Chemistry issues. How to identify, chemical reaction identification

First, the flame color reaction was used to distinguish potassium and sodium salts.

Potash appears purple through blue cobalt glass.

Sodium salt is yellow.

Sodium salt plus hydrochloric acid has gas is sodium carbonate, and sodium chloride is not.

Potassium salt is also added to hydrochloric acid, and potassium carbonate is potassium carbonate that has gas, and potassium sulfate is not added.

First, it is burned to distinguish between potassium and sodium salts by a flame reaction. Dilute hydrochloric acid is then added to the potassium and sodium salts. You can tell the difference.

Color test reflection. Yellow with dilute hydrochloric acid is added to the gas is sodium carbonate and the other is sodium chloride.

Through the blue cobalt glass, the purple color is added with dilute hydrochloric acid, and the gas is potassium carbonate, and the non-sulfate is potassium sulfate.

You can also reflect the dilute hydrochloric acid and color test.

Nitrate acidified barium chloride-potassium sulfate.

A solution for the identification of potassium persulfate is added to silver nitrate-sodium chloride.

Sodium carbonate and potassium carbonate. This one. Use potassium pyroantimonate or sodium perchlorate to identify. This is not a junior high school problem.

2b I read someone else's reply and forgot that there was a flame reaction.

First, the glass rod was dipped in 4 kinds of solutions for flame color reaction, the flame color was yellow, containing sodium ions, sodium carbonate, sodium chloride; Through the blue cobalt glass is purple, it contains potassium ions, then potassium carbonate, potassium sulfate, divided into two groups, and then tested separately, adding hydrochloric acid, the first group of solutions have bubbles to produce sodium carbonate, and one is sodium chloride, and the second group of solutions have bubbles to produce potassium carbonate, and one is potassium sulfate.

First of all, a small amount of nitric acid is added to each substance, Na2CO3 and K2CO3 are bubbled, and the solution is dipped for flame color reaction, and sodium carbonate is yellow flame. The other is potassium carbonate.

The remaining two substances are added with barium nitrate. There is a precipitate of potassium sulfate. The rest is potassium chloride.

The first question, A correctly uses the different solubility at different temperatures, and B heats to convert sodium bicarbonate into sodium carbonate without achieving the purpose of "separation". c. Apply the method of dialysis.

The second question, B is correct, there is no phenomenon of sodium sulfate, there is a white precipitate and then disappears is aluminum sulfate, and there is a white precipitate of magnesium sulfate.

The third question, a is correct, because silica does not react with hydrochloric acid and is insoluble in water. B. If the purpose of separation is not achieved, C should be a filtration operation, and D. Ferrous ions react with nitric acid.

1. Answer: B. Sodium bicarbonate is thermally decomposed to produce sodium carbonate. The purpose of separation can be achieved.

2. Answer: BThe sodium hydroxide solution and sodium sulfate do not react, so there is no phenomenon; and aluminum hydroxide into aluminum hydroxide precipitation, and then further reaction to form soluble sodium metaaluminate, the precipitate has begun to dissolve; Sodium hydroxide and magnesium sulfate produce magnesium hydroxide precipitates.

3. Answer: a. Silica does not react with acids, whereas ferric oxide can react with acids and gradually dissolve to get silica after filtration.

Sodium sulfate is now used.

Aluminum sulfate. Magnesium sulfate is a colorless solution of three kinds of magnesium sulfate, which can be distinguished by one reagent, which is.

b Sodium hydroxide solution.

The following impurities removal kit main operations are correct.

Substance. Impurity.

Reagent. Main operation.

a. Silica.

Fe2O3 hydrochloride. Filtration. b

Sodium bicarbonate. Sodium carbonate.

Excess carbon dioxide. cfe

Al sodium hydroxide. Dispensing. d

fe(no3)2

Barium nitrate. Sulphuric acid. Filtration.

Only one reagent is used to distinguish the three colorless solutions of nano3, agno3 and na2co3, which cannot be used (b c).

a Hydrochloric acid. Hydrochloric acid does not react with sodium nitrate, reacts with silver nitrate to form white precipitate, and reacts with sodium carbonate to form colorless gas, so it can be identified.

B NaCl solution.

Sodium chloride does not react with sodium nitrate and sodium carbonate, and reacts with silver nitrate to form a white precipitate, so it cannot be identified.

C CaCl2 solution.

The reaction of calcium chloride with silver nitrate and sodium carbonate all forms a white precipitate, which does not react with sodium nitrate, so it cannot be identified.

DBA(NO3)2 solution.

Barium nitrate and sodium nitrate do not react, and react with silver nitrate to form a yellow precipitate, and react with sodium carbonate to form a white precipitate, so it can be identified.

a。Optional. There is no phenomenon of hydrochloric acid in case of sodium nitrate, white precipitate in case of silver nitrate, and bubbles in case of sodium carbonate.

b。Optional. Sodium chloride can identify silver nitrate. The identified silver nitrate can identify sodium carbonate.

c。No way. Calcium chloride can only identify sodium nitrate because sodium nitrate does not react with calcium chloride. The other two and calcium chloride respectively obtain two white precipitates of silver chloride and calcium carbonate, and the phenomenon is the same and cannot be distinguished.

d。Optional. Sodium nitrate can be reacted with sodium carbonate to obtain barium carbonate precipitate, so that sodium carbonate can be identified, and silver nitrate can be identified with sodium carbonate.

bc。Nano3, Agno3 and Na2CO3 react in pairs of three colorless solutions, and there is no precipitation phenomenon of Nano3, so as long as Agno3 and Na2CO3 can be distinguished, the three substances can be identified, B has no phenomenon with these two substances, and C and these two substances produce precipitation (if the shape of these two precipitates is not considered, these two substances cannot be distinguished).

