Excessive iron powder is added to the mixed solution of CuSO4 ZnSO4 H2SO4, and the solution after th

d. After adding excess iron powder, sulfuric acid will react with iron powder to release hydrogen gas to form ferrous sulfate.

Copper sulphate reacts with iron to form elemental copper and ferrous sulphate.

Due to the excess of iron powder, both copper sulphate and sulphuric acid are reacted off.

But in the table of metal activity sequences, zinc comes before iron, so iron cannot replace zinc from the salts of zinc.

So zinc sulfate was born.

So what must be contained is d

d Copper is replaced, and sulfuric acid is reacted with iron.

d Because iron reacts with CuSO4 first, the oxidation of Cu2+ is stronger than that of H+, and FeSO4 is formed after the reaction of excess Fe, but Fe does not react with ZnSO4.

Because the oxidizing properties of iron ions are less than those of zinc ions.

Therefore, iron powder will not react with zinc sulfate, but can only react with copper sulfate and sulfuric acid.

After the reaction, copper and hydrogen are replaced, and the solute becomes ferrous sulfate.

After the reaction, the only thing left in the solution is ferrous sulfate and zinc sulfate.

D. should be chosen

Do it with elimination:

Metal activity: Zn>Fe>Cu

Ionic oxidation: Cu2+>H+>Fe2+>Zn2+ are arranged according to ionic oxidation.

Iron first reacts with H+ to form hydrogen.

a, b are excluded.

And because of the liveliness of metal.

The copper separation will be replaced.

c error. The answer should be d

fe2+ can be sought, sleepy [(.]

The sum of the concentrations of CuSO4 and H2SO4 in the original mixed solution is, but there is no way to find out how much they are.

The concentration of Fe2+ is not available, and the concentration of CuSO4 and H2SO4 cannot be found, and the conditions are insufficient.

The conditions are insufficient, and it cannot be answered!

Add excess iron powder to H2

SO4 and CUSO4

In the mixed solution, the reaction occurs as Fe+CuSO4 FeSo4

cu；②fe+h2

so4═feso4

H2, add excess iron powder, H2

SO4 and CUSO4

All reacted completely

A. Add excess iron powder, H2

SO4 and CUSO4

All of them are completely reacted to form FeSo4, so the filtrate only contains FeSo4, so A is correct;

b. Add excess iron powder, H2

SO4 and CUSO4

All of them are completely reacted to form FeSo4, so the filtrate only contains FeSo4, so B is incorrect;

c. Add excess iron powder, H2

SO4 and CUSO4

Both completely react to release hydrogen and replace copper, so the solids obtained by filtration are Fe and Cu; Therefore c is correct;

d. The reaction is Fe+CuSO4

feso4cu；②fe+h2

so4═feso4

h2 , both are displacement reactions; Therefore d is correct;

Therefore, choose B

Choose A. Because of the excess iron, as long as the solutes CuCl2 and HCl that can react with iron are completely consumed to form FeCl2, while ZnCl2 does not react with Fe, therefore, the solutes contained in the solution after the reaction must be FeCl2 and ZnCl2, so A is chosen

It can replace Cu ions, not Zn, and can also react with HCl. Chin this.

Solution: According to the meaning of the topic, let the amount of Cl- be Xmol and the amount of So42- be ymol, then. 13

solution, x=y

Therefore, the amount of substances in solution SO2- to Cl- is 1:1

Note: According to the title, the meaning here is "at this time, the concentration of Fe2+ in the filtrate is 3:1 compared to the amount of Cu2+ in the original mixed solution (the volume change before and after the reaction is negligible)."", from which an equiquantitative relationship is derived.

Choose C. The condition is given in the question that the quality of the filter residue and the quality of the iron powder are the same, from which we can know that the mass of Cu and the mass of Fe are the same, we might as well set the mass of Cu to be 64g, because the relative mass of Cu is 64, which is easy to calculate.

Well, the second equation is like that.

fe + cuso4 = feso4 + cu

x y 64g

It can be found that the mass of x is 56g and the mass of y is 160g

Because the mass of Fe is equal to Cu, the above equation consumes Fe 56g, so the mass of Fe consumed by the following program is 64-56g = 8g

fe + h2so4 = feso4 + h2↑

8g zz=14g

The mass ratio of CuSO4 to H2SO4 is 160:14=80:7

It is CCUSO4+FE=FEso4+Cu, H2SO4+FE=FESO4+H2

The iron valence is +56, the copper valence is +64, and the hydrogenation valency is +1, but the hydrogen gas flies away after the reaction, and the mass is not counted in the solid.

Therefore, the mass of iron and copper sulfate increased by 8 units after the reaction, and the mass of iron and sulfuric acid decreased by 56 after the reaction, so the mass ratio of copper and hydrogen is 7:1

7 times 160 (copper sulfate mass unit) ratio 1 times 98 (sulfuric acid mass unit) = 80:7

Because in the presence of H+, H+ (excess), NO3-, Fe3+, Cu2+ ions, H+ oxidation is the weakest, therefore, there are four reactions in succession of the above particles, and theoretically the last reaction is (

The answer is d

Because hydrogen ions have the lowest oxidizing properties.

The answer is wrong, right? How could Fe2 exist in the end?!

Because there is an excess of hydrogen ions... The rest has been reflected, h+, and there are a lot of them.。。。 And cuso4 didn't precipitate directly = =

From the sequence table of metal activity, it can be seen that zinc, iron, hydrogen, copper can be seen from the title, adding excess iron powder to the mixed solution containing CuCl2ZnCl2HCl, iron can react with copper chloride to form ferrous chloride and copper, and can react with hydrochloric acid to form ferrous chloride and hydrogen Since iron tardination is excessive, CuCl2 in the solution

HCl is all reacted and cannot react with zinc chloride Therefore, the solute in the solution has Fel2ZnCl2 from the above analysis

a. The solute that must be contained in the solution is FeCl2

ZnCl2 therefore a is correct;

B. Due to the excess of iron powder, the solution does not contain CuCl2, so B is wrong; C. Due to the excessive amount of iron powder, the solution does not contain HCl, so C is wrong;

D. If the amount is excessive when the iron powder is dry, the solution does not contain HCl, so D is wrong, so choose A