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1) 44*73*10%/(

2) NaCl and HCl

3) Na2CO3 has 106*73*10% (, NaCl has, and the reaction produces NaCl, so the mass fraction of the solute is.

Fe2+ :Fe3+ ==1 :2 and 480gFe2O3, i.e. 3 mol Fe2O3 in 3*2==6 mol Fe

Then Fe2+ has 6*1 (1+2)==2 mol and Fe3+ has 6*2 (1+2)==4 mol

After the reduction of one Fe3+, there is only one Fe2+ to produce 2 mol Fe2+, i.e., 2mol FeO, or 1 molFe2O3

According to C+2 Fe2O3=4FeO, ten CO2 C is mol i.e. 6 g

It doesn't have to be that complicated.

It is calculated that 480gFe2O3 contains Fe6mol, because the amount of Fe2+ and Fe3+ is 1; 2, then, there must be 2mol of Fe involved in the reduction reaction, that is, 1mol of Fe2O3

According to the proportion, calculated.

C+2 Fe2O3=4FeOTen CO2

x 1x=, too.

Your steps are too messy, and there are typos, I don't understand much, but 12g is definitely not right, it should be 6g.

1.It is required that 2-valent Fe and 3-valent Fe are 1:2, so 160g of 480g of raw material (Fe2O3) is reduced to 2-valent by C, and the remaining 320g does not react and is still 3-valent.

2.The c required to calculate the reduction of the 160g principle is 6g according to the reaction equation.

You've got yourself in it.

The ratio of the amount of ferrous ion Fe2+ to iron ion Fe3+ in the catalyst is 1:2, which is 1 3 in 480 gFe2O3, that is, 160 g.

C+2 Fe2O3=4FeOTen CO2

m（c）=160g*12/320=6g

Solution: (1) After adding dilute hydrochloric acid for the second time, the mass of "beaker + residual substance" is: so the answer is:

2) The mass of carbon dioxide gas produced in the reaction is 50g + 100g + and the mass of calcium carbonate in the shell sample is x

caco3+2hcl=cacl2+h2o+co2↑100 44x

100x=x=45g

Mass of calcium carbonate in shell sample: 45g50g 100%=90%A: The mass fraction of calcium carbonate in shell sample is 90%.

3) The mass of carbon dioxide gas produced after the third experiment is assumed that the mass of calcium carbonate consumed is x, and the mass of calcium chloride produced is ycaco3+2HCl=CaCl2+H2O+CO2 100 111 44x y

100x=x=30g

111y=y=mass fraction of solute in solution:

Answer: The mass fraction of the solute in solution.

1,x=,Answer: Because the same amount of 30g of hydrochloric acid is added each time, then the amount of each time is also incremental, the third subtraction and the second increase each time, 2) The mass fraction of calcium carbonate in the shell sample is 90%.

Answer: If the amount of hydrochloric acid increased by hydrochloric acid for the last time is large, then the hydrochloric acid is excessive, and the CaCO3 reaction is complete. The CO2 generated by the reaction runs away.

Then CO2 mass = 50 + 100 + is mole.

The caco3 consumed is also moles. CaCO3 has a molecular weight of 100The mass of CaCO3 is 45 grams.

Mass fraction of calcium carbonate = 45 50 =

3) The mass fraction of solute in the solution obtained after the third experiment =

When water is added, there is no color and no precipitation. (The color is Cu2+ is blue, and the precipitate is Ca2+ and [CO3]2- to generate CaCO3).

Therefore, (1) the powder must not contain ---CuSO4 and CaCl2, and the name of operation 1 is --- filtration.

p,s.Is operation 1 to separate the precipitate from the solution after adding BaCl2 to form a precipitate? ,,It's not very clear.,It should be filtering.。。

2) In order not to interfere with the judgment of the next experimental phenomenon, it is best to choose dilute nitric acid (HNO3) for reagent A

The reason why sulfuric acid cannot be used is because the next step is to add BaCl2, where the addition of H2SO4 will introduce the [SO4]2- interference test, 3) To check whether KCI exists, the method that can be used is to add AGno3 solution, there is a precipitate and there is Cl-, and it has been proved that there is no CaCl2, so if there is a precipitate, it is KCL

Adding water is colorless, there is definitely no CuSO4 (blue) Adding reagent A can produce colorless gas, it can only be CO2, so there must be Na2CO3 So there must be no CaCl2 in the original mixture Otherwise, a white precipitate will be produced Adding barium chloride to produce a white precipitate can confirm the presence of sulfate.

So it must contain Na2SO4 Na2CO3

Must not contain CuSO4 CaCl2 may contain KCl Operation 1 is filtration.

A is better to choose HCL

Determine the potassium ion with the flame color reaction and confirm the Cl with Agno3

Colorless solution: no CuSO4;

Colorless gas: contains Na2CO3, does not contain CaCl2;

There is a precipitate with the addition of BaCl2, containing Na2SO4;

Filtration; Nitric acid; influencing the determination of Na2SO4;

Color reactions.

First, dissolved in water to obtain a colorless solution, it proves that there is no CuSO4, because the aqueous solution of copper ions is blue;

Then, after adding a, there is a colorless gas, which proves to contain Na2CO3, because only carbonate ions in the remaining four substances can produce the colorless gas carbon dioxide; At the same time, calcium chloride can also be excluded, because calcium ions and carbonate ions can produce precipitation (there is no turbidity when dissolved with water in the front);

In the third step, the addition of barium chloride produces a white precipitate, which seems to prove the existence of sodium sulfate; However, don't ignore A, the title doesn't say A is not sulfuric acid!

Therefore, only Na2CO3 can be determined, and only CuSO4 and CaCl2 are non-existent. The other two substances cannot be determined.

a. It is best to use dilute hydrochloric acid; KCl was tested by flame color reaction.

(1) CuSO4, CaCl2 filtration (colorless solution, no Cu2+ CO3 2- and Ca2+ do not exist at the same time, and because there is gas in the next step, there is Na2CO3, no CaCl2).

2) HCl introduces SO4 2-, which affects the test results.

3) Add AgNO3 solution, if there is a precipitate, there is KCL

Stir thoroughly.

2.The HCl solution was added to introduce the reaction of SO4- and Ba+ ions, which affected the experimental results.

3.Take a small amount of powder, dissolve, add dilute nitric acid and add silver nitrate to see if there is a precipitate, if there is a precipitate, then it contains, and if there is no precipitation, it does not contain.

1. Copper sulfate (aqueous divalent copper solution is blue) dissolved.

2. Nitric acid (the gas produced can be known to be the addition of acid, but it cannot be considered that the addition of chloride ions and sulfate ions, and the artificial addition of sulfate will have an impact on the inspection process in one step.) ）

3. The flame color reacts, and the flame of potassium ions is purple.

1) The powder must not contain CuSO4, sedimentation filtration, (2) Hno3 solution, if dilute sulfuric acid is used, sulfate is introduced, resulting in 3-step misjudgment and interference. (3) Add silver nitrate solution to the filtrate filtered in 3 steps, if there is a white precipitate, it means that there is KCI

o3+o=202 The above statement is correct.

There is element N in the reactant and G and H2 in the product, indicating that there must be element N in G. The molecular mass of 81 and the boron mass fraction are known to be 3, while hydrogen is also known to be 6

1 NH3 gives 2 H2, and G is obtained, and there is no other element, so it is B2H6

The equation of the total reaction o3 + o = 2 does not change the nitrogen oxides in this reaction, so it plays a catalytic role.