How to solve the overall idea of this chemistry problem?

Sodium sulfite decomposes to produce sodium sulfide, which is known to be disproportionated, and the other solid is sodium sulfate.

In the chloride ion environment, barium sulfite precipitate cannot be formed, and it can only be oxidized. Magenta is used to monitor sulfur dioxide, and copper sulphate is used to absorb sulfur dioxide.

The hydrogen sulfide produced by the reaction of sodium sulfide with hydrochloric acid can undergo a neutralization reaction of sodium sulfite.

The parsing has been spoken. AB is in the same row, the second period, BE is in the same column, ACD is all non-metallic, A is in the second row, CD is in the third row, the number of electrons in the outer shell of CD is twice that of B, Cd only has silicon and sulfur, D E, it cannot be chlorine, E is chlorine, and Launch A is nitrogen, and B is fluorine.

1. This question A is (NH4)2S

2. When X is a strong acid, B is H2S, C is S, D is SO2, E is SO3, and F is H2SO4

d e 2 SO2 + O2 = 2 SO3 (conditions: heating, catalyst).

3. When X is a strong base, B is NH3, C is N2, D is NO, E is NO2, and F is HNO3

e→f 3 no2 + h2o = 2 hno3 + no

For this, you have to start with A, because A can react with both strong acids and strong bases, and this class of substances can be summarized. For example, Al(OH)3, Al, ammonium sulfide, ammonium hydrosulfide, ammonium bicarbonate and so on. In this topic, it is (NH4)2S, and continuous oxidation is also regular, you can summarize it yourself, generally H2S---S---SO2---SO3

nh3---n2---no --no2---hno3,na---na2o---na2o2,c --co --co2

Under normal circumstances, if the inorganic is tested, there is more sulfur and nitrogen

f is a strong acid, and from the perspective of transformation, it must be an oxygenated acid.

Therefore, there are H2SO4 and HNO3 that are compliant

nh3---n2---no---no2---hno3

h2s---s---so2---so3---h2so4

X is a strong base, i.e. B is NH3

1) A is: (NH4)2S

2) When X is a strong acid, A, B, C, D, E, F all contain the same element, and F is: H2SO4

The chemical equation for d e is: 2SO2 + O2 = catalyst heating = 2SO3

The ionic equation for the reaction introduced into dilute nitric acid is: 3SO2 + 2H2O + 2NO3 - = 3SO4 2- +2NO + 4H+

3) When X is a strong base, A, B, C, D, E, F all contain the same element, and F is: Hno3

The chemical equation for E f is: 3NO2 + H2O = 2HNO3 + NO

b have a surplus i.e. hydrochloric acid is completely consumed h2 from hydrochloric acid.

All are equal to ditto.

Speaking of magnesium, aluminum, zinc, and iron?

(1) Choose B, because the hydrochloric acid consumed is equal, so the hydrogen produced is equal, and it has nothing to do with the rest!

2) Equal, and (1) the same!

All are equal, and the mass of hydrogen produced depends on the amount of hydrochloric acid.

b.a=b=c, excess metal, calculated by the amount of hydrochloric acid, hydrochloric acid is equal, and the hydrogen produced is equal.

The principle is the same as above, the amount of hydrogen produced is equal.

Group 1 10g is obviously insufficient Ca to produce 2g precipitate, 5g 1g precipitate when the carbonate is sufficient, while group 30g 30g should produce 6g, but only 5g (maximum precipitation), 25g 5g precipitate, it can be known that 25g CaCl2 solution is just completely precipitated, group 2 (Ca is insufficient) 20g 4g precipitate, m = 4

10g + 40g 5g according to this count the carbonate ions, and then it's done.

By the way, sodium carbonate is Na2CO3

You have a problem with the topic.

It is conditionally known that as long as CaCl2 is greater than 30g, the CO3 root can completely react to form CaCO3 (because the precipitate mass does not increase when CaCl2 is added to 40g), but the precipitate is only 5g, and the amount of the substance is also known by the reaction formula, that is to say, as long as CaCl2 can completely react with the CO3 root in the mixture, and the 10g CaCl2 in the first group can also completely react with Na2CO3 in the mixture, how can 2g of precipitate be generated, I don't understand it.

Also, you said that you added a certain quality fraction of cacl2 in your question, but you only gave quality.

4g。When it is 30g or 40g, it can not be increased, indicating that CaCl2 has been overdosed. Obviously the first group is insufficient, if the second group is sufficient, then m = 5, which is not possible because 10 g of CaCl2 emulsion can produce up to 2 g of precipitate.

Therefore, the second group is not excessive, and m 4 grams can be obtained proportionally.

m=4g

Because of the comparison of this test, after the amount of calcium chloride increases, the amount of precipitation increases, indicating that the first calcium chloride reaction is complete, so that the mass ratio of calcium chloride and the precipitate is 10:2, so the second time with 20 grams of calcium chloride can generate 4g of precipitate.

Data 3 and 4 indicate that the CaCl2 solution is excessive, and the amount of precipitate generated in data 1 is less than 1 2 of the complete precipitate, so m = 4

n(no)=

According to the ionic equation: 3Cu + 8H+ +2NO3- =3Cu2+ +2NO+4H2O

After the first reaction, the remaining NO3-: in the solution, and then add dilute sulfuric acid to complete the remaining NO3- reaction, and consume Cu: 3Cu + 8H+ +2NO3- =3Cu2+ +2NO+4H2O

m(cu)=

Answer If you have any questions about choosing A, please follow up.

3cu+8hno3=3cu(no3)2+2no+4h2o8mol

n(hno3)

n(hno3) =

3Cu + 8H + + 2NO3-= 3Cu2 + + 2NO+ 4H2O (dilute sulfuric acid to introduce H+, which is equivalent to nitric acid again).

192g 2mol

m(cu)m(cu)=select a

Both CO and CO2 are available.

Qualitatively: 12gc, if all generate co, 16go2 is required; If all CO2 is generated, 32go2 is required. And the actual consumption is 20go2, so the product must have both.

Quantitative calculation: If the amount of the substance involved in the formation of C in the CO reaction is x, then the amount of the substance involved in the formation of C in the CO2 reaction is 1-x. From this, the amount of substances consuming O2 = x 2 + 1-x = 1-x 2 = is obtained, and the products are 21 gco and 11 gCO2

[Analysis].

The corresponding method of a typical binary mixture is a system of binary equations, and the calculation is very simple if the unknown is the quantity of matter.

First calculate the reaction with hydrochloric acid.

The amount of OH- is expanded by a factor of 10 to be the total amount of matter of the original mixture, and then the sum of the masses and the sum of the quantities of the matter are a series of equations.

Solution] n(hcl)=

original mixture.

n(oh-)=

Let the quantities of substances of naoh and koh be respectively.

xmol，ymol

x＋y＝0．0525

40x ＋56y ＝2．5

x ＝0．0275

y＝0．025

m（naoh）＝40g／mol＊0．0275mol＝1．1gm（koh）＝0．025mol＊56g／mol＝1．4g

Let the masses be xy

The amount of substances for each substance OH can be found.

The total amount of substances from hydrochloric acid to which OH is introduced.

and xy and for.

A system of equations emerges simultaneously.

Just solve the system of equations.