element in the chemical formula determines the number of measurements before the chemical formula;

3).If a fraction appears, the denominator is removed.

For example: Fes2 O2 Fe2O3 SO2

Analysis: In the formula, the chemical formula Fe2O3 is the most complex, and its stoichiometric number is 1; Then, through observation, the measurement number 2 is matched before FES2, the measurement number 4 is matched before SO2, and the measurement number 11 2 is matched before O2;Finally, multiply the measurement numbers before the chemical formula on both sides of the equation by 2, remove the denominator of the measurement number before O2, and the chemical equation is balanced.

That is, a substance (atom) is one, balanced, and the coefficient before other substances is a fraction or integer cxhy and o2, and the coefficient before cxhy is 1; There are x c's, and the coefficient of CO2 is x; There is y h, and the coefficient of h2o is y 2

cxhy+o2==xco2+（y/2）h2o

Count the number of o, divide by 2, and write before o2.

cxhy+（x+y/4）

o2==xco2+（y/2）h2o

The adjustment factor is that it becomes an integer, multiplied by 4

4cxhy+（4x+y）

o2==4xco2+（2y）h2o

Chemical Trim Methods and Techniques:

a) Least common multiple method.

1) Find the atom with a large number of atoms and appear once on each side of the reaction formula, and find its least common multiple.

2) The coefficients of each molecule are introduced.

2) Observational method.

1) Estimating the stoichiometric number of the chemical formula of each reactant and the stoichiometric number of the product from a product with a complex chemical formula.

2) According to the stoichiometric number of the obtained chemical formula, find the stoichiometric number of other chemical formulas until the trim.

3) Odd-even matching.

1) Find the element with the highest number of atoms at the left and right ends of the chemical equation and find their least common multiple.

2) Divide this least common multiple by the original number of atoms on the left and right sides, and the quotient is the stoichiometric number of their chemical formula.

3) According to the stoichiometric number of the determined chemical formula of the substance, derive and find the stoichiometric number of the chemical formula until the chemical equation is balanced.

4) Normalization Law.

1) Find the key chemical formula in the chemical equation, set the number before the chemical formula as 1, and then balance the stoichiometric number before the other chemical formula according to the key chemical formula.

2) If there are measured numbers as fractions, multiply each measured number by the same integer to turn the fraction into an integer.

a) Least common multiple method.

This method is suitable for common chemical equations that are not too difficult. For example, KCLO3 KCl+O2 is in this reaction.

If the number of oxygen atoms on the right is 2 and 3 on the left, the least common multiple is 6.

Therefore, the coefficient before KCLO3 should be matched with 2, and O2 should be matched with 3, and the formula becomes: 2kclo3 KCl+3O2, and since the number of potassium atoms and chlorine atoms on the left becomes 2, then KCL should be matched before the system.

Number 2, ** is changed to an equal sign, indicating the condition i.e

2kclo3==2kcl+3o2↑

2) Odd-even equalization.

This method is suitable for multiple occurrences of an element on both sides of a chemical equation where the total number of atoms of that element on both sides is odd and one.

Even, for example: C2H2+O2 CO2+H2O, the equation is balanced with the oxygen atom that appears the most first. O2.

2 oxygen atoms, regardless of the number of coefficients before the chemical formula, the total number of oxygen atoms should be even.

Therefore, the coefficient of H2O on the right should be matched with 2 (if other molecular coefficients appear as fractions, it can be matched with 4), from which it is inferred that the first 2 of C2H2 becomes: 2C2H2+O2 CO2+2H2O, from which it can be seen that the coefficient before CO2 should be 4, and the final element O2 is 5.

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

c) Observational balancing.

Sometimes there is a substance with a complex chemical formula in the equation, and we can extrapolate other substances from this complex molecule.

The coefficients of the formula, for example: Fe + H2O - Fe3O4 + H2, Fe3O4 chemical formula is more complex.

Obviously, FE3O4 FE** is in elemental FE, O comes from H2O, then FE is preceded by 3, H2O is preceded by 4, then the formula is: 3Fe+4H2O Fe3O4+H2 thus deduces that the H2 coefficient is 4, and the conditions are specified, ** can be changed to an equal sign: 3Fe+4H2O==Fe3O4+4H2

4) Normalization Law.

Find the key chemical formula in the chemical equation, set the number before the chemical formula to 1, and then trim according to the key chemical formula.

Stoichiometric number before other chemical formulas.

