Methods to identify CH4, CO, H2, three colorless gases!

The gas is ignited, a dry beaker is covered on the flame, no water mist is CO, if there is water mist, then the gas generated by combustion is passed into the clarified lime water, there is a white precipitate to generate CO2, and if there is no H2

BCH4: There are liquid beads on the walls of the beaker, and the lime water becomes cloudy.

CO: There are no liquid beads on the beaker wall, and the lime water becomes turbid.

H2: There are liquid beads on the wall of the beaker, and the lime water is not turbid.

The three phenomena can be separated, choose b

The rest doesn't.

b If it is CH4, the phenomenon is: there are water droplets at the bottom of the beaker, and the lime water becomes turbid.

In the case of CO, the phenomenon is: no water beads appear at the bottom of the beaker, and the lime water becomes turbid.

If it is H2, the phenomenon is: there are water droplets at the bottom of the beaker, and the lime water does not become turbid.

The three gases are ignited through a needle-nose tube, and a small dry beaker is covered above the ignited gas, and then the changes in the three small beakers are observed, and the clarified lime water is immediately introduced into the small beaker, and then the phenomenon is observed.

There are small water droplets in the small beaker, and there is no phenomenon when the clarified lime water is introduced, indicating that the gas is H2.

There are no water droplets in the small beaker, and it becomes turbid when clarified lime water is introduced, indicating that the gas is CO.

There are small water droplets in the small beaker, which become turbid when the clarified lime water is introduced, indicating that the gas is CH4.

CH4 + 2O2 = ignition = CO2 + 2H2O

2CO+O2=Ignition = CO2

2h2 + o2 = ignition = 2h20

From 2, it can be seen that there are one or two reactions of H2 or CH4; The mass of the hydrogen element is determined by 3, and one or two reactants have CO or CH4; If the mass of the carbon element is, the ratio of the number of carbon atoms to hydrogen atoms is::4

From this, it can be seen that there are three cases of results:

1. This gas is CH4

2. This gas is a mixture of Co and H2, and the volume ratio is 1:23, and this gas is Co, H2, and CH4, but the volume ratio of Co to H2 must be 1:2

To sum up, choose D.

bLucky beginnings About money With money you can buy a house but not a home With money you can buy clocks but not time With money You can buy a bed but not enough sleep With money ...

Therefore c:h=1:4

Option d The other options are either absolute or incorrect.

Solution: From 2, it can be known that there are one or two reactions of H2 or CH4; The mass of the hydrogen element is determined by 3, and one or two reactants have CO or CH4; The mass of the carbon element is so m(c):m(h)=:1

Then the ratio of carbon atoms to hydrogen atoms is: 1:4 (so the 1:8 you calculated above is wrong).

From this, it can be seen that there are three cases of results:

1. This gas is CH4

2. This gas is a mixture of Co and H2, and the volume ratio is 1:23, and this gas is Co, H2, and CH4, but the volume ratio of Co to H2 must be 1:2

co2h2o

So the CH ratio is (:4

Therefore, it is possible that there are two possibilities: CO, H2 or CO, H2, and CH4, and if there are two possibilities, the ratio of the latter two will be less than 1:4 and may be 1:8

c:h=1:4 may contain only ch4 in the first place

Secondly Co:H2=1:2 can also make C:H=1:4 gas made up of Co H2.

Finally at co:h2=1; Adding any amount of CH4 to 2 makes C:H=1:4 gas composed of Co H2 CH4.

The ratio of hydrocarbon atoms to hydrogen atoms is 1 to 4

CH4CO, H2 (1:2 ratio of molecules).

co、h2、ch4

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.

a. Hydrogen and carbon monoxide are both colorless gases Therefore A is false B. Hydrogen and carbon monoxide are both odorless gases So B is false C. Under heating conditions, hydrogen and copper oxide react to form copper and water, and carbon monoxide and copper oxide react to form copper and carbon dioxide.

d. The carbon dioxide generated by carbon monoxide combustion can make lime water turbid, and the water generated by hydrogen combustion cannot make lime water turbid, so d is correct

Therefore, d

a. The three gases have no color, and they cannot be identified by observing the color, so the option is wrong B. The three gases are insoluble in water, and cannot be identified according to the solubility, so the option is wrong C. The three gases are flammable, hydrogen combustion produces a light blue flame, methane and carbon monoxide combustion produce a blue flame, and the three gases cannot be identified at one time, so the option is wrong D. The hydrogen combustion product is water, the carbon monoxide combustion product is carbon dioxide, and the methane combustion product is water and carbon dioxide. Their combustion products are different, and three gases can be identified according to the combustion products, so the option is correct: D

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.

Answer] C Answer Analysis] Test Question Analysis: "Some mixed gases are passed through a sufficient amount of clarified lime water, and white turbidity appears", indicating that the mixed gas contains carbon dioxide gas; "The remaining gas is exported, dried and ignited, and water beads are formed on the inner wall of the cavity of the shry beaker that covers the flame", indicating that the gas is burned to produce water. Therefore, it is impossible for the mu to be c.

Therefore, C. Test site: Gas inspection.

Answer] D Answer Analysis] a, hydrogen and carbon monoxide are reducible, through the hot copper oxide red solid dry reed sedan car can turn red, the phenomenon is the same, so do not choose;

b. The three gases are all odorless gases, and the method of smelling the smell cannot identify the three gases, so they are not selected;

C, H2, Co, CH4 all emit blue flames when ignited, the phenomenon is very similar, can not be identified, so do not choose;

d. The products of the three gases after combustion are not the same, and the method of "testing the combustion products after ignition" can be used to identify the three gases