A question about alkanes and olefins Please ask me Thank you

The density is a multiple of the density of hydrogen under the same conditions, which means that the average relative molecular weight of alkanes and alkenes is 28 for the simplest olefin ethylene, which has been exceeded, then the relative molecular weight of alkanes must be less than, so that the average molecular weight of the two hydrocarbons will be equal. , while the alkanes with less molecular weight are only methane, so the alkane should be methane.

Alkenes react with bromine water, while alkanes do not, so bromine water adds weight to the quality of the olefins. The mass of methane is the quantity of the substance, and the quantity of the substance of the alkene is n mol, then (n+, the calculated n = so the relative molecular weight of the olefin is propylene.

So choose C

The average molar mass is 28 for the simplest olefin ethyleneSo there must be methane, and the mass is so the amount of methane is , let the olefin be n mol, then 9 (, get n = so the molar mass of the olefin m = 42....for propylene.

So choose C

The ratio of the density of the gas is equal to the ratio of m, so the average m of the gas mixture is, according to the mean method, so there must be a molar mass greater than and less than, and the simplest olefin ethylene molSo there must be methane, the mass of which is.

So the amount of methane is , let the olefin be n mol, then 9 (, n = from the previous known mass is so the molar mass of the olefin m = 42....I choose C for propylene

The answer is c. Analysis: The density is a multiple of the density of hydrogen under the same conditions, indicating that the average relative molecular mass of the gas mixture is, and the relative molecular mass of ethane and olefin is the smallest 28, so if you want to average the relative molecular weight of methane, there must be methane; The weight of bromine water indicates that the olefin is and methane is, then methane is; From the total mass of the mixed gas is 9g, it can be seen that the total gas is 9, then the olefin is, so the relative molecular mass of the olefin is:.

So if it is propylene, then the answer is c

Option c The average molecular weight is smaller than 22,5 only methane, so there must be methane, the amount of total gas substances 9 mol olefin 4,2g methane is so olefin 0,. 1mol is the molecular weight, and 48 is propylene.

The molecular weight is pushed to be . 717*

That could only be methane CH4.

Because methane is the organic substance with the smallest molecular weight.

Push the molecular weight.

and water, which contains hydrogen.

The carbon-to-ch ratio is 2:1

Binding molecular weight.

So the molecular formula is C2H4 ethylene.

At first glance, it is methane, and the amount of gaseous substances is.

The molecular formula h is: 2*

H is 4C: so C is 1 so CH4

The density of the air is.

He's very close to air, with a molecular weight of about 28.

Because the air is 29

c: The ratio of the number of h is .

c=h=2*

The ratio of c:h is 1:2

And the molecular weight is about 28

So c2h4

Since the density under the standard condition is that the relative molecular mass of the gaseous hydrocarbon is known to be 1mol, and then the number of carbons and hydrogen are known, the molecular formula C4H8 experimental formula CH2

Anhydrous calcium chloride absorbs water, the mass of water is grams, the amount of the substance is, calcium hydroxide absorbs carbon dioxide, the mass is grams, and the amount of the substance is. The mass of organic matter is, the mass of carbon in organic matter is the mass of hydrogen, organic matter only contains carbon, hydrogen two elements (the sum of the mass of hydrocarbon is equal to the mass of organic matter). So, the amount of organic matter:

The amount of carbon in the substance: the amount of the hydrogen in the substance = 1:1:

4.So, the chemical formula is CH4

In the same way, CO2=

H2O=:H=1:2, the chemical formula is CNH2N, because the density is, so, it should be C2H4 (I'm not sure about this, sorry,)

Let the chemical formula of the alkane be CNH2N+2

Therefore, it is known from the inscription that the volume of the gas is 2L, and all of it is CO2, CNH2N+2+(3N2+1 2)O2--- NCO2+(N+1)H2O

13n/2+1/2n4

2 So: 4 2=(3n 2+1+1 2) n So: n=3

Alkanes are C3H8

In the case of naphthenes, it is set to CNH2N

cnh2n+3n/2o2---nco2+nh2o13n/2n4

24/2=(3n/2+1)/n

n=2, i.e., c2h4. Not alkanes.

So it can only be c3h8

The volume change is half of the original, that is, the product CO2 after the reaction of alkanes and oxygen is half of the volume of the two, and the reaction formula is as follows:

c3o8+5o2=3co2+4h2o

Whereas other gaseous alkanes are not, so they are propane.

Suppose the alkane chemical formula is c(x)h(y), then.

c(x)h(y)

x+y/4)o2=x

co2+y/2h20

So [1+(x+y 4)] x=4 2

1+(x+y/4)=2x

So 1+y 4=x

x = 1 then y = 4 (methane).

x = 3 then y = 8 (propane).

There are no other answers.

Solution: Set.

If CO2 is Nmol, then H2O is (N+1)molN(C)=N(CO2)=N

moln(h)=2n(h2o)=2(n+1)mol because the amount of hydrocarbon a is 1mol

Then: the general formula of n(c):n(h)=n (2n+2)a is cnh(2n+2).

At this point, saturation has been reached, so the hydrocarbon should be a saturated alkane.

If you have any doubts, you will hi me again.

cnh2n+2 ->nco2 + n+1)h2ocnh2n ->nco2 + nh2o

The ratio of CO2 volume to initial volume is the number of carbon atoms in alkane or alkene, n = 21 volume C2H6 to generate 3 volumes of water vapor.

1 volume of C2H4 produces 2 volumes of water vapor.

Let the volume fraction of alkane be a: 3a + 2(1-a) = a =

The volume fraction of alkene is: =

Volume ratio of alkanes and olefins in gas mixtures: = 2:3

The volume of this gas mixture is completely burned in oxygen, and the volume of CO2 and the volume of water vapor are generated under the same condition, then each mole of substance contains 2 moles of C atoms and 2* moles of hydrogen atoms. It should be ethane and ethylene.

Set alkanes x molar and olefin y mol.

x+y=16x+4y=

The volume ratio of alkanes and olefins in the gas mixture is 2:3

Let x ml of H2, Y ml of C2H6, z ml of CO21molH2 be burned to produce 1mol of water, and 1mol of C2H6 to be burned to produce 2mol of CO2 and water.

H2: Gas, all turned into water.

C2H6: Gas becomes 2molCO2 and water.

Gas Reduction:

And because after removing the water, the soda lime can only absorb CO2, then the amount of CO2 is:

2y+z=100（ml）

Finally, the original gas was 100ml

x+y+z=100

Simultaneous equations, get.

x=25ml；y=25ml；z=50ml

Finally, it was verified that 25ml of oxygen consumed by hydrogen and 25ml of C2H6 consumed oxygen, for a total consumption of 100ml of oxygen, less than 300ml.

The molar volume of the gas under the standard condition is , so it should be, since both gases are alkanes, so the number of hydrogen atoms is 2n+2 and 2(n+2)+2 respectively, let the mol number of one of them be x, and the other is, so x*(2n+2)+ (simplify the formula, and then let n=1,2,3,4,。。 See which one holds, and that's it.

The answer is c. Analysis: The density is a multiple of the density of hydrogen under the same conditions, indicating that the average relative molecular mass of the mixed gas is, and the minimum relative molecular mass of ethane and alkene hydrocarbons is 28, so if you want to average the relative molecular weight of methane, there must be methane; The weight of bromine water increases, indicating that the olefin is and methane is, then the methyl is methane; From the total mass of the mixed gas is 9g, it can be seen that the total gas is 9, then the olefin is, so the relative dispersion molecular mass of the olefin is:.

So if it is propylene, then the answer is c