If the measured number is a fraction, and then the measured number is multiplied by the same integer, and the fraction is turned into an integer, this balancing method in which the key chemical formula measurement number is 1 is predetermined, which is called the normalization method. Method: Select the most complex chemical formula in the chemical equation, let its coefficient be 1, and then infer it in turn.

Step 1: Let the coefficient of NH3 be 1 1NH3+O2 - NO+H2O

Step 2: The N and H atoms in the reaction are transferred to No and H2O respectively, by.

Step 3: Push the O2 coefficient from the total number of oxygen atoms at the right end.

Chemical formula balancing methods and techniques are: 1, least common multiple method 2, odd spouse method 3, fraction balancing method This method can balance chemical reactions with elemental participation in the reaction or elemental generation 4, algebraic method (also known as the undetermined coefficient method) 5, observation balancing.

1. Least common multiple method. Applicable conditions: The matching atom appears only once on the left and right sides of the equation, and this method is suitable for common chemical equations with little difficulty.

2. Odd spouse method. Applicable conditions: It is applicable to the multiple occurrences of an element on both sides of the chemical equation, and the total number of atoms of the element on both sides is odd and even.

Applicable conditions: such as the reaction of most hydrocarbons or compounds containing hydrocarbon and oxygen with oxygen, as well as certain decomposition reactions.

3. Fractional balancing method: This method can balance chemical reactions with elemental participation in the reaction or elemental generation. Specific steps: first trim the atoms of each element in the compound; Equalize the atoms of elemental elements with fractions; Remove the denominator so that the stoichiometric number after trim is an integer.

4. Algebraic method (also known as undetermined coefficient method). Applicable conditions: There are many types of reactants or products, and the trim does not know where to start with the more complex reactions.

5. Observation method balancing. Applicable conditions: Sometimes there will be a substance with a complex chemical formula in the equation, and we can use this complex molecule to deduce the coefficients of other chemical formulas.

a) Least common multiple method.

This method is suitable for common chemical equations that are not too difficult. For example, 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, then the least common multiple is 6, so the coefficient before kclo3 should be matched with 2, and the coefficient before O2 should be matched with 3, and the formula becomes: 2kclo3 KCl+3O2, since the number of potassium atoms and chlorine atoms on the left becomes 2, then the coefficient 2 before KCL, ** is changed to equal sign, indicating the condition is:

2kclo3==2kcl+3o2↑

2) Odd-even equalization.

This method is suitable for multiple occurrences of an element on both sides of the chemical equation, and the total number of atoms of the element on both sides is odd and even, for example: C2H2+O2 - CO2+H2O, and the balance of this equation starts with the oxygen atom with the highest number of occurrences. There are 2 oxygen atoms in O2, and the total number of oxygen atoms should be even, regardless of the number of coefficients before the chemical formula.

Therefore, the coefficient of H2O on the right should be matched with 2 (if other molecular coefficients appear as fractions, it can be matched with 4), from which it can be deduced that the first 2 of C2H2 becomes: 2C2H2+O2==CO2+2H2O, from which it can be seen that the coefficient before CO2 should be 4, and the final coefficients with elemental O2 are 5, and the conditions can be specified:

2c2h2+5o2==4co2+2h2o

c) Observational balancing.

Sometimes there will be a substance with a more complex chemical formula in the equation, we can deduce the coefficients of other chemical formulas through this complex molecule, for example: Fe + H2O - Fe3O4 + H2, Fe3O4 chemical formula is more complex, obviously, Fe3O4 Fe** in the elemental Fe, O comes from H2O, then Fe is preceded by 3, H2O is preceded by 4, then the formula is: 3Fe + 4H2O Fe3O4 + H2 This deduces that the H2 coefficient is 4, indicating the conditions, ** Change to an equal sign:

3fe+4h2o==fe3o4+4h2↑

This method is suitable for the valence state of all elements in the reducing agent, which is difficult to determine, but it is an elevated redox reaction. In a substance that is difficult to determine the valence, all elements can be balanced as zero valent.

Such as: FeSi + Hno3—H4SiO4 + Fe(No3)3 + No + H2O

Assuming that the two elements in Fesi are both zero valent, and the corresponding products in Fe are +3 valence and Si are +4 valence, the total increase is 7, and the valence state of N element in Hno3 is reduced (+5 +2) to 3 valence, and the least common multiple is 21, so there is a measurement number 3FeSi + 7Hno3, and then the 9 NO3- that have not changed in 3Fe(NO3)3 are returned to Hno3, so the measurement number of Hno3 is 16, and the balance is observed

3fesi+16hno3=3h4sio4+3fe(no3)3+7no↑+2h2o

It's just one letter instead of all.