How to balance chemical equations? 15 bounty

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 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↑

I usually look for electronic gains and losses to match.,The last two sets of scores are divided by a line.,I don't feel that it's difficult ==.

Mr. Bai Hechou | 2010-01-20

Specific topics are analyzed on a case-by-case basis.

Atoms are conserved, mass is conserved.

First with the complex, then with the simple.

Observe! According to the conservation of electrons.

Chemical equations are very simple in junior high school, but complex equations in high school and are not easy to match with observation. All of this requires a new approach:

The most common is trimming with a redox reaction. First of all, find out the price increase and fall of the elements, which elements have changed before and after the reaction, and how much has changed, all of which need to be marked. Then use the least common multiple method to match the previous coefficients, and then observe the rest of the trim.

For example: Cu+Hno3 (dilute)--Cu(NO3)2+NO+H2O first observe the change of the valency of copper and nitrogen: copper increases the 2 valence, and the nitrogen element decreases the 3 valence, so Cu and Cu(NO3)2 are matched with 3 in front of NO, and 2 is assigned before NO, and then the nitrogen in HNO3 is divided into Cu(NO3)2 and NO, so 8 is assigned before HNO3, and 4 is matched before water

There are several methods of trimming, the most common being observation. When encountering complex chemical formulas, the most complex ones are standard. After matching one, immediately chase another element in the chemical formula, and after chasing the next one, the straight is finished.

Theoretically, it is very laborious, as long as you match a few more, you will have experience. Generally speaking, as long as the equations you have learned can be balanced, it is no problem, and you can expand on others outside of class.

1.Conservation of Atoms 2Valence changes of oxidizing and reducing agents.

Cu changes from +2 to +1, and s changes from -1 to -2 and +6

Let's look at the price reduction part first, generating 1mol of Cu2s, and the total number of transferred electrons is 2*1+1=3E-.

And s changes from -1 to so42-, transferring 7e-. So it is obvious that the Cu2S coefficient is 7, and there are 3 S for the price increase.

The price of S has been reduced by 7 and the price has increased by 3, a total of 10 have changed, so the FES2 coefficient is 5. This means that out of 5*2=10 s, 7 have a reduction reaction and 3 have an oxidation reaction.

The atom is conserved, the FeSO4 coefficient is 5, the CuSO4 coefficient is 14, the H2SO4 coefficient is 12, and the H2O coefficient is 12.

14cuso4+5fes2+12h2o=7cu2s+5feso4+12h2so4

14cuso₄ +5fes₂ +12h₂o = 7cu₂s + 5feso₄+ 12h₂so₄

1 : n =14 : 14+7）

n = answer A

Analysis: Among the 5 moles of Fes, 7 moles of Fes decreased from -1 to -2, and 3 of them increased from -1 to +6

Total valency reduction: 14 + 7 = 21

Total increase in valency: 6 (-1) 3 = 21