A colorless gas may contain one or more of CO, CO2, H2, HCL

D, your teacher is right.

1. The original gas first passes through the clarified lime water, and no turbidity appears, but the gas volume is significantly reduced. --There must be HCl, but CO2 may have.

2. The residual gas of the second step is passed through the calcination copper oxide powder, and the copper oxide powder turns from black to red,-- there may be H2 or CO or both.

When we continue to pass through the anhydrous copper sulfate powder, the anhydrous copper sulfate turns blue--- there is H2O here, but it may be carried out by lime water, so it is still uncertain.

Co CO2 H2 The volume of clear lime water in HCl decreases, indicating that there is HCl, CO2, and no turbidity appears, indicating that there is no CO2.

Then the residual gas of the second step passes through the calcination copper oxide powder, the copper oxide powder turns red from black, and the copper oxide powder can be oxidized from black to red, and there are CO, H2.

This gas has gone through step 1 before, and the gas may be mixed with water vapor, so it is not certain whether H2 is present or not, so the answer is D.

Chemistry teacher. Water may be carried by gas through clarified lime water.

To prove that it must contain H2, it is necessary to add concentrated sulfuric acid to remove the disturbance of the water before calcing the copper oxide powder.

The chemistry teacher's water of choice may be carried by the gas through the clarified lime water.

It should not be certain whether there is H2 or not, and you want to add a drying device when you want to dry the gas coming out of the clarified lime water? There must be a lot of similar topics about dry and wet gases.

Because the clarified lime water contains not much solute calcium hydroxide, if the mixed gas HCI is more, the calcium hydroxide will be completely converted into calcium chloride, and the substantial test of CO2 has been lost, and the possibility that the mixed gas contains CO2 exists. Then the residual gas of the second step is passed through the calcination copper oxide powder, and the copper oxide powder turns from black to red, which may have H2 or CO or both.

When we continue to pass through the anhydrous copper sulfate powder, the anhydrous copper sulfate turns blue--- there is H2O here, but it may be carried out by lime water, so it is still uncertain.

Answer: It does not necessarily contain co, the answer is c

The landlord should pay attention to the paragraph in the title "Carry out the following experiments in turn", after the step the volume becomes larger, CO2 has been converted into CO, and then the step is entered, the copper oxide may turn red, and there may not be CO in the original gas.

ps: When doing this kind of similar question, you should pay attention to the completeness and rigor of the topic

In fact, you ignore a phenomenon that returns to its original state after passing through the red-hot layer of carbon, and the volume of the gas increases. It means that there must be a reaction C+CO2==High temperature==2CO, so the original gas can be produced without CO.

In addition, I analyze that you may not have read the question.

Perform the following experiments in order to show that after passing through the first time, the original gas is no longer the original composition, and you understand that the following experiments are carried out separately! So here's the reason for not understanding!

Wrong. After passing through the red-hot carbon layer, it returns to its original state, and the volume of the gas increases.

C+CO2==High temperature==2CO, in this process there is CO generation, which can make copper oxide red.

C+CO2==High temperature==2CO is good, I watched it for a long time.

Through soda lime, the gas volume becomes smaller. It can be assumed that the gas contains one or both of CO2 and H2O vapors.

When passing through the red-hot copper oxide, the solid turns red. It can be assumed that the gas contains one or both of the COs and H2s.

When passing through anhydrous CuSO4 powder, the powder turns blue. It is certain that there is water vapor passing through, and this water vapor must have come from , which means that the reaction with copper oxide produces water, so that there is H2.

When passing through the clarification of lime water, the lime water becomes turbid. It means that there is CO2 gas passing through, because there is no CO2 generation in it, and CO2 cannot be absorbed, if there is CO2 in the original gas, it will be removed when passing through (because the reaction is complete in each treatment), so CO2 must come from , that is, there is CO that reacts with copper oxide to produce CO2.

However, the presence or absence of CO2 and water vapor in the gas mixture has no effect on the whole process, so it is not possible to judge whether they exist or not.

Soda lime is generally used as a desiccant to absorb H2O and CO2 at this time, the volume of the gas decreases, so it should be one or both of these two gases are absorbed, but it is not certain that there may be H2O CO2

When passing through red-hot copper oxide hydrogen. Carbon monoxide reacts with copper oxide to form water or CO2

Anhydrous copper sulphate is generally used to detect whether there is H2O in the gas, because the H2O in the original gas is absorbed, so there must be H20 There must be H2

In the same way, the CO2 detected in the clarified lime water is post-generated, so there must be CO in the original gas.

