Some good examples of how to level chemical equations are introduced

Valence lifting method.

Are you in junior high school or high school? If you're in high school, just look at the textbooks, and if you're in junior high school, ask your high school seniors.

Trim chemical equation examples, as follows:

1 First of all, find the element with the largest number of atoms at the left and right ends of the reaction formula and find their least common multiple.

2 Secondly, divide this least common multiple by the original number of atoms on the left and right sides, and the quotient obtained is the coefficient of their chemical formula.

3 Then, according to the coefficients of the chemical formula of the substance that has been determined, the coefficients of its chemical formula are deduced and found until the equation is balanced.4 Finally, the reaction formula is verified. Whether the trim is correct or not.

The following is an example: the reaction equation Fe3O4 for co-combustion of flat magnetite (Fe3O4) and aluminum powder Fe3O4 Decal FeDecaAl2O3

(1) The number of oxygen atoms in this reaction is large and only occurs once on both sides, so the least common multiple of the number of oxygen atoms on both sides is first found: 4 3 12.

2) According to the law of conservation of mass, in order to make the number of oxygen atoms on the left and right sides of the reaction equation equal, the coefficient of Fe3O4 must be 3 (12 4) and the coefficient of AI2O3 must be 4 (12 3), i.e., 3Fe3O4 Ten A1 Fe Ten 4A12O3

3) Rematch Fe and Al atoms. From the matching 3Fe3O4 and 4Al2O3, it can be seen that when the coefficients of Fe and Al are 9 and 8 respectively, the atomic numbers of Fe and Al on the left and right sides are equal, so there is: 3Fe3O4+8Al=9Fe+4Al2O3

4) In the final test, the complete chemical equation is: 3Fe3O4+8A=9Fe+4Al2O3

1. Take "na+h2o=naoh+h2" as an example.

2. Parity balancing method, 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 has an odd and even. The hydrogen atom is matched first, there are two hydrogen atoms on the left, and there is only one hydrogen atom on the right Naoh, there are two hydrogen atoms in the hydrogen, and the total number of hydrogen atoms should be even, so the coefficient before NaOH is 2, and the number of oxygen atoms on the right is 2, and so on, the coefficient before H2O is 2, and the coefficient before NaOH is also 2.

3. Electron gain and loss method, find out the oxidant and reducing agent from the reaction formula, and show that the atoms of the oxidized element or the reducing element change in valency before and after the reaction, so as to determine the number of electron gain and loss. Sodium loses one electron and the hydrogen atom gains two electrons, so the coefficient before sodium is 2, and according to the law of conservation of mass, the coefficient of NaOH and H2O is 2.

Chemical equation.

It is an important part of chemistry learning, and after everyone has mastered the way of writing chemical formulas, the balancing of chemical equations has become a difficult point in learning chemistry. To balance the chemical equation, it is necessary to match the appropriate coefficient in front of each chemical formula of the chemical equation, so that the total number of atoms of each element on the left and right sides of the equation is equal, so that it is the correct way to balance the chemical formula, so how to balance the chemical equation?

The chemical equation of the trim should conform to the law of conservation of mass faith.

Correctly represent the mass ratio between reactants and products, providing accurate relationships for chemical calculations. There are several ways to do this:

1. Least common multiple method: suitable for the balancing of some relatively simple chemical equations.

2. Observation.

Applicable to some simple equations, the key to balancing with observation is to understand the essence of the reaction.

3. Pending coefficient method.

For some more complex chemical equations, the least common multiple method cannot be used, and it is difficult to use the observation method, so the undetermined coefficient method can be considered. With the undetermined coefficient method, not every chemical formula is preceded by an unknown, and the fewer the number of unknowns, the better.

There are many ways to balance chemical equations, in addition to the least common multiple method, the observation method, the undetermined coefficient method, there are also odd 2 times method, the coefficient is "1" method, etc.

(1) 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↑

Conservation according to the number of atoms.

Example: ch4+o2==

The first step is to judge the product, carbon element oxidation to form CO2, hydrogen element oxidation to form H2O, the second step is a methane has a carbon atom, so the generated CO2 has one, in the same way, two H2O are generated, and finally the carbon and hydrogen are balanced, and the number of oxygen is balanced.

ch4+2o2==co2+2h2o

According to the conservation of electrons.

Example: ch4+o2==

In the first step, the carbon element is oxidized to form CO2, the hydrogen element is oxidized to form H2O, and the carbon in the second step is -4 valence.

The carbon in CO2 is +4 valence, that is, after the reaction, eight electrons are lost, there must be atoms to get eight electrons, oxygen in oxygen is 0 valence, oxygen in CO2 and H2O is -2 valence, that is, each oxygen atom gets two electrons, so four oxygen atoms are needed in the reactant, that is, 2 O2, the reactants are balanced, and then the number of products can be known according to the conservation of the number of atoms.

ch4+2o2==co2+2h2o

Chemical Equation Balancing: The equation of chemical reactions strictly adheres to the law of conservation of mass.

First find out the elements that appear only once on both sides of the = sign in the chemical equation, first match the slag and then find the = good difference between the two sides, and sell the elements that accompany the quietly, and so on in the order of trim, until the trim.

Method of Trimming of Chemical Equations:

Least common multiple method: The starting point of the trim of the element that appears once on the left and right sides and has a larger number of atoms.

Ask the brother to come up with their least common multiple.

Divide the least common multiple by the number of atoms of the selected element, and the resulting quotient is the dust measurement number of the corresponding chemical formula.

And then launch others.

Examples: Aluminum and ferric tetroxide.

Odd spouse method: An element appears multiple times on both sides of a chemical equation, and the total number of atoms of that element on both sides is odd and even.

In this chemical formula where the number of atoms is odd, add an even number (2, 4) and then push the others.

Examples: ferrous sulfide and oxygen.

Seeing the chemical reaction in general, you can divide him into two categories. One is that the element does not have valency rise, that is, it does not involve redox, and the other is that the element has valency rise, that is, redox occurs. For the first type, just use the observation method to find the least common multiple of the element and balance it.

For the second category, the first thing to figure out is which elements have risen and fallen in valence, how much each has risen and fallen, find the least common multiple of rise and fall, first balance the gain and loss electrons, and then observe whether the number of atoms is flat, and the general number of gains and losses is balanced The atoms are also balanced, unless they are mixed with those atoms that are incomplete in redox, then refer to the first type to balance both sides through observation, be careful not to destroy the number of gains and losses that have been balanced, and remember to double on the left and on the right. In addition, if it is an ionic reaction, it is necessary to take into account that the number of positive and negative ions on both sides of the reaction should be symmetrical.

It's actually very simple, just remember that formula and everything will be solved.

Three Conservations: Conservation of Atoms, Conservation of Electrons, Conservation of Valency.

First with non-metallic elements, then with metal, then with oxygen.

Ninety-nine to one method, the number of atoms is conserved.