There is a sachet of powder that may contain one or more of CaNO32, NaSO4, Na2CO3, Fecl3, KCL

1) No FeCl3

2) There is Na2CO3 and no Ca(NO3)2

3)na2so4

4) There may be KCL

na2so4+bacl2==baso4↓+2naclna2co3+bacl2==baco3↓+2naclbaco3+2hno3==ba(no3)2+co2↑+h2o

A: 1There is no fecl3

2.There is Na2CO3 and no Ca(NO3)2

3.There may be Na2SO4

4.There is definitely Na2SO4 and Na2CO3 and possibly KCl

The chemical equation is:

In step 3: BACl2+Na2CO3=BAC3+2NaCl

bacl2+na2so4=baso4↓+2nacl

In step 4: 2HNO3+BAC3=BA(NO3)2+H2O+CO2

Analysis: Because the FeCl3 solution is yellow, the resulting solution is colorless as known from step 1, therefore, there is no FeCl3 in this package of powder.

Because Na2CO3 reacts with hydrochloric acid to form CO2 gas, and Na2CO3 reacts with Ca(NO3)2 to form CaCO3 precipitate, therefore, it is known from experimental step 2 that there is Na2CO3 and no Ca(NO3)2 in this package of powder.

Because both Na2SO4 and Na2CO3 can react with BaCl2 to form white BaSO4 and BaCO3 precipitates, and BaCO3 reacts with HNo3 to dissolve and generate CO2 gas, BaSO4 is insoluble in HNo3, therefore, according to the experimental procedure, there must be Na2SO4 and Na2CO3 in this package of powder, and there may be KCL.

Good luck with your studies!

1."The solution is colorless"Explain that there is no FeCl3 because the solution containing Fe3+ is yellow; "There is no sediment"This indicates that Ca2+ and CO2- cannot coexist, i.e., Ca(NO3)2 and Na2CO3 cannot coexist.

2."Hydrochloric acid is added to create bubbles"It is stated that there is CO3 2-, that is, there is Na2CO3, and it was said earlier that Na2CO3 and Ca(No3)2 cannot exist at the same time, so there is no Ca(No3)2

3.There are CO32- and SO42- that produce white precipitates with Ba2+, and since it has been determined earlier that there is CO32-, there may be SO42-.

4.Since the BaCO3 precipitate is soluble in acids, the Baso4 precipitate is insoluble in acids. Since it is"The precipitate is partially dissolved"(i.e., part of the precipitate is dissolved, and part of the precipitate is insoluble) so the precipitates obtained earlier are BaCO3 and Baso4, indicating that there is Na2SO4 in the solution

na2so4+bacl2==baso4↓+2nacl

The mixture definitely has Na2CO3 and Na2SO4, certainly no FeCl3 and Ca(NO3)2, and possibly KCl

ca(no3)2

It does not react with Na2CO3, otherwise it violates the principle of strong acid HCl to weak acid H2CO3).

because he is yellow when dissolved in water.

2.There are bubbles, which means that there is gas coming out, so there must be Na2CO3, since there is Na2CO3, there must be no Ca(NO3)2, otherwise calcium carbonate precipitate will be generated.

3, it shows that there is Kenneng and Na2SO4.

4.The precipitation dissolves and bubbles are generated, which should be reflected in the BAC3 and nitric acid generated in step 3, but as I said before, there must be Na2CO3, so I think this question is a bit problematic.

Equation: Na2CO3 + BACL2 = 2NaCl + BaCO3 (white precipitate) Na2S04 + BaCl2 = 2NaCl + BaSO4 (white precipitate) BaCO3 + 2Hno3 = Ba (Hno3) 2 + H20 + CO2 (gas).

1.The solution is colorless and transparent, indicating no Fe3+ (yellow), no precipitate, no Ca(No3)2 or no Na2CO3

2.If there are bubbles, it means that there is Na2CO32 and there is no Ca(NO3)23. A white precipitate is produced, indicating that Na2SO4 may be present

4.The precipitate is partially dissolved, and the dissolved is BaCO3, and the remaining precipitate is Baso4BaCl2+Na2CO3=BaCO3 (precipitate) + 2NaCl, and BaCl2+Na2So4=Baso4 (precipitate) + 2NaCl

BaCO3 + 2Hno3 = Ba(NO3)2 + H2O + CO2 (gas).

