How do you balance a chemical equation with a cluster of atoms? 10

Al(OH)3+H2SO4--- AL2(SO4)3+H2O considers AL2(SO4)3 as a whole, where the ratio of AL3+ to SO42- is 2:3

Al(OH)3 is preceded by 2, H2SO4 is preceded by 3, and finally H is conserved, H2O is preceded by 6, in fact, this equation can be balanced by the reaction substance, and the essence is acid-base neutralization, that is, H++OH-=H20

1molAl(OH)3 provides 3molOh- and 1molH2SO4 provides 2molH+, so 2molAl(OH)3 provides 6molOh- and 3molH2SO4 provides 6molH+ to produce 6mol water, and 1molAl2(SO4)3 is generated at the same time

The equation is: 2al(oh)3+3h2so4=al2(so4)3+6h2o

Another equation is to use electron conservation balancing, i.e., the loss of electrons by the reducing agent is equal to the gain of electrons by the oxidant.

zn+hno3=zn（no3）2+no+h20

1molZN loses 2mol electrons, oxidizing nitric acid 1mol gives 3mol electrons, and according to the conservation of electrons, the ratio of the amount of matter between the two is 3:2

That is, 3molZN oxidized 2molHN3 reacts to generate 3molZN(NO3)2 (ZN conservation) and 2molNO(n conservation), and 6mol acidic HNO3 is required to generate 3molZN(NO3)2, so the total number of HNO3 participating in the reaction is: 2+6=8mol, and finally H is conserved, H2O4mol

The equation is: 3zn+8hno3=3zn(no3)2+2no+4h20

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 a chemical equation where 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 are introduced, the fraction can be matched with 4), from which it is inferred that the first 2 of C2H2, the formula 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

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

c) Observational balancing.

Sometimes there will be a substance with a 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 from which the H2 coefficient is 4, indicating the conditions, ** Change to an equal sign:

3fe+4h2o==fe3o4+4h2

Al(OH)3+H2SO4--- AL2(SO4)3+H2O considers AL2(SO4)3 as a whole, where the ratio of AL3+ to SO42- is 2:3

Al(OH)3 is preceded by 2, H2SO4 is preceded by 3, and finally H is conserved, H2O is preceded by 6, in fact, this equation.

It can be leveled with the reaction parenchyma, which is acid-base neutralization, i.e., h++oh-=h20

1molAl(OH)3 provides 3molOh- and 1molH2SO4 provides 2molH+, so 2molAl(OH)3 provides 6molOh- and 3molH2SO4 provides 6molH+ to produce 6mol water, and 1molAl2(SO4)3 is generated at the same time

The equation is: 2al(oh)3+3h2so4=al2(so4)3+6h2o

Another equation is to use the electron conservation balance, i.e. the reducing agent.

The loss of electrons is equal to the gain of electrons by the oxidizing agent.

zn+hno3=zn（no3）2+no+h20

1molZN loses 2mol electrons and is oxidizing.

1mol of nitric acid gives 3mol of electrons, according to the electron code, the amount of both substances.

The ratio is 3:2

That is, 3molZN oxidizes 2mol Hno3 to generate 3molZN(NO3)2 (ZN conservation) and 2molNO(n conservation), and 6mol acidic HNO3 is required to generate 3molZN(NO3)2, so the total HNO3 of Samson defeat reaction is: 2+6=8mol, and finally H is conserved, H2O4mol

The equation is hand this: 3zn+8hno3=3zn(no3)2+2no+4h20

A very easy to use trimming method that can simplify chemical equations: the priority atomic cluster method, the balancing method that you regret after watching it for a while, the balancing method of chemical equations, if you don't master the method, you encounter more chemical formulas, or chemical formulas with more elements, and your thinking is easy to mess up. Hurry up and learn and clear your mind.

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

First of all, you made a mistake when you sent this question, it should be:

Al(OH)3+HCl——AlCl3+H2O This question should belong to the reaction of acid-base neutralization, as long as you notice that the essence of acid-base neutralization is OH+H2O, and it is a one-to-one reaction, then the reaction can be balanced.

There are 3 oh in the front, then it takes 3 h in the back, and the void is attacked into 3 parts of water.

So the trim factor should be: Al(OH)3+ 3 HCl—AlCl3+ 3 H2O. That's it.

When the manuscript is funny, I guess you should be a junior high school student brother, the content about acid-base neutralization is the knowledge of the second half of the third semester of junior high school, you may not understand it for the time being.

Acid-base reaction: OH reacts with H to form water.

Balancing; Let H and OH be equal first, and then match the others.

3 zn+ 8 hno3= 3 zn（no3）2+ 2 no+ 4 h20

It can be used, but the valency change of redox is better.

zn+ hno3= zn（no3）2+ no+ h20

x 2y x 2y-2x y

6y=6x+2y-2x+y

x/y=3/4 x=3 y=4

3 zn+ 8hno3= 3zn（no3）2+ 2 no+ 4 h20

zn+ hno3= zn（no3）2+ no+ h20

Variation 2 Variation 3

3 zn+ 2*3+2 hno3= 3 zn（no3）2+ 2 no+ 4 h20

3 zn+ 8hno3= 3zn（no3）2+ 2 no+ 4 h20

Oxygen is generally put last.

zn+ hno3= zn（no3）2+ no+ h20

x 2y x 2y-2x y

I level the number of oxygens.

6y=6x+2y-2x+y

x/y=3/4 x=3 y=4

3 zn+ 8hno3= 3zn（no3）2+ 2 no+ 4 h20

There are no clusters of atoms here.

zn 0 valence change +2 change 2

n +5 valence change +2 (no) change 3

Least common multiple 6.

ZN with 3, NO with 2, others with rematch.

3zn+8hno3=3zn（no3）2+2no+4h2o