A mixture of iron and copper powder is put into a mixture of silver nitrate and zinc nitrate

All three metal elements are available.

Ferrous nitrate Zinc nitrate.

fe cu ag

Ferrous nitrate Zinc nitrate.

agno3+cu=cuno3+ag

cuno3+fe=feno3+cu

Fe + 2HCl = 2H2 rise + FeCl2

It was copper that was first reflected in the replacement of iron.

There must be an excess of Fe in the title, because only Fe and Zn can be reflected with HCL, but Fe cannot be replaced with Zn, so it can only be reflected with Fe and HCL.

So FE must be excessive.

Reducibility, Zn>Fe>Cu>Ag, the filter residue is gas-generated, there must be Fe, and the filtrate must be full of Fe and Zn.

There must be Fe in the filter residue and ferric nitrate in the filtrate.

Iron + silver nitrate = iron nitrate + silver.

Copper + silver nitrate = copper nitrate + silver.

Iron + copper nitrate + ferric nitrate + copper.

Because iron powder and copper powder are mixed, iron and copper will react with silver nitrate, but gas is generated when hydrochloric acid is dripped into the filter residue, indicating that there is still iron in the filter residue (if there is too much iron, the silver is completely replaced, and the copper is equivalent to no change, even if there is a reaction, it will eventually be replaced by iron). Copper does not react with hydrochloric acid; Zinc nitrate also does not react with iron and copper.

Iron, copper, silver.

Ferrous nitrate, the equation itself looked at and wrote.

There is absolutely too much iron, otherwise how would you explain the formation of bubbles?

Add a certain amount of zinc powder to the mixed solution of silver nitrate and copper nitrate, and the residue on the filter paper may be Cu and Ag.

According to the order of the activity of the three metals from strong to weak: zinc, copper and silver, when zinc powder is added to the mixed solution of copper nitrate and silver nitrate, the silver is replaced first, and the copper can be replaced after the silver is completely replaced; A certain amount of zinc powder may be replaced with all copper and silver, or only all or part of the silver.

Silver nitrate and copper nitrate are two different salts, the reactive metal can replace the inactive metal in the solution, and in the metal activity order table, the more reactive the metal is the more difficult to be replaced when it forms a salt.

Add a certain amount of zinc powder to the mixed solution of silver nitrate and copper nitrate, and fully react If a small amount of zinc powder is added, the zinc first reacts with silver nitrate and displaces part of the silver, and the filter residue only contains silver; If a certain amount of zinc powder is replaced with silver, and the remaining zinc powder will be replaced with copper, then the filter residue contains copper and silver; If there is a surplus of excess zinc powder after replacing all the silver and copper, then the filter residue contains copper, silver and zinc.

The highly active metal is placed in a mixed salt solution of the less active metal, and the more active metal displaces the least active metal from its salt solution first, and then the less active metal.

If there are bubbles after adding dilute hydrochloric acid, it means that there is still excess Fe, so the Ag+ and Cu2+ in the solution have all been replaced by Fe. Fe(NO3)2 and Zn(NO3)2 were present in the filtrate, and Fe, AG and Cu were present in the filter residue

If there are no bubbles after adding dilute hydrochloric acid, it means that there is no residual Fe, which should be classified and discussed

1) If Fe happens to react completely with Ag+ and Cu2+, only Fe(No3)2 and Zn(No3)2 are in the filtrate, and Ag and Cu are in the filter residue;

2) If Fe can replace all Ag+ but not Cu2+, there are Cu(No3)2, Fe(No3)2 and Zn(No3)2 in the filtrate, and Ag and Cu in the filter residue;

3) If Fe cannot replace all Ag+, there are AgNO3, Cu(NO3)2, Fe(NO3)2 and Zn(NO3)2 in the filtrate, and only AG is in the filter residue

Iron is more active than copper, it can replace copper in solution, but it cannot replace silver with zinc, copper does not react with dilute hydrochloric acid, and iron can react with dilute hydrochloric acid to form hydrogen.

There are bubbles because there is still a surplus after the iron replaces all the copper in the solution, so the filter residue is copper and iron, and the filtrate is silver nitrate, iron nitrate, and zinc nitrate.

There are no bubbles, there are two situations: one is that the iron just replaces all the copper in the solution, the filter residue is copper, and the filtrate is silver nitrate, ferric nitrate, and zinc nitrate: the other is that the iron does not replace all the copper in the solution, and there is still a part of copper nitrate in the solution, and the filter residue is copper, and the filtrate is silver nitrate, iron nitrate, zinc nitrate, and copper nitrate.