1, FeCl3 yellow.

Ca(NO3)2 NaCO3 and HCl react to form CO2, and NaCO3 and Ca(No3)2 cannot coexist.

Na2SO4 precipitates with barium chloride.

There is no effect on the reaction.

3 BaCl2 + Na2CO3 = BaCO3 (precipitate) + 2NaClBaCl2 + Na2SO4 = Baso4 (precipitate) + 2NaCl

There is a precipitate, and after adding dilute hydrochloric acid, all dissolve, and at the same time produce colorless gas, the precipitate must be barium carbonate, from which it can be seen that there must be barium nitrate in the raw powder, at least one of sodium carbonate and potassium carbonate, and there must be no sodium sulfate and potassium sulfate.

The filtrate is used for flame color reaction, and the purple flame is observed through the blue cobalt glass, so there must be potassium carbonate in the raw powder, and it is uncertain whether there is sodium carbonate.

Sodium chloride is also not certain, therefore.

The mixture must contain (barium nitrate, potassium carbonate).

Must not contain (sodium sulfate, potassium sulfate).

It is not certain whether it contains (sodium carbonate, sodium chloride).

1) From the experiment, it can be seen that there must be no FeCl3 in the original powder, which can indicate that there is no MgCl2 in the original powder

Experiments have proved that there is no Na2CO3 in the original powder

2) According to the above experiments, it can be concluded that the substances that may be present in the original powder are NaCl and Na2SO4.

3) From the experiment, it can be seen that there must be NaCl in the original powder

From the experiment, it can be seen that there must be no Na2SO4 in the original powder

Try to write out the chemical formulas of the solutes in solutions A and B used in the two experiments: A Agno3 and B BaCl2

From the experiment, it can be judged that the powder must be contained

Copper sulphate CuSO4

Barium chloride. bacl2

Must not contain calcium carbonate.

CaCO3 may contain:

Potassium nitrate Kno3, sodium sulfate Na2SO4

In general, insoluble strong acid salts are insoluble in acidic solutions, i.e., insoluble in nitric acid.

For example, silver sulfate.

Wait a minute. But at the secondary school level, it is generally only considered.

Silver chloride and. Barium sulfate is used for these two substances.

A should be chosen

Because the serum in the lower layer is colorless, there is an option for cuso4 to be ruled out first, because the white precipitate is partially dissolved.

Among the above substances, only BaSO4 can be composed of insoluble dilute nitric acid, so there must be Na2SO4 and BACL2

Then it is also said that there is a colorless gas produced.

Of the above-mentioned substances, only CaCO3 is insoluble in water and reacts with dilute nitric acid to produce bubbles, and it is impossible to determine whether Kno3 is left or not.

Since the supernatant is blue, only the solution of Cu2+ is blue, and there is CuSo4 directly due to the white precipitate insoluble in dilute nitric acid, only Baso4 is insoluble in dilute nitric acid in the combination of all ions, and there must be BaCl2 (SO42 - from CuSO4).

The white precipitate is insoluble in dilute nitric acid, and caCO3 is a white precipitate soluble in dilute nitric acid, which is not in line with the topic, so there must be no caco3

The remaining Na2SO4 and KNO3 cannot be judged and there is no characteristic response.

Answer: There must be CuSO4, BACL2

There must be no caco3

There may be Na2SO4, KNO3

1. From experiment (1), it can be seen that there must be no FeCl3 and CuSO4 in the original powder

2. From experiment (2), it can be shown that the original powder must not have Na2CO3, and the reason is explained: if there are bubbles

3. From experiment (3), it can be seen that the original powder must not have NH4Cl, and there must be MGCL2

4. From experiment (4), it can be seen that there may be NaCl in the original powder (fill in one of "certainly, definitely, maybe").

