A high school chemistry calculation problem A mixture of methane, carbon dioxide and nitrogen.

The total amount of matter of the three gases is:

Let the amount of methane be x mol using the difference method.

CH4+4CuO==CO2+2H2O+4Cu Poor solid mass.

1mol 64 (solid mass reduced by 4o).

x molx = so, methane is. The CO2 generated is.

CO2 + Ca(OH)2=H2O + CaCO3 (precipitation) 1mol 100g

The total amount of CO2 is, the amount of the substance of the original CO2:

The amount of nitrogen is 1).

The amount of substance of this gas mixture is.

The mass of the rigid glass tube is reduced, and the mass of the reduced mass is o, that is, the amount of copper oxide substance that is reacted is Thus the amount of methane substance is calculated according to the reaction formula.

According to the formation of precipitation, it can be seen that the amount of CO2 after the reaction is so the amount of CO2 in the original gas is the amount of nitrogen in the original mixed gas.

First, starting from the precipitation in the back, the amount of carbon dioxide is obtained from the weight of calcium bicarbonate through the stoichiometric formula, and then the amount of methane is obtained through the known stoichiometric formula, converted into volume, and then nitrogen is obtained according to the volume ratio of the air under the standard condition.

Suppose the gas mixture is 1

mol, composition 1

molar mass of each gas of the mixed gas.

The sum of the product of the volume it occupies is given in g

In mol, Wu Lubi is the average molar mass of the gas mixture. The average Sicker molar mass applies not only to gases, but also to solids and liquids.

For example, the volume ratio of N2 and O2 in air is known to be 4:1, and the average relative molecular weight of air is obtained.

Solution: Let n2 be 4

mol,O2 is 1

moln1/n2=v1/

v2m=mmixed nmixed=(n1m1+n2m2)(n1+n2)

m=(4 24+1 32)g (4+1)mol=A: The average relative molecular mass of air is.

Calculate x as the total mass.

That is, the n2 quality can be known.

That is, the volume of H2.

The above calculations are the stupid nuclear answers to the questions you are talking about.

After a while, the quality of H2 is 30%, and it is 70% on the horse manuscript bucket.

Divide the gas into 10 parts, hydrogen gas accounts for 3 parts, N2 accounts for 7 parts, and the gas is set up for a total of 10 grams.

n(n2)==

moln(h2)

The volume ratio of N2 and H2 is 1:6

In total, the gas is divided into seven parts. n2 is.

molh2.

The mass of moln2 is.

2 grams. The volume of h2 is under standard conditions. =

After a while, the mass of H2 is 30%, and immediately it is 70% again, how to calculate it, divide the gas into ten parts, hydrogen accounts for three parts, N2 accounts for seven parts, and the gas is a total of 10 grams N(N2) == 7 28 == mol N(H2) == 3 2 == mol

So the volume ratio of N2 and H2 is 1:6

In total, the gas is divided into seven parts.

N2 is mol H2 is 6* molN2 has a mass of 28* == 2 grams.

The volume of h2 is 6* == under standard condition

Calculate x as the total mass, i.e., we can know the n2 mass.

This is the volume of h2.

The above calculations are the answers to the questions you are talking about.

1 molar ethylene C2H4, after combustion, it generates 2CO2+2H2O, consuming 3 moles of oxygen.

1 mole of ethanol can be written as C2H4+H2O, and after combustion, 2CO2+3H2O is generated, consuming 3 moles of oxygen.

So choose D

Hello! The density is 1g l

Then the molar mass is 1g l * = g mol cross method. Methane 16

The average CO28 ratio of the amount of matter = 7-8

Mass ratio 7*16 8*28 = 1 2

Solution ideas:

Density 1g l=

Then the amount of methane is x and carbon monoxide is y

x+y=116x+

Finally, the mass ratio is calculated, please ask if you don't understand.

Mixed moles, carbon dioxide = moles; Water = moles.

That is, 1 molar mixture contains molar C, 4molar H. So.

1 mol c, 4 mol h; ethylene 1 mol C, 2 mol H; Acetylene argues for 1 mol c, 1 mol h.

aThe average molecular formula of the gas mixture is: C2H4.

B, D excess hydrogen, C carbon excess.

Therefore, it is item a.

If the amount of carbon dioxide and water is the same, the amount of the respective substance can be calculated first, and the average chemical formula of the mixed hydrocarbon is CXHY

n(co2)=

n(h2o)=

Then x=n(co2) n(cxhy), y=2*n(h2o) n(cxhy).

That's the basic idea.

Classification Discussion If the remaining gas is O2, then react 10ml of gas.

4NO2 + O2+2H2O==4HNO3 volume reduction.

x 10 x = 2ml so there is a total of 4ml of oxygen

If the remaining gas is NO

Then 3 no2 + h2o ==2hno3 + no

x 2 x = 6 indicates an excess of 6 ml of NO2.

The remaining 6ml was carried out according to the first reaction, then the O2 consumed is the answer or 4 A option.

If you don't understand, you can ask and adopt!