1 Iron, copper, silver, ferric nitrate, ferrous nitrate, because iron is not as active as zinc, iron cannot replace zinc, and there are bubbles with hydrochloric acid, indicating that there is too much iron, if there is too much iron, then all the copper and silver are replaced, so there is iron, copper and silver in the filter residue, and zinc nitrate and ferrous nitrate in the filtrate.

2 (1) silver, copper nitrate, zinc nitrate, ferrous nitrate.

2) Silver, copper, copper nitrate, zinc, ferrous nitrate.

3) Silver, copper, zinc nitrate, ferrous nitrate.

Because there are no bubbles, there is not enough iron, or it happens to react perfectly, because iron displaces copper, and copper replaces silver, so when there is very little iron, there is only silver in the filter residue, and when there is more iron, it can displace some of the copper, and when the iron happens to react completely with copper, there is silver and copper in the filter residue.

Answer: This kind of problem can be solved according to the metal activity order table, the metal activity oil is strong to weak in order: potassium, calcium, sodium, magnesium, aluminum, zinc, iron, tin [h] copper, mercury, silver, gold, we can understand it this way, the more backward the metal, the stronger their metal cations.

There are 4 cations in the solution: silver ions, copper ions, zinc ions, and hydrogen ions ionized by water; Zinc is more reducible than iron, so zinc ions cannot react with iron to remove zinc. Copper and silver are arranged after hydrogen, which can be precipitated by iron reduction.

There are bubbles in dilute hydrochloric acid dropwise later, indicating that there is still a residual iron element, and copper and silver do not react according to the activity of the metal.

Question 1: There are three elements in the filter residue: iron, copper, and silver. The filtrate contains: zinc nitrate, ferrous nitrate (there can be no ferric nitrate, ferric can react with three elemental elements, until the 3-valent iron is consumed), ferrous chloride (no ferric chloride, the reason is the same as ferrous nitrate).

The second question, adding hydrochloric acid and no gas is produced, then the iron is just after the reaction. There is copper and silver in the filter residue. Zinc nitrate and ferrous nitrate in the filtrate (it is not possible to contain ferric nitrate because there are no hydrogen ions in the filtrate).

This type of question examines the activity of metals, as well as the 2- and 3-valent properties of iron.

There are bubbles to illustrate, there is iron powder. Because the only metals that iron can replace are silver and copper, but it does not react with dilute hydrochloric acid. So there are ferric nitrate and zinc nitrate in the filter residue.

In solution, copper, iron, silver. If there are no bubbles, there are many cases, such as silver in the filter residue, silver nitrate, copper nitrate, zinc nitrate, and ferric nitrate in the solution. There are several more cases.

I won't go into details. I am studying chemical engineering now, and I can discuss this aspect of the problem in the future. The question you asked was something from high school, and some of the questions may not be as thoughtful as they were before.

Just recall.

From the activity of metals, it can be seen that zinc is more active than iron, so iron cannot replace zinc, only copper and silver.

After adding iron powder to the solution, the first thing that is replaced is the silver ion, then the filter residue must contain silver Ag, and the dropwise addition of dilute hydrochloric acid in the filter residue produces gas, which proves that the iron powder has completely replaced Ag and Cu, and there is a margin, so there are Cu and Fe in the filter residue.

The solution after the reaction contains zinc ion Zn2+, ferrous ion Fe2+, nitrate ion No3-, and a small amount of H+ and OH-.

After the hydrochloric acid reaction, Cu and Ag will not participate in the reaction, and the filter residue contains Cu and Ag. There are Cl- and Fe2+ and very little H+ and Oh- in the solution.

I choose BFirst of all, it is necessary to understand that adding iron powder to this mixed solution can only replace silver and copper, and magnesium is more reducible than iron, so it cannot be replaced. In addition, since silver is more reducible than copper, silver is replaced first.

Let's analyze the options, item A, add less iron powder, put a part of the silver in the code line, and all the iron powder reacts, which is in line with the situation. In item b, if copper is to be replaced, the silver ions in the mold solution must be replaced, so there can be no silver ions in the solution if there is copper precipitation. B false.

In item C, the amount of iron powder added is excessive, and all the silver is replaced, but the amount is not enough to precipitate all the copper, or the amount of iron powder added happens to replace all the silver and copper, and it also happens to reflect it, C is right. Item D, the excess iron powder added, all the silver and copper are replaced, and the iron powder has not yet finished reacting, D is right. Complete.