1. All soluble in dilute nitric acid, indicating that there is no Baso4, but if there is a requirement for precipitation, the precipitation is BaCO3, so there are Na2CO3 and BaCl2. And because there are three substances, plus sodium chloride.

2. Partially soluble in dilute nitric acid, there are both Baso4 and BaCO3, NaSO4, Na2CO3 and BaCl2.

3. Needless to say, it's about the same as the first one.

4. Of course, it's NACL, it's very simple, you can give it a try.

,bacl2，nacl

naco3,bacl2

Your sodium sulfate is written incorrectly, and condition 2 and condition 3 seem to mean the same thing.

1. Since the precipitate can be completely soluble in acid, it means that there is no sodium sulfate, that is, the components of a are Na2CO3, BaCl2, and NaCl;

2. Incompletely soluble in acid, indicating that there is carbonate and sulfate and barium, indicating that there is no sodium sulfate, that is, the A component is NaSO4, Na2CO3, BACL2;

3. Question, right? It should be a precipitate that is completely insoluble in acid, which means that there is no sodium carbonate, which means that there is no sodium sulfate, that is, the component A is NaSO4, BACl2, NaCl;

4 Obviously, from the previous three questions, due to the presence of precipitation, it means that there is no sodium sulfate, that is, the composition of A must contain BaCl2

[Answer]:

Problem solving knowledge points and key points: Barium sulfate is insoluble in dilute nitric acid, but barium carbonate is completely soluble in dilute nitric acid, and it is white. After a chemical reaction occurs to form a precipitate, another soluble substance is either sodium sulfate, sodium chloride, or the original sodium chloride.

Therefore, sodium chloride is possible in any case, and it cannot be determined, what can be determined is that barium chloride must be present so that a precipitate can be generated.

Analysis: If the precipitate is partially soluble in dilute nitric acid, it means that there must be sodium sulfate, sodium carbonate and barium chloride; If it is completely dissolved, it means that there must be sodium carbonate and barium chloride, and there must be no sodium sulfate; If it is completely insoluble in dilute nitric acid, it proves that there must be sodium sulfate, barium chloride, and no barium carbonate.

Summary: Master the special phenomena in the chemical reaction, conduct a single analysis, and then comprehensively analyze the reaction situation to determine.

In fact, in high school, it is clearly written in the textbook that potassium permanganate solution is not used for carbonate testing.

Carbonate ions.

Hydrochloric acid is to be used. Acidic potassium permanganate solution.

Clarification of lime water, adding hydrochloric acid will produce bubbles, and the resulting gas will pass through the acidic potassium permanganate solution (to exclude sulfite ions) and then through the clarified lime water to produce white precipitates can prove that there are carbonate ions.

1 If B can be fully soluble in dilute nitric acid, the composition of A is (Na2CO3, BaCl2, NaCl)-B must be BaCO3, that is, there is no Na2SO4, so it is the other 3.

2 If B is only partially soluble in dilute nitric acid, the composition of A is (NaSO4, Na2CO3, BACl2)--B is a mixture of BACO3 and BASO4.

3 If B is not fully soluble in dilute nitric acid, the composition of A is (NaSO4, Na2CO3, BaCl2) - cannot be fully dissolved, that is, partially dissolved, as above.

4 The solute that must be present in solution C is (NaCl)-there is a precipitate formation, that is, there must be BaCl2 in A, and no matter which one is left, there is NaCl in the solution

It is distinguished by the differences in the properties of each substance.

1. Precipitation method.

In the solution of the substance to be identified, a certain reagent is added to observe whether there is a precipitate and identify.

Example: How to distinguish between Na2SO4 solution and NaNO3 solution.

Identification was performed using BaCl2 solution. The ionic equation of the reaction: Ba2++SO42-=BaSO4

2. Gas method.

According to the addition of a certain substance to a reagent, whether there is gas generated, it is identified.

Example: There are two bottles of missing labeled solids, they are Na2CO3 and NaCl, how to identify.

Identification using hydrochloric acid. Reactive ion equation: CO32-+2H+=CO2 +H2O: Ask how to turn this Vajra Bodhi red.

Ask if there are any chemicals that turn red quickly.

First look at the color, the yellow one is FeCl3, and then take a sample, add the other four substances to FeCl3, and the reddish-brown precipitate is NaOH (the chemical equation is: FeCl3 + 3NaOH = = Fe(OH)3 +3NaCl).

After the above two substances are obtained, the other three substances are sampled again, and they are added to the NaOH taken out, and the white precipitate is MgSO4 (the chemical equation is: MgSO4 + 2NaOH = = Na2SO4 + MG(OH)2), and then the BA(NO3)2 and KCl that are not separated are also sampled, and they are added to the samples of MGSO4, and the white precipitate is BA(NO3)2 (the chemical equation is: MGSO4 + BA(NO3). )2==BaSO4 +mg(NO3)2), then the last thing left is KCl