Nine chemical acid base salt knowledge points also have the color of all solutions and the valence o

Iron oxide generally refers to ferric oxide, in which the iron is trivalent, the hydrated ion of the ferric ion is yellow, and the aqueous solution is also yellow.

It depends on the reagent that reacts with it and the amount of reagent.

Reagent of the reaction (oxidation dependent).

For example, iron reacts with dilute hydrochloric acid and dilute sulfuric acid to form ferrous iron, iron reacts with chlorine and bromine to form ferric iron, iron ignites in oxygen to form Fe3O4 has both 2 and 3 valents, and iron reacts with elemental sulfur to form Fes, 2 valent iron.

The amount of reagent used.

Iron reacts with nitric acid, and excess iron forms 2-valent iron, and insufficient iron forms 3-valent iron.

For junior high school students, as long as they pay attention to the reaction of iron elements with dilute acids, divalent iron ions will be formed, and only when iron oxide reacts with dilute acids will trivalent iron ions be formed.

I remember that the valency is -2 valence for oxygen, because oxygen can be combined with a variety of substances.

(1) Hydrochloric acid, carbonic acid, hydrogen sulfate ion (H+).

2) K+, OH one, H2O H+ and OH one.

3) Carbonate ions cannot be formed into precipitate, gas or water because the ions cannot be combined with each other.

4) Na2CO3+2HCl==2NaCl+CO2+H2O (other carbonate and acid reactions in the table can also be used) metathesis reaction.

h2co3 h2so4;h+

k+;oh- h+

No, because no new substance is produced.

Up arrow +h2o; Metathesis reaction.

1 (hydrochloric acid, carbonic acid, sulfuric acid) (hydrogen ion).

2(oh-·k ion) (oh-, h+).

3（co32-）

(1) HCl, H2CO3, H2SO4 hydrogen ions (H+) (2) K+, OH a, H2O H+ and OH a.

3) Carbonate ions cannot be formed into precipitate, gas or water because the ions cannot be combined with each other.

4) Na2CO3+2HCl==2NaCl+CO2+H2O (other carbonate and acid reactions in the table can also be used) metathesis reaction.

1. Hydrochloric acid, carbonic acid, sulfuric acid; h+

2. K+, OH- (these two are in junior high school).

3、co32-；If it is a sodium chloride solution, it can react with many substances, but it cannot react with the first two, because it can neither produce gas, nor precipitate and water (so in junior high school).

4、caco3 + 2 hcl= cacl2 + co2↑ +h2o

1. Hydrochloric acid, carbonic acid, sulfuric acid; h+

2、k+、oh-,h2o

3、co32-；No, can not produce gas, precipitation with water.

4、caco3 + 2 hcl= cacl2 + co2↑ +h2o h2o

na2co3+2hcl==2nacl+co2↑+h2o

Solution: n(cl2)=(56x10 -3).

2fe2+cl2==2fe3+2cl-

n(fe2+).2/n(fe2)=1/>n(fe2+)=5x10^-3mol

and n(hcl)=50x10 >n(cl-)=

2N(FeCl2) + 3N(FeCl3) = obtained from the conservation of Cl- in HCL).

and n(fe2+)=5x10 -3mol=n(feCl2).Obtained according to the conservation of Fe2+).

2x5x10^-3+3n(fecl3)=>n(fecl3)=n(fe3+)=

n(fe2+)：n(fe3+)=5x10^-3：：4

The oxides formed are: FeO·2Fe2O3 or Fe5O7

The answer is C

From the amount of hydrochloric acid consumed, it can be seen that the sample contains o=1 2(h+)=35*10 (-3)mol

From the consumption of Cl, it can be seen that the sample contains Fe2+=2*N(Cl)=5*10 (-3)mol

That is, Fe2+:O=1:7 in the sample

The rest of FE exists at +3 valence.

Substituting the answer shows that the chemical formula of the original compound is ferric hepoxide.

Your option C should be changed.

Solution: n(cl2)=(56x10 -3).

2fe2+cl2==2fe3+2cl-

n(fe2+).2/n(fe2)=1/>n(fe2+)=5x10^-3mol

and n(hcl)=50x10 >n(cl-)=

2N(FeCl2) + 3N(FeCl3) = obtained from the conservation of Cl- in HCL).

and n(fe2+)=5x10 -3mol=n(feCl2).Obtained according to the conservation of Fe2+).

2x5x10^-3+3n(fecl3)=>n(fecl3)=n(fe3+)=

n(fe2+)：n(fe3+)=5x10^-3：：4

The oxides formed are: FeO·2Fe2O3 or Fe5O7

The answer is C

From the amount of hydrochloric acid consumed, it can be seen that the sample contains o=1 2(h+)=35*10 (-3)mol

From the consumption of Cl, it can be seen that the sample contains Fe2+=2*N(Cl)=5*10 (-3)mol

That is, Fe2+:O=1:7 in the sample

The rest of FE exists at +3 valence.

Substituting the answer shows that the chemical formula of the original compound is ferric hepoxide.

Your option C should be changed.

Solution: n(cl2)=(56x10 -3) +cl2==2fe3+ +2cl-2....1n(fe2+).

Finch tease n(Fe2+)=5x10 -3mol and n(HCl)=50x10 according to Cl-Shouqing in HCL and sell constant) and n(Fe2+)=5x10 -3mol=n(FeCl2).According to the conservation of Fe2+, the segment rot is obtained) 2x5x10 -3+3N(FeCl3)=:N(Fe3+)=5x10 -3::

4 The oxide formed is: FeO·2Fe2O3, i.e. Fe5O7 Answer: C

From the amount of hydrochloric acid consumed, it can be seen that the sample contains O=1 2(H+)=35*10 (-3)mol, and the consumption of Cl can be known that the sample contains Fe2+=2*N(Cl)=5*10 (-3)mol, that is, Fe2+:O=1:7 in the sample, and the rest of Fe exists in +3 valence, and the substitution is in the answer file, and the chemical formula of the original compound is ferric hepoxide, and your C option should be changed.