In addition, whether the question is wrong or you make a mistake, iron will form ferrous ions after reacting in this series of reactions in the mold hailstone, and will not form iron ions.

You can use the metal mobility order table.

In the metal activity sequence table.

Electron loss ability of metal: mg > Fe > Cu > ag ions to gain electrons defeat good ability: mg2+ ag>cu, and mg cannot be replaced.

The answers a, c, and d are incorrect, and they should not be iron ions (Fe3+), but ferrous ions (Fe2+).

B is false, when there is Ag+ in the solution, Cu2+ has not been replaced, and there can be no copper on the lead and filter paper.

You just have to figure out that there are metals that are in front of the activity, and there are no ions in the solution that are behind them.

b, the reaction order is silver and copper, magnesium does not react, so there will be no reaction of copper and silver ions.

In the order of metal activity, zinc, hydrogen, copper, silver, zinc can react with silver nitrate to form zinc nitrate and silver, and can react with copper nitrate to form zinc nitrate and copper, so the solution must contain zinc nitrate, and copper can react with silver nitrate to form copper nitrate and silver Dilute hydrochloric acid is added to the filter residue, and no gas is generated, indicating that zinc all participates in the reaction, and whether silver nitrate and copper nitrate contain can not be determined From the above analysis, it can be known

a. Add dilute hydrochloric acid to the filter residue, and no gas is generated, indicating that all zinc participates in the reaction; According to the reaction law, adding zinc powder must be able to replace metallic silver, so the filter residue must contain silver, whether it contains copper, and whether there is any surplus after the zinc powder is replaced with metallic silver; Therefore, the filtered solid must contain silver and possibly copper, so the statement A is correct;

b. According to the reaction law, adding magnesium powder will definitely be able to replace the metallic silver, so the filter residue must contain silver, but not necessarily copper; If it contains copper, it proves that Ag+ has been completely replaced, then there will be no silver chloride precipitation when dilute hydrochloric acid is added, so copper and silver chloride cannot coexist, so B's statement is wrong;

c. The filtrate must contain zinc nitrate, which may contain copper nitrate, and if the hydrochloric acid is excessive, it may also contain dilute hydrochloric acid; Therefore, C's statement is correct;

d. The filtrate must contain zinc nitrate, the amount of zinc is very small, and only part of the silver nitrate is replaced, so it may contain copper nitrate and silver nitrate, so D is correct

Therefore, choose B

A few questions for you:

The sequence of metal activity is fe cu ag

Fe must replace the least reactive metal first, and if there is no Cu, Cu will continue to replace Ag

Depending on the conditions, the white precipitate must be AGCL

1.How can you tell if FE is insufficient? It says a certain amount, how to judge it if it is envy and buried?

If the Fe is sufficient, there will be no AG in the solution, and there will be no white precipitate AGCl2Silver nitrate reacts with hydrochloric acid to produce a white precipitate.

There is no question of the order of activity of the metals, because he does not generate H2, and does not accompany the change of valence, and does not generate H2 elemental burial.

Rather, AgNO3 + HCl = AGCL (precipitate) + HNO33Depending on the conditions, the white precipitate must be AGCL

I don't understand hi

3 Copper chloride is not white so there is no Feng Nai Cu, so the amount of Fe is insufficient.

If there is enough Fe, Yin Chongchun will not only have white precipitated AGCL, but also copper chloride.

So FE is exhausted, and CU is not summoned to judge Min - so it's AG

I think; 1.The title says that a certain amount of iron powder is fully reacted means that one of the two reactants has been reacted and one has not reacted, if the iron difference is sufficient, then the iron reaction is finished, and now it is said that a certain amount means that the amount of iron is not enough, and other substances have not reacted.

2. Hydrochloric acid is poured into the filtrate, and hydrogen is used to replace silver.

3. Only silver chloride is a white precipitate in the product.

Metal, lively order: Iron, Copper, Silver.

Therefore, iron powder and copper powder are put into silver nitrate, and the silver will be replaced.

If there is more silver nitrate, and less copper powder and iron powder, or both are just enough to complete the reaction, then the copper powder and iron powder completely react to replace the silver, then after the reaction, what is left is solid silver The other two solid reactions are finished.

If there is less silver nitrate and more copper and iron, there are two situations after the reaction of silver nitrate, first, the iron powder is not reactive, and the rest is three solids, that is, iron, copper and silver. Second, after the iron powder is reactive and the copper reaction is partly done, then the remaining solids are, copper and silver.

So the answer A definitely has silver CIron D. may be presentThere may be iron and silver.