Chemical problems to detail the process Chemical problems to the process

1) The ratio of the atomic weights of each element in ammonium bicarbonate is n:h:c:

o=1:5：1：

3. The ratio of the masses of each element is n:h:c:

o=14:5：12：

48. The percentage of hydrogen in ammonium bicarbonate is h%=5hh4hCO3*100%=5 79*100%=

2) [The chemical formula of urea is Co(NH2)2].

The ratio of the atomic weights of the four elements in urea is c:o:n:

h=1:1:2:

4. The ratio of the mass of the four elements in urea is c:o:n:

h=3：4：7：

13）m=5000t*90%=4500t h'%=4h/nh4no3*100%=4/80*100%=5%

m'=h'%*m=5%*4500t=225tA: "5000t, the fertilizer contains 225t

Pay attention to the formatting.

I wish you to break out of your own world.

1. Ammonium bicarbonate, NH4HCO3, atomic weight ratio n:h:c:o=14:1:12:16

Elemental mass ratio n:h:c:

o=14：5：12：

48h percentage content: h%=(5h NH4HCO3)*100%=(5*1 14+4*1+1+12+3*16)*100% is approximately equal.

2. The ratio of atomic weight c:o:n:

h=12:16:14:

1 element mass ratio c:o:n:

h=6:8:14:

13. In NH4NO3, h%=(4HNH4NO3)*100%=(4 14+4*1+14+3*16)*100%=5% (just right).

5000t*90%*5%=225t

You have to give me points.

c:n:h:o=1:1:5:3

c;h:n:o=1:4:2:1

Mass ratio = 12 * 1:4 * 1:1 * 14 * 16 * 1 = 3:1:7:44 (4 + 14 * 2 + 16 * 3))) * Next, do the math yourself.

The mixture of co, hydrogen and oxygen happens to be completely reacted under certain conditions, and the first product is 101 kPa 120 carbon dioxide and water vapor. The average molar mass of the gas mixture gas is.

29g/mol

g mol is set to the amount of carbon dioxide and water vapor substances in amol and bmol, respectively (44g mol

amol+18g mol bmol) (a+b)mol=g mol is calculated as a b=3:1

That is, the volume ratio of carbon dioxide to water vapor is 3:1, and the volume ratio of carbon monoxide to hydrogen is 3:1 from 2H2+O2=2H2O and 2CO+O2=2CO2.

Assuming that the volume of hydrogen in the original gas mixture is m ml, oxygen m 2 ml is required; Then carbon monoxide is 3m ml, and oxygen needs 3m 2ml.

m+m 2+3m+3m Qiaoda 2=60

m=10, then the volume fraction of hydrogen in the original gas mixture is 10 and the number in 60=1 6

After the reaction. h2o

The form of existence is:

Water vapour. The average mass of the gas after the reaction is .

mol/l2h2+o2=2h2o

2CO+O2=2CO2 because. co

h2 with. O2 is the same as the reaction ratio.

2:1, so. coh2

The gas after the O2 reaction is. CO2 with.

The H2O molecular weights are:

And. Therefore, it can be obtained by multiplying the Qing clan method with the cross. co2

h2oco2h2o

From this, coh2co2 can be known

h2o=3 because. H2 is.

So the corresponding o2 is.

CO2 is. 3, the corresponding o2 is.

So the volume fraction of hydrogen is.

Thank you, Gao Chaling.

1）5co+i2o5═i2+5co2

Which tremor 2) according to 5CO+I2O5 i2+5CO2i2+2s2o2-3 2i-+s4o62-

According to I2—2S2O3, the excess S2O3 solution is 2 10-5mol, and the total excess is 1 10-4mol because only one-fifth of the solution is taken

The S2O3 solution that actually reacts with Li Liang I2.

According to i2-2s2o3, the amount of I2 substance is the amount of CO in the gas sample that can be known according to i2-5co, and the amount of CO in the gas slag sample is =

Therefore, 1L of gas sample contains milligrams of harmful gases.

Concentration of CO:

I haven't written this kind of question for a long time, I forgot the format, I'll write the idea!

2h2o2 ==2h2o + o2↑

First, understand that manganese dioxide is a catalyst, and the mass will not change! Therefore, the remaining substances in the beaker are filtered and washed. After drying, 2g of filter residue indicates that the mass of manganese dioxide is 2g, and the mass of H2O after reaction is .

Then according to the law of conservation of mass, oxygen mass = is produced

Since oxygen is, then the H2O2 consumed is.

Then the mass fraction of the solute in the hydrogen peroxide solution =

mna2o2=80%*975=780g

2Na2O2 relative molecular mass = 156, O2 relative limb closed molecular mass wild hunger hole amount = 32 set to O2 xg

156\780=32\x

x=160g

Answer: 160 g

Solution: Let the mass of the generation line Bu Oxygen answer Qi Qi be Qing Dai Ling X

2na2o2+2co2=2na2co3+o2↑975g*80% x

156/32=975g*80%/x

x=160g

A: It can produce 160 grams of oxygen.

2.Increase (endothermic reaction The temperature equilibrium shifts to the right, and the product naturally increases)3How to solve this question?Is it wrong?No, at least not yet.

4.Neither will this question.

LZ is a ** person?Beijing doesn't seem to have such a thing == I'm sorry, just ask the first two questions.

Add 208g of BaCl2 solution with a certain mass fraction, and the filtrate obtained after coarse filtration is sodium chloride solution.

Solution: Na2SO4 + BACl2= BaSO4 + 2Nacix y Z

142：x= 233： x= ； 208：y =233 ： y= ；

233： = 117：z z=

cl---nacl

w：：w w=

m(naci) = in the original solid mixture

Mass fraction of the resulting solution: NACI%=

Solution: Let the mass of BaCl2 in the sludge large BaCl2 solution be X.

Na2SO4 + BACL2 = BaSO4 letter base vertical + 2Nacix

208:233 = x=

The fraction of solute matter in the added BaCl2 solution: 100% 10%.

It is obtained, indicating that the sulfate happens to be completely anti-hunger, and the solution is a mixed solution of BaCl2 and NaCl, and the barium ion is, so it is good to dig BaCl2=, and the chloride ion is, so the chlorine element in sodium chloride is, so the sodium chloride is =, and it is rotten.

So, w%(bacl2)=

w%nacl=

At the same temperature and pressure, the ratio of the density of the gas is the ratio of the molecular weight.

This is because the density ratio of the gas mixture to hydrogen is 8:1

So the relative molecular weight of the gas mixture is 2*8=16

Let hydrogen account for x and oxygen for 1-x

then 2x+32(1-x)=16

The solution is x=8 15

So hydrogen accounts for 8 15 and oxygen accounts for 7 15

So the volume ratio of hydrogen to oxygen is 8:7

At the same temperature and pressure, if the density of hydrogen is 1, the density of oxygen is 16Let the volume ratio of hydrogen to oxygen be x. Then it can be seen from the meaning of the title: (16*1+1*x) (1+x) 1=8

Solve the equation to obtain: x=8 7

From the chemical formula: Na2CO3+BACl2=2NaCl+BaCO3, it can be seen that:

The molar ratio of BaCl2 and Na2CO3 at the time of reaction should be 1:1.

The questions given are BaCl2 and Na2CO3, i.e., Na2CO3 is more than BaCl2

So BaCl2 is the limiting reactant.

bacl2.The molar ratio of the two substances in the equation is 1:1, and when feeding, N(Na2CO3) N(BaCl2), and Ba2+ is a toxic substance, so BaCl2 is a limiting reactant

The Great Britain above should be right.