5 6 g of iron reacts well with a solution of nitric acid containing 0 2 moles, at least losing elect

Analysis: n(fe) = g 56 g mol = mol

n（hno3）= mol

1) When it is dilute nitric acid, there is N(HNO3):N(Fe) 8:3, so the reaction is incomplete (Fe excess):

3Fe + 8Hno3 (dilute) 3Fe (NO3)2 + 2NO + 4H2O

At this time, calculated according to the amount of Hno3 (diluted), we get n (electron) = mol 3 4) = mol

2) When it is concentrated nitric acid, (passivation), it can react after heating

At this time, there is n(hno3):n(fe) 4:1, so the reaction is incomplete (fe excess):

Fe + 6Hno3 (concentrated) - Fe (No3)3 + 3NO2 + 3H2O

fe + 2fe（no3）3 → 3fe（no3）2

Fe + 4Hno3 (concentrated) - Fe (No3)2 + 2NO2 + 2H2O

At this time, according to the amount of Hno3 (concentrated), we get n (electron) = mol 1 2) = mol

So at least mol electrons are lost.

The chemical equation of this reaction is: Fe + 4Hno3 = Fe (No3) 3 + No + 2H2O

According to the reaction equation, it is not difficult to find that nitric acid can react with iron to produce iron nitrate, and the number of electrons lost is electrons.

In the reaction, iron (Fe) is oxidized to ferric ions (Fe2+), nitric acid (HNO3) is reduced to nitric oxide (NO) and water (H2O), and the reaction equation is as follows: Pishen.

fe + 2hno3 → fe(no3)2 + h2o + no↑

According to the reaction equation, each iron atom loses 2 electrons because the oxidation state of the Fe atom changes from 0 to +mole of iron is 10 23 = 10 22, so the total number of electrons lost is:

Halo, regardless of whether nitric acid is concentrated or dilute (as long as it is excessive), reacts with Fe, and trivalent iron should be formed (of course, Fe should be passivated in cold concentrated nitric acid), which is common sense! From the question, only n (nitrogen) is required, n(n) = n(no) + n (nitrate), and n(no) can be found by applying valence rise and fall conservation to the whole process of this reaction systemColumn:

n(fe)*2+ n(i)=n(no)*3, and n(no)= is conserved by the elements: only fe(no3)2 is left after the reaction, so n(nitrate) = n(fe)*2 = in summary: n(nitric acid) = nitric acid) = mol per liter.

Iron in small amounts, and iron in excess.

Equivalent to: Fe + 4Hno3 = Fe (No3) 3 + No+ 2H2 OFe + 2Fe (No3) 3 = 3Fe (No3) 2 Nitric acid has strong oxidizing properties and directly oxidizes iron to +3 valence.

Due to the excess of iron, Fe+2Fe(No3)3=3Fe(No3)2 becomes +2 valence.

Iron. Equivalent.

Iron. According to the chemical equation of the return of excess nitrate and insufficient iron, Fe + 4Hno3 = Fe (NO3) 3 + No + 4H2O

So the number of electrons transferred is.

The number of electrons lost by the reaction of gram iron with a sufficient amount of hydrochloric acid is (pcs). The amount of the substance that consumes hydrochloric acid is (.

n(fe)=m/m=

fe+2hcl=fecl2+h2↑

1 Fe loses 2 electrons in this reaction, so Fe loses electrons, and the number is one.

The first problem, the second problem itself has a problem, I don't know the concentration of HCl, I can't solve it.

No. Because there are several scenarios for this.

1. If there is an excess of nitrate, then it is ferric nitrate that is generated, and the total number of transferred electrons is 2, if it is an excess of iron, the iron generated will oxidize the iron into ferrous ions, and the total number of electrons transferred at this time is.

3, in between.

I don't know how to ask.

The number of electrons lost by the reaction of gram iron with a sufficient amount of hydrochloric acid is (pcs). The amount of the substance that consumes hydrochloric acid is (n(fe)=m m=

fe+2hcl=fecl2+h2↑

1 FE loses 2 electrons in this reaction and shouts Ji Cha Deficiency, so FE loses Zheng Wujin electrons, and the number is one.

Iron. Equivalent.

Iron. According to the chemical equation of the return of excess nitrate and insufficient iron, Fe + 4Hno3 = Fe (NO3) 3 + No + 4H2O

So the number of electrons transferred is.

At room temperature, iron is passivated in concentrated sulfuric acid and concentrated nitric acid, and a dense oxide film is formed on the surface, so it should not be reflected.

Heated case:

Iron and sufficient concentrated nitric acid to form ferric nitrate.

Sufficient amounts of iron and concentrated nitric acid produce ferric nitrate and hydrogen.

Acid does not become dilute is not what you understand, unless the acid you add is originally dilute nitric acid, otherwise even if you react a part, the remaining part should still belong to concentrated nitric acid, so the method you want to do is not possible, if you want to generate ferrous nitrate, that is only:

Iron and sufficient amounts of dilute nitric acid to form ferrous nitrate.

Sufficient amount of iron and dilute nitric acid to produce nitric acid, with iron and nitrogen dioxide and water, dilute nitric acid is more oxidizing than concentrated nitric acid, so it is easy to lose electrons.

Therefore, the valency of iron should be increased.

Iron and dilute nitric acid will be increased by two valents.

Iron and concentrated nitric acid will increase trivalent.

Gas and water are generated.

The analysis of the first reed body is calculated by chemical equations.

At room temperature, Fe is passivated with concentrated nitric acid.