Take 6 80 g of liquid alcohol CxHyOz and burn in 10 liters of excess O2 standard condition .

You're on the wrong topic! It should be:

The volume decreases by concentrated sulfuric acid and then by CAO. Then what is the amount of substances C, H, and O in this gram of alcohol?

Excess O2 means that no carbon monoxide is generated, only carbon dioxide is produced.

The volume decreases in liters after passing the concentrated sulfuric acid, indicating that the water produced is molar.

Then through the CAO to reduce the liters, indicating that the carbon dioxide produced is moles.

That is: water is twice as much as carbon dioxide.

2CH4O+3O2==Ignition ==2CO2+4H2OC4O32 grams per mole.

The ratio of the number of atoms of carbon, hydrogen and oxygen in methanol CH4O is 1:4:1, so:

The amount of carbon is: moles.

The amount of the substance of hydrogen is: 4 moles.

The amount of oxygen is: moles.

This question is very simple, you just ask the class who is slightly better at learning.

Please see the ** I uploaded, I don't know if I can help you. To add, the second term of the reactant is missing oxygen.

Idea analysis] is calculated according to the equation of oxygen.

Problem solving process] Let the amount of XO2 formed by the reaction be x molxy2(l)+3o2(g)=xo2(g)+2yo2(g) x 3x x 2x

1) The amount of O2 is 3x, and the amount of total matter on the right side of the equation is 3x, because they are all gases, so the volume of O2 is 627ml

2) The molar mass of xy2 is conserved according to the mass, m(xo2)+m(yo2)-m(o2)] n(xy2)m(xo2)+m(yo2) can be calculated by multiplying the volume by the density m(o2), then the amount of oxygen can be multiplied by its molar mass m(xy2)=76

3) m(x)=3 (3+16) 76=12, m(y)=1 2 16 (3+16) 76=32, so x is carbon and y is sulfur.

1) If the volume of gas before the reaction is 3 units, since the volume of this reaction does not change, the number of moles of gas (g) is equal, so the volume of O2 before the reaction is equal to the volume of the mixed gas after the reaction, that is, 672ml

2) With the volume of oxygen, it can be calculated that the number of moles of oxygen = 672 1000

Because it happens to be a 1:3 reaction, xy2 =, and according to the law of conservation of mass, the total mass after the reaction (that is, the mass of the mixed gas) = the mass before the reaction (the mass of xy2 + the mass of oxygen).

So the mass of xy2=

Molar mass of compound xy2 = because molar mass is mostly in the form of an integer).

3) If the mass ratio of the elements in the XY2 molecule is 3:16, let the mass of X be 3a, then Y is 16a

3a+16a=76, a=4, so m(x)=12, m(y)=64 2=32

So it's CS2, carbon disulfide (in fact, you don't need to calculate it, if you have experience, you can come out, the liquid compound XY2, which is completely burned in oxygen to produce XO2 (g) + 2YO2, which is CO2 and SO2 that have been learned).

1）672 ml

Before and after the reaction, the number of gas molecules did not change. At the same pressure and temperature, the ideal gas with the same number of molecules (or the amount of matter) occupies the same volume, so the gas volume remains the same before and after the reaction.

2）76 g/mol

The amount of gaseous substances after the reaction n (gas) = , the amount of the species of xy2 before the reaction n (xy2) = according to the law of conservation of mass, the mass m (xy2) = the molar mass of xy2 m (xy2) = m (xy2) n (xy2) = according to the law of conservation of mass

CXHyoz+(x+y 4 Z2)O2=XCo2+(Y2)H2O is calculated based on the conservation of oxygen atoms.

1.In sufficient amount of O2, the combustion products should be CO2 and H2O2Excess Na2O2 indicates that the product is well absorbed.

The amount of increase should be the amount of h in the water.

The CO2 reaction is a bit more complicated because it can react with both NaOH and Na2O2.

2NaOH CO2 Na2CO3 H2O But the resulting water can further react with excess Na2O2, which is actually equivalent to the following equation:

Na2O2 CO2 Na2CO3 1 2O2 increases the amount of CO, then the BG increased by Na2O2 is the amount of H in the original substance, and the amount of C converted to CO, if x=z, then A=B

If x>z, then ab

The answer should be A, if you don't understand it, tell me, I haven't studied chemistry for a long time, and I may not be able to speak clearly.

The answer is x=z The teacher told us.

XY2+ 3O2-->XO2+ 2YO2 happens to be completely combusted, indicating that the product 672ml is XO2 and YO2 gas, that is, XO2 has 224ml and YO2 has 448ml. Substituting the equation yields:

xy2+ 3o2-->xo2+ 2yo2

1mol 22400ml

The amount of 224ml of compound XY2 substance is:

Let the relative atomic masses of the elements x and the elements y be a and b, and we can get a:2b=3:16, i.e., a:b=3:8

My first feeling is that XY2 is CS2. The mass ratio that follows is also confirmed. So I'll just write cs2.

CS2 + 3O2 = ignition = CO2 + 2SO2

After cooling, the volume of the product measured under standard conditions was 672 ml, i.e. The ratio of carbon dioxide to sulfur dioxide is 1:2, so CO2 and SO2 are available. Then CS2 has.

x:2y=3:16，x/y=3/8.

When you do the question, write xy2.

1) The amount of compound is SC2 substance is 76

2) X is carbon c and Y is sulfur s, and the relative atomic mass ratio of the two is 3 8

D, x=z, the substance M g is burned in O2 and introduced into Na2O2, and the solid mass increases M G. Since O2 is released later, the reaction can be combined to treat O2 as a catalyst-like existence.

Yield: CXHyoz + x+Y2)Na2O2 = XNa2CO3 + YnaoH + z-x) 2]O2

It can be seen that in order to maintain the equilibrium before and after the reaction, there is no relationship between x and y, and x must be equal to z to maintain the equilibrium before and after.

Calculated with volume difference:

cxhy(g)＋(x＋y/4)o2(g)=xco2(g)＋y/2h2o(l)

The volume coefficient before the gaseous hydrocarbon is 1, then v before v after v = 1 + y 4 then 1 10 = (1 + y 4) (10 + 50-35) then y = 6

Gaseous hydrocarbons. c2h6 c3h6