Satisfied, thank you.

The first step has been to get rid of any H2O and CO2 that may be present

Later, CuSO4 turned blue and lime water became turbid, indicating that H2O and CO2 were produced in the second step, indicating that there must be H2 and CO

There are at least one of CO2 and H2O, but it's not clear if either is both.

Hence choose D

If carbon dioxide gas passes through the red-hot carbon layer, carbon monoxide gas is generated, which returns to its original state, and the volume of the gas increases, and the volume before and after introduction remains unchanged, indicating that the original mixture has no CO2.

When passing through the scorching Cuo, the solid turns red, indicating that there is at least one of Co and H2, and the phenomenon of white CuSO4 powder and lime water indicates that CO2 and water vapor are generated, so the original mixture must contain Co and H2.

Answer: d

Then the original gas mixture must contain (NO, NH3, CO2) gas skin, and its volume ratio at the same temperature and pressure is (2:1:2) This gas mixture must not contain (HCL, O2) gas, the reason is that [containing ammonia gas (through concentrated H2SO4, the volume is reduced),—then it does not contain HCl, because if it contains, it will react; If the pants contain no (the gas under the residual weight changes color immediately after contact with the air), it does not contain oxygen - it will react to the discoloration. 】

There must be H2, CO2, HCl.

There may be co.

Reason: There must be CO2 in the white precipitate produced by clarifying the lime water in step 2, because it is not possible to generate CO2 in step 1.

In the 3 steps, the copper oxide powder turns from black to red, indicating that there must be something reducing, that is, CO and/or H2, and then it is said that anhydrous copper sulfate turns blue, then there must be H2, and there may be CO.

HCl must be present because the CO2 can be made in step 1 without turbidity in the clarified lime water, that is, HCl generates hydrochloric acid in the water, resulting in the inability to precipitate CaCO3.

I don't know if it's right, it's been many, many years since I left school.

If there is an obvious phenomenon in device 1, gas a must contain (H2O);

If there is an obvious phenomenon in device 2, gas A must contain (CO2);

If there is an obvious phenomenon in device A, gas A must contain (CO or H2);

If there is an obvious phenomenon in device B, gas A must contain (H2), so if there is an obvious phenomenon in device, A and B, gas A must contain (CO2, H2, H2O), and must not contain (CO) The effect of device 3 is (absorption of excess CO2).

Then gas A must contain (CO2, H2 and H2O) and must not contain (CO) The function of device 3 is (to absorb excess H2O and CO2).

When passing through soda lime, the volume of the gas becomes smaller; Soda lime absorbs water and carbon dioxide, and the reduction in volume proves that at least one of them is absorbed, and all of them are absorbed after passing through soda lime.

When passing through the red-hot copper oxide, the solid turns red. There may be CO reduced copper oxide, hydrogen reduced copper oxide, or both.

When passing through the white copper sulfate powder, the powder turns blue; It is proved that there is water generation, and this part of water intelligence** is generated when hydrogen reduces copper oxide, so there must be hydrogen.

When the lime water is clarified, the solution becomes cloudy as evidence of carbon dioxide, and carbon dioxide** reduces copper oxide to carbon monoxide, so carbon monoxide must be used.

3: We know that anhydrous copper sulfate can be used to test water. Note the first step:

Is there any water left over from soda lime (Naoh, CAO)? No! It's all eaten by CAO!

So where does the water come from? The second step is the response. How do you react to the second step?

Hydrogen reduces copper oxide! Because of hydrogen, water can eventually come back again.

4: In the same way, the first step has been to eat the CO2 clean, and it is impossible to have it. Therefore, there must be CO in order to react to form CO2.

Let's take a look first: only CO2 can react with carbon, but if there is CO2, then a complete reaction between one volume of CO2 and carbon will form two volumes of CO, (C+CO2===2CO).

However, the title says that the gas does not change before and after, so it can be concluded that there is no CO2. Judging by , the original gas may be Co, H2, Co and H2.

Look: Both CO and H2 can redden a hot cuo. But the product is different, if it is a single CO, CO2 will be generated. If it is a single H2, water will be generated. If it is a mixture of the two, there will also be two products: CO2 and water.

However, the condition in says that no water will be produced, so the gas can only be CO, not H2, or a mixture of the two.

And the conditions also verify the previous judgment.

The inferential answer is: the gas is a single CO.

Looking at the options again, ABC is all wrong.

d is correct, because it proves that the original gas does not contain CO2, and it proves that CO2 is present in the gas, so it can be determined that CO2 must be produced during the experiment.