A high score will be rewarded for a chemical calculation question, and a chemical calculation proble

The conditions for this problem are insufficient, and the conditions are insufficient to solve it!

1）2kclo3=2kcl+3o2

In A, oxygen is produced as 6

So the mass of potassium chlorate can be found to be 245

So manganese dioxide in the 6g mixture is 6

Answer: potassium chlorate, manganese dioxide.

2) In B, there is potassium chlorate in 12g of mixture.

There is manganese dioxide.

The mixture in B is 2 times that of A, and the gas released by complete heating should be 2 times that of A, and if it is, there should be 12 left, and the actual remainder.

Or if you look at the gas produced in B as 12 and in A, it is obvious that the solid in B is not completely reacted. Therefore, it can be seen that the solid mixture is not completely reacted.

The remaining solids are a mixture of potassium aluminate, potassium chloride and manganese dioxide.

The oxygen produced in B is 12

It is exactly the same as in A, so the remaining potassium chlorate in B is.

The manganese dioxide mass is:

The mass of potassium chloride is.

2kclo3=2kcl+3o2

The reduction in mass is due to the amount of oxygen.

So kclo3 is.

mNO2: KCLO3 in 12g

The reduction in the mass of mno2 is due to the amount of oxygen.

So the reaction.

KCl is generated, so the remainder is mNO2

Manganese dioxide is used as a catalyst.

Solution: Mixture disturbs the nucleus, m(c):

12=m（cuo）：

The product of Cuo being reduced by C is Cu2O (Buyu red).

Cuo is all reduced to Cu2O's chemical equation Slow Mining:

4cuo+c→2cu2oco2xx=

First of all. Cuo is reduced by c is the chemical equation of the celery-producing mass erection copper cu (red) column:

Precipitate 2g (is calcium carbonate) divided by the relative molecular mass of calcium carbonate.

Multiply by the relative molecular mass of CO2 (carbon dioxide).

Let the number be x (x is the mass of carbon dioxide generated).

Columnize the University or Lead Equation:

x divided by the relative molecular mass of CO2 (carbon dioxide).

Multiply by the relative molecular mass of Cu.

That's it.

2/3 means that 2 FE needs 3 O's, so 1 FE needs "2/3" O's.

Three-quarters means the same thing.

In other words, feo and o are (1:1) in feo, fe2o3 and fe and o are (2:3).

Fe3O4 where Fe and O are (3:4) The problem now requires that Fe have the same mass, so they are both reduced to 1, then the corresponding O becomes 2/3 and 3/4

NaHCO3 = (heating) = Na2CO3 + CO2 + H2O while sodium carbonate is non-reactive, that is, the mass reduced is the mass of carbon dioxide and the mass of water = (1) You can think of water and carbon dioxide as a whole.

NaHCO3 = (heating) = Na2CO3 + CO2 + H2OX

x= Because it is divided into two equal parts (

2) The second question only requires the amount of the corresponding substance, remember that the sodium ion of sodium carbonate should be *2, and the sodium bicarbonate should not be used. The two can be added up, and the amount of sodium ions can be found directly, and then the concentration of sodium ions can be found directly with c=n v.

It doesn't matter if it's a problem or a method, what is the use of learning it. It's enough to keep this trick in mind: a system of equations.

Let the amount of sodium carbonate in the original sample be x mol and the amount of sodium bicarbonate be y mol.

The first equation is based on "Take a mixed powder sample of sodium carbonate and sodium bicarbonate".

106x+84y=

The second equation: according to the "after cooling, its mass is called."

x= y=

The second question is c=(2x+y).

Remember: don't be frightened by any difference method, in fact, the difference method comes from the idea of eliminating elements when solving a system of equations, if you can't understand it, just list the equation to solve the equation, and you don't need any "method".

There is no chemical equation for this, because no chemical reaction takes place.

The mass fraction of nitrogen in this nitrogen fertilizer sample = 90%*Mass fraction of n in NH4NO3 = 90%*(14*2 80)=

80 of them is the relative molecular mass of ammonium nitrate.

**Correct solution

Solution: If the mass of the ammonium nitrate sample is 100, then the mass of the ammonium nitrate is 90. The molecular mass of ammonium nitrate is 100, then.

The mass fraction of nitrogen in ammonium nitrate is 28 100=, then the mass fraction of nitrogen in ammonium nitrate is x 90 100 x100%=

That's it. I don't know if I'm satisfied!

You don't need an equation for this question! All you have to do is take the percentage of nitrogen in the chemical formula and multiply it by 90 percent.

No equation calculation is required, just the amount of N in NH4NO3 can be calculated: