Urgent! How to distinguish methane, ethylene, acetylene?

1. Methane cannot fade potassium permanganate solution and bromine water; Ethylene and acetylene can discolor potassium permanganate solution and bromine water; When ignited, acetylene carbon content is high and black smoke is emitted; In addition, acetylene can form a white precipitate with silver nitrate.

2. CXHY+(X+Y4)O2=XCo2+Y2H2O1 X+Y4 x Volume change 1+(X+Y 4)-X=1+Y 410 Volume change 10+60-45=25 Y=6, that is, the chemical formula of gaseous hydrocarbons is CXH6, so BEthane and dAcrylic is fine.

1.If potassium permanganate is used, both ethylene and acetylene will form CO2 gas and fade the solution.

Methane combustion is a blue flame, ethylene is a yellow flame, and acetylene is a yellow flame with black smoke.

CH4 + 2O2 = CO2 + 2H2O CH4 reaction is complete, leaving 40ml of oxygen to generate 10ml of CO2, which is not feasible.

The reaction of C2H6+ C2H6 is complete, leaving 25ml of oxygen to generate 20ml of CO2, which is feasible.

C3H8 + 5O2 = 3CO2 + 4H2O C3H8 reaction is complete, leaving 10ml of oxygen to generate 30ml of CO2, which is not feasible.

The C3H6+ C3H6 reaction is complete, with 15 ml of oxygen remaining to generate 30 ml of CO2, which is feasible.

With potassium permanganate, methane, bubbling does not discolor, ethylene does not bubble, fade.

Acetylene bubbling (carbon dioxide) fades.

1. First identify alkanes: olefins and alkynes can have an addition reaction with bromine, which makes the carbon tetrachloride solution of bromine fade, while alkanes cannot have an addition reaction. Alkanes can therefore be identified by reacting with a carbon tetrachloride solution of bromine.

2. Secondly, there are two ways to identify olefins and alkynes.

First: Although both alkenes and alkynes can react with the carbon tetrachloride solution of bromine to cause them to fade, the speed of discoloration is different. Olefins can cause the carbon tetrachloride solution of bromine to fade immediately, while alkynes take a few minutes to fade.

Olefins and alkynes can be judged by the speed of fading of the carbon tetrachloride solution of bromine.

Second: alkynes are passed into silver ammonia solution or cuprous ammonia solution, and white and red-brown alkynes can be precipitated respectively, while olefins cannot. Therefore, the gas can be passed into the silver ammonia solution or cuprous ammonia solution, and the phenomenon can be observed by static closure and rest, if the white precipitate or reddish-brown precipitate is precipitated, it is an alkyne, otherwise it is an alkene.

Extended Materials. The general formula of single-stranded olefin molecule is CNH2N, and C2-C4 is gas at room temperature, which is a non-polar molecule, insoluble or slightly soluble in water. The double bond group is a functional group in the olefin molecule, which is reactive, and can undergo addition reactions such as hydrogenation, halogenation, hydration, halohydrogenation, hypohalogenation, sulfate, epoxidation, polymerization, etc., and can also oxidize the cleavage of double bonds to generate aldehydes, carboxylic acids, etc.

The physical properties of olefins can be compared to those of alkanes. The physical state is determined by the mass of the molecule. In standard or room temperature, ethylene, propylene, and butene are gases in simple olefins, linear olefins containing 5 to 18 carbon atoms are liquids, and higher olefins are waxy solids.

Under standard conditions or at room temperature, C2 and C4 olefins are gases; C5 and C18 are volatile liquids; C19 or above solids.

In n-olefins, the boiling point increases as the relative molecular weight increases. The boiling point of n-olefins with the same carbon number is higher than that of alkenes with branched chains. For the alkenes of the same carbon frame, the double bond moves from the end of the chain to the middle of the chain, and the boiling point and melting point are increased.

Olefins are chemically stable but more reactive than alkanes. Considering that the carbon-carbon double bond in alkenes is stronger than the carbon-carbon single bond in alkanes, most of the reactions in olefins have the double bond broken and two new single bonds formed.

1. Through the acidic potassium permanganate, ethane cannot make it fade. (Reason: Double and triple bonds can fade potassium permanganate solution).

2. The remaining two gases are passed into the silver nitrate solution, and acetylene can produce a white precipitate with silver nitrate. (Reason: Silver nitrate can react with triple bonds to form a white precipitate).

Distinguishing ethaneethylene acetylene is divided into the following two steps:

1. The three gases are respectively passed into a small amount of acidic potassium permanganate solution The faded is ethyleneacetylene, and the non-faded is ethane.

2. The other two gases are respectively passed into the ammonia solution of silver ammonia solution or CuCl, and the precipitate is acetylene, and the rest of the pie is ethylene.

Potassium permanganese oak acid is an oxidizing agent, which is used for organic synthesis, disinfection, oxidation, etc. Contact with ether, ethanol, sulfuric acid, sulfur, hydrogen peroxide, etc.**; Burning such as Hehe, glycerin immediately decomposes and burns strongly.

The ammonia aqueous solution of silver nitrate is added, and there is a white precipitate to produce acetylene, and there is no other phenomenon. Add acidic potassium permanganate solution, there is no phenomenon of ethane and cyclopropane, potassium permanganate purple red disappears ethylene and 1,3-butadiene, but it is 1,3-butadiene that is seen to be produced by gas, and ethylene is not formed by gas. The identification of ethane and cyclopropane is to add bromine water, and the red and black color of bromine disappear from cyclopropane, and there is no phenomenon of ethane.

Summary. 1. Add Toren reagent, acetaldehyde and propionaldehyde are attached to a layer of metallic silver on the inner wall of the test tube, and acetone is not visible;

2. Add iodine and sodium hydroxide, and the yellow precipitate produces acetaldehyde, and the non-phenomenon is propionaldehyde.

The silver mirror reaction is a qualitative experiment used to detect aldehydes and reducing sugars, mainly used to detect the presence of aldehyde groups (i.e., -cho), that is, to distinguish aldehydes and ketones (except -hydroxy ketones);

Iodoform reaction (haloform reaction) is a methyl ketone compound or a compound that can be oxidized by sodium hypohalide to a methyl ketone, which can detect compounds with CH3Co- structure, that is, to distinguish acetaldehyde or methyl ketone.

Methane, ethylene, and propylene are chemically identified.

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<>First of all, these three gases are introduced into Br2 water, and it is methane that cannot fade the bromine water (reason: methane is a saturated hydrocarbon, there is no unsaturated bond, and the addition tartarder reaction with bromine cannot occur. However, in light it is possible to substitute with halogens.

Next, ethylene and propyne are ignited separately, and the thickest black smoke is propylene (reason: the carbon content of propyne is greater than that of ethylene.

Uh-huh. It may take about 20 minutes.

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Than] is added to the silver ammonia solution, and the reeds of the silver mirror reaction are acetaldehyde. Secondly. The addition of NaHSO3 has a precipitate to produce acetone. Finally, after heating into steam and oxidizing with oxygen, the liquid can undergo a silver chain microscopic reaction is n-propanol.

1. Add Toren reagent, acetaldehyde and propionaldehyde are attached to a layer of metallic silver on the inner wall of the test tube, and acetone is not present; 2. Add iodine and sodium hydroxide, and the yellow precipitate produces acetaldehyde, and the non-phenomenon is propionaldehyde. The silver mirror reaction is a qualitative experiment used to detect aldehydes and reduced sugars, mainly to detect the presence of aldehyde groups (i.e., -cho), that is, to distinguish aldehydes and ketones (except -hydroxyketones); Iodoform reaction (haloform reaction) is a methyl ketone compound or a compound that can be oxidized by sodium hypohalide to a methyl ketone, which can detect compounds with CH3Co- structure, that is, to distinguish acetaldehyde or methyl ketone.

3 is the synthesis of dienes, 4 is the Friedel-Kle alkylation, and 5 is the reaction properties of benzene and potassium permanganate.

A small amount of acidic potassium permanganate solution is respectively introduced into ethylene acetylene, which is not faded into ethane, and then ethylene acetylene is respectively passed into the ammonia aqueous solution of silver nitrate, and the white precipitate (generation of acetylene silver) is acetylene, and ethylene is not obvious.

The equations are c2h4 2br c2h4br2, c2h2 2br c2h2br2.

Potassium permanganate is an oxidizing agent, which is used for organic synthesis, disinfection, oxidation, etc. Contact with ether, ethanol, sulfuric acid, sulfur, hydrogen peroxide, etc.**; In case of glycerin, it immediately decomposes and burns strongly.

The three are respectively passed into the CCL4 solution (or potassium permanganate acidic solution) of BR2, and the discolored one is acetylene, and the non-fading one is ethane.

The remaining two are respectively passed into the ammonia solution of silver nitrate, and the white precipitate produces acetylene, and the non-phenomenon is ethylene.

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The specific operation steps are as follows: 1. Through the acidic potassium permanganate, ethane cannot make it fade. (Reason:.)

Double bonds and triple bonds can fade potassium permanganate solution) 2. The remaining two gases are passed into the silver nitrate solution, and acetylene can form a white precipitate with silver nitrate. (Reason: Silver nitrate can react with triple bonds to form white precipitate)

Ethane structure Ethylene structure Acetylene structure.

A: 1Methane: CH4

Structural formula: <>

Electronic: <>

2.Ethylene: C2H4

Structural formula: <>

Electronic: <>

3.Acetylene: old C2H2

Structural formula: <>

Electronic: <>

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