There is a mixture of oxygen and carbon dioxide, and its mass percentage of oxygen element is 60

1) If the total mass of the mixed gas is 100kg, then the oxygen element accounts for 60%, that is, the total mass of the oxygen element is 60kg, then the total mass of the carbon element is 40%, that is, 40kg.

2) The molecular formula of carbon dioxide is: CO2, that is, 44kg of carbon dioxide, the mass of oxygenated elements is 32kg, and the carbon element is 12kg.

3) Let the total mass of carbon dioxide be x, then there is 12:44=40:x

Find x= kg 100kg, that is, the original question is wrong.

In the case of sulfur dioxide, the above calculation process is still as follows:

1) If the total mass of the mixed gas is 100kg, then the oxygen element accounts for 60%, that is, the total mass of the oxygen element is 60kg, then the total mass of sulfur element is 40%, that is, 40kg.

2) The molecular formula of sulfur dioxide is: SO2, that is, 64kg of sulfur dioxide, the mass of oxygen-containing elements is 32kg, and the sulfur element is 32kg.

3) Let the total mass of carbon dioxide be x, then there is 32:64=40:x

x=80 kg

That is, SO2 accounts for 80% of the mass of the gas mixture.

It is impossible to calculate the amount of carbon if it is pure carbon dioxide.

If the topic is carbon monoxide, the mass ratio of carbon monoxide can be calculated.

Oxygen is 100% oxygen, carbon dioxide is oxygenated 32 44, how can it be 60% combined?

Carbon dioxide mass a = (16 60) * 40 (12 44) oxygen mass b = 16 60 -a

Mass of carbon dioxide and oxygen ratio = b a

Addendum: Your title is not very clear.

Set the mixture to a total of 1 gLet the mass of carbon monoxide be x, then the mass of carbon dioxide is 1-x

12x/28 + 12(1-x)/44 =x =

So the mass fraction of oxoxane clump is.

The mass fraction of the selected oxygen element is 16*2 44*100%=

The mass fraction of oxygen in CO is 16 28*100%=

Now the mass fraction of oxygen is 65%, which is in between, so the gas mixture is a mixture of CO and CO2, choose D.

1. Let Co be Xmol and Co2 be Ymol

2. Oxygen accounts for 64%, that is, carbon accounts for 36%, the mass ratio of carbon and oxygen is 36:64=9:16, and the ratio of carbon atoms to oxygen atoms is c:o=9 16:16 16=3:4

3. So there is (x+y):(x+2y)=3:4; 28x+44y=104, the solution is x=, y=

5. According to Ca(OH)2+CO2=CaCO3+H2O, it can be known that 10g of calcium carbonate precipitation is generated.

I hope it helps you and I wish you happiness in your studies.

co 57%

CO2 destroys the content of both types of oxygen at a lower rate than 78%. Therefore, there must be a wheel and a void o2A: D

First, the mass fraction of element C in common gases containing C and O elements was calculated, and then the mass fraction of element O was inferred based on the known condition that "the mass fraction of element O is 78%".

If the gas is carbon monoxide gas; Then the mass fraction of element O in CO is 16 12 + 16 = 57%.

If the gas is carbon dioxide gas; The mass fraction of O element in CO2 is: 16 12 + 16 * 2 = 36%.

Both of these types of oxygen contain less than 78%. So it must contain oxygen.

So choose D

After passing the scorching copper oxide, CO is converted into CO2O%=64%.

c%=1-64%=36%

It contains imitation m(c)=20*36%=

co2--caco3

m(caco3)=

The mass fraction of C can be found to be 24%.

Because the mass fraction of C in CO2 is 12 44, there is 88% CO2

Subtract it to find the O2 mass fraction of 12%.

c→co2

12 44xx=

The mass fraction of oxygen in the gas mixture is =

Let the mass of carbon dioxide be x, then the mass of oxygen is 100-xx*12 44*100%=

x=then the mass fraction of oxygen in the gas mixture is (

Let CO2 be XG and O2 be YG, according to the title:

x+y=1g

The mass of the element carbon is and the mass of the element oxygen is .

carbon) to find the CO2 mass as: 1 60mol 44g mol = 11 15g

The oxygen mass is: 1-11 15 = 4 15g

Percentage of oxygen content (%)4 15 1 100% = The answer is.

But the upstairs is simpler

Because the carbon elements are all in carbon dioxide, according to the mass fraction of carbon elements, the mass fraction of carbon dioxide can be calculated, and then the mass fraction of oxygen can be calculated.

There is carbon dioxide x gram in the town of Shiling, and the brigade is hidden in one foot, and the carbon oxide is 10g-xg

x*32/44+16*(10-x)/28=10*64%

x = gram The mass of carbon dioxide contained in the mixture is.

Let the Co content band be a percentage, and the phenom x CO2 is (1-x) (mol content), and the equation is (16x+32(1-x)), 28x+44(1-x))=64%.

The solution is x=2 3

Therefore, 10g contains CO2 10 (44 (44+28 2)) = cross is too troublesome, and it is not as fast as the envy equation.