A chemical calculation problem, ask for a solution, and trouble to write down the process clearly

16g before the reaction and after the reaction, according to the equation, the weight reduced after the reaction is the mass of oxygen produced by the reaction.

So 1. The reaction produces oxygen.

2kmNO4==Heating ===K2mNO4 + MNO2 + O2 x y z

According to the equation, the reaction produces oxygen in grams, so let the potassium permanganate consumed by the reaction be , and the potassium manganate generated is y grams, and manganese dioxide z grams are generated.

Solve x=g y=g z=

Since there is a total of 16 grams of potassium permanganate before the reaction, there is a residue after the reaction:

kmno4: gram k2mno4: gram mno2: gram.

Satisfied

The team will answer you as follows:

Originally, there were 16g, but later it became, indicating that 16 oxygen escaped.

2kmno4==k2mno4+mno2+o2↑x y z

Solution: x= y= z=

So the last solid contains ;;.

2kmNO4==K2mNO4 + 02 + MMO2 oxygen is.

The remaining substances are K2mNO4 and MNO2

You use the ratio of the molecular weight to the total molecular weight, respectively.

1) M (pure H2SO4) = 80g 10% = 8g (the black dot in the following process is the meaning of the space, don't get me wrong).

·m(pure H2SO4)·· 8g

M concentrated H2SO4 = ———this is the fraction line) = ——— originally Answer: To configure 80g of dilute hydrochloric acid with a mass fraction of 10 solute, a concentrated sulfuric acid with a mass fraction of 98 is required.

2) m(pure H2SO4) =

Solution: Let the mass of ammonium bicarbonate in the fertilizer sample be x.

2nh4nco3+h2so4=(nh4)2so4+2co2↑2h2ox ··

=———x (Originally.)

x··· A: The mass of ammonium bicarbonate in this fertilizer sample is about .

m(n nitrogen) = originally was.

The mass fraction of nitrogen in the fertilizer = ———100% Originally··· 5G A; The mass fraction of nitrogen in the fertilizer is about yes, the topic is a bit problematic, it is not sodium bicarbonate, it should be ammonium bicarbonate (NH4HCO3), and sodium bicarbonate does not contain nitrogen, not nitrogen fertilizer. There is also the first question, which should be dilute sulfuric acid, not dilute hydrochloric acid).

1) 100g % hydrochloric acid, then we first require the mass of the solute, that is, m(HCl) = 100g*, the amount of the substance = mass relative molecular mass. Then n(HCl) = divided by the relative molecular mass of HCl, the answer is 1mol

2) The method is the same as question 1, m(naoh)=200*10%=20g, n(naoh)=20g 40 g mol=

3) The reaction formula of sulfuric acid and sodium hydroxide is: H2SO4 + 2NaOH = 2H2O + Na2SO4. That is, 2mol NaOH is required to neutralize 1mol of sulfuric acid. Then 2mol is needed.

4) Solute mass Solution concentration = solution mass. Then 160g 10% = 1600g. 1600g of sodium hydroxide is required.

Belch.. Typing is tiring.

（1）n=m×w=100g×

2) The formula is the same as above = 200g 20% = 40g

3）2naoh+h2so4=na2so4+2h20n： 2 1

4mol 2mol so naoh 4mol is needed

m=m n=40g, mol 4mol=160g(3) from the above question, m naoh=160g, so m naoh solution = m naoh w naoh=160g 10% = 1600g

Uh, it's easy, but it's hard.

（1）mcl=100* n=m/m=

2)200*10%=20(g) n=m/m=20/40=(3)h2so4+2naoh=(na)2so4+2h2o1 2

2 n n=2*2=4(mol) m=nm=2*40=80(g)

4) The conditions are insufficient or the meaning of the topic is unclear.

(1) In 100 grams of hydrochloric acid, the mass containing HCl is: 100g The amount of substances containing HCl is: per mole) = 1mol (2) In 200 grams of 10% sodium hydroxide solution, the mass containing NaOH is:

200g×10%=20g

The amount of substances containing NaOH is 20g 40 (per mole) = (3) H2SO4 + 2NaOH = 2H2O + Na2SO4 From the above, it can be seen that 2 moles of H2SO4, the amount of substances that require sodium hydroxide is 4 moles. The mass of sodium hydroxide is:

4×40=160g

4) If 10% sodium hydroxide solution is used for neutralization, the mass of sodium hydroxide solution is required: 160g 10% = 1600g

Question 1: 100 times divided by hydrochloric acid.

2 （200* = mol

Equal to 4mol mass is 4 * 40 = 160 grams 4 160 = 1600 grams.

2fe(oh)3=fe2o3+3h2o

Red Fe2O3 powder.

Calculate the fe for.

It turned out to be added.

Then there is Fe in the original compound

Fe2+ is completely precipitated.

2Fe2+ +4OH- +O2=2Fe(OH)3 is calculated to have Fe

Then "only a pure oxynate (non-double salt) without crystalline water can be obtained, and there is no Fe2+ in the original solution, then the salt is K2SO4

Or the consumption is calculated by Fe2+ just completely precipitated.

So there's potassium from compounds.

Suppose x:fe:k=1:2:4 The quality of the calculation does not match the problem.

fe:k=1:2 then x=fe y=k

It can be calculated that the molar mass of the compound is 198

198-56-78) 4=16g mol z=o The compound is K2FeO4

The calculation is performed using the relational method.

The quality of the required ammonia is x

nh3---hno3

98%×94%x

The column equation is solved to x=

Do it according to the conservation of the quantity of n.

Hno3 with a mass fraction of 50%.

There is n in mol

Then need no need nh4

Its mass is .

1. so2---s so3---s

x=128 64 y=80 32

Mass ratio of sulfur dioxide and sulfur trioxide: 128:80

2.Let the mass of the fertilizer sample be m

The mass of ammonium nitrate is m*90%.

Mass of nitrogen: m*90% (28 80).

The mass fraction is = m=

3.The mass of the fertilizer sample was m

Mass of nitrogen element: m*

Mass of ammonium nitrate: (m*

Mass fraction of ammonium nitrate = < (m*> m=50%.

4.The mass fraction of ammonium nitrate nitrogen is: 28 80 = 35% A The mass fraction of nitrogen element is 28 60 =

b The mass fraction of nitrogen element is 14

The mass fraction of nitrogen is 28 132=

The mass fraction of d nitrogen is 14 79=

So choose A

Don't ask for online answers Go to the office tomorrow morning and find me.

Set the total quality of this batch of urea. For. m

Yes. The N content in this batch of urea is Changzhong.

m 45, then the mass of urea is.

m×45％/

Therefore, the content of urea is resistant to the mountain.