5. Write the chemical equation NaCl+AGNo3==AGCL +Nano3 that generates the white precipitate in experiment (4).

Analysis: (1) Sampling is dissolved with water to obtain a colorless permeability discrimination solution, from which it can be judged that there must be no CuSO4, because the copper ions are blue in the solution;

2) take a small amount of the obtained solution, add NaOH solution dropwise, observe, there is no obvious phenomenon, from which it can be judged that there must be no MgCl2, because the reaction of hydroxide ions and magnesium ions produces white precipitates;

3) Take a small amount of powder, add dilute hydrochloric acid dropwise, and observe, there are bubbles, which proves that there must be Na2CO3.

Based on the above experiments, the conclusions that can be drawn are:

1. The substance that the original powder must contain is (Na2CO3), and the substance that may contain it is (NaCl). Write the chemical equation (Na2CO3+2HCl=2NaCl+CO2+H2O).

2. If you want to prove that the substances may be contained, how will you carry out the experimental operation (take a small amount of powder, dissolve it in the water, add silver nitrate solution dropwise, and produce a white precipitate, then it proves that there is sodium chloride).

1. The substance that the original powder must contain is (Na2CO3), and the substance that can contain bending is (NaCl). Write out the chemical equation about nano (Na2CO3+2HCl=2NaCl+CO2+H2O).

2. If you want to prove that the substances that may be contained, how will you perform the experimental operation (adding silver nitrate with discoloration precipitation indicates that there is sodium chloride).

There must be NaSO4 and NaCO3 in this package of solids, and there is no CuSO4 and CaCl2 in the -fixed answer beam, which may slow the source of KCL

Na2SO4 + BA(NO3)2= BASO4 + 2nano3Na2CO3 + BA(NO3)2 = BACo3 + 2nano3BACo3 + 2HNO3 = BA(Qing Nayun NO3)2 + CO2 + H2O

CaCO3, NaSO4 BACL, CaCO3, NaSO4, NaCO3 BaClCAC3, NaSO4 BACL NASO4 is known from (1) There is no CuSO4, because CuSO4 itself is blue, and the white precipitate is partially dissolved, indicating that there is CaCO3 and acid-insoluble precipitate BASO4 to produce, then there must be NaSO4 and BACl, because these two form BaSO4 in water

Hope it helps.

CuSO4 is blue and can be ruled out, the precipitate is partially dissolved, indicating that there are two kinds of precipitate, and one is insoluble in dilute nitric acid, which should be barium sulfate, and the one given in the question can be obtained by the reaction of Na2SO4 and BaCl2, and there is also a soluble precipitate, which is CaCO3, so the answer is CaCO3 Na2SO4 BaCl2 Well, you made a mistake.

(1) From the experiment, it can be seen that there must be no MgCl2 in the original powder; (Because only magnesium carbonate is insoluble in water).

Experiments can show that there is no FeCl3 in the original powder; (Because only Fe3+ reacts with OH- to form a precipitate).

Experiments have proved that there is no Na2CO3 in the original powder(Because only CO3- reacts with H+ to form CO2 gas).

2) According to the above experiments, it can be concluded that the substances that may be present in the original powder are NaCl and Na2SO4.

To determine whether the original powder is a pure substance or a mixture, proceed with the following experiments:

take a little of the obtained solution, add solution A dropwise, and a white precipitate is generated; (The most typical precipitations in high school chemistry are silver chloride and barium sulfate.) Therefore, here we can infer that the added solution A is AGNO3, which generates a silver chloride precipitate. Why not barium sulfate?

Continue to add solution A dropwise until there is no white precipitate generated, let it stand, take a little supernatant and add solution B dropwise, and there is no obvious phenomenon. (Continue to add solution A dropwise until no white precipitate is formed, indicating that AG- has completed the reaction of Cl- in the solution.) At this time, we can add BaCl2 dropwise to check whether there is a white precipitate of BaSo4.

There is no obvious phenomenon, indicating that there is no Na2SO4 in the original powder. The inference here is in line with the meaning of (3) below, and if agno3 and na2so4 are interchangeable, it is not in line with (3).

3) From the experiment, it can be seen that there must be NaCl in the original powderFrom the experiment, it can be seen that there must be no Na2SO4 in the original powder

Try to write out the chemical formulas of the solutes in solutions A and B used in the two experiments: A Agno3 and B BaCl2