Urgently, please help me take a look at the chemical ion equation problem

The universality and reducibility of Fe(OH)2 alkali each have an ion equation.

As follows +h+ =h20

o2 + 4h+ = 4fe3+ +2h2o

The general and strong oxidation of Hno3 acid each have one ion range.

1. cu+ 2no3- +4h+ =cu2+ +2no2+2h20

2. h+ +oh- = h2o

The reduction products of SO2 gas with Fe3+ are sulfate and Fe2+

Write out the ion equation for the reaction of Na2SO3 dilute solution with I2.

h2o+so32- +i2 = so42- +2i- +2h+

Write the ionic equation for potassium iodate potassium iodide under dilute sulfate acid.

io3- +5i- +6h+ = 3i2 + 3h2o

And then there's the special properties of nitric acid that you mentioned.

I don't know if you're talking about it. When it reacts with more reactive metals, it generally does not produce hydrogen, but nitrogen oxides.

This reaction is the embodiment of oxidation.

I'll write you another one.

3zn+8hno3 (dilute) ====3zn(NO3)2+2NO(gas) + 4H2O

Zn + 4Hno3 (dilute) ====Zn (NO3) 2 + 2NO2 (gas) + 2H2O

The general and reducing properties of Fe(OH)2 alkali each have an ion equation as follows + H+=Fe2++H20

O2 + 2H2O = 2Fe(OH)3 + 2H2OHno3 acid has one ion range each for the general and strong oxidation of the acid.

1. h+ +oh- = h2o

2.Cu+ 2NO3- +4H+ =Cu2+ +2NO2+2H20 An ionic equation for the special properties of nitric acid Cu+ 2NO3- +4H+ =Cu2+ +2NO2+2H20

The reduction product of SO2 gas and Fe3+ is sulfate and Fe2+, write the ion equation for the reaction of Na2SO3 dilute solution with I2, H2O+SO32- +I2 = SO42- +2I- +2H+, write the ion equation for potassium iodate in dilute sulfate acid.

io3- +5i- +6h+ = 3i2 + 3h2o

fe(oh)2+hcl=

fe(oh)2+o2+h2o=

hno3+naoh=

cu+hno3=

HNO3 = decomposition.

cl2+so2+h2o=

fecl3+fe=

Let's check the book by yourself in detail.

It's too much trouble, and I'll be happy to type......It's good to get 2 points......

First of all, we need to define the concept of ionic reactions and ionic reaction equations. An ionic reaction is simply a reaction in which ions are involved or formed in a solution (or in a molten state). In middle school, we generally think that the reaction equation that expresses the ionic reaction in solution with ion symbols is called the ion equation.

The combination of ions to produce micro-or refractory substances or gases or refractory or volatile substances, that is, metathesis reactions, is only one type of ionic reaction. Ionic reactions also include redox reactions with the participation of ions, and so on.

For the two ionic equations asked. The reaction between CO2 and H2O to form H2CO3 cannot be written as an ionic reaction equation, because when writing the ionic reaction equation, the substances in the chemical equation that are easily soluble in water and can be completely ionized are dismantled into anionic and cationic symbols; Substances that are insoluble in water, gaseous substances and water are still expressed by chemical formulas; Therefore, whether it is CO2, H2O or H2CO3, it can only be written in the form of molecules.

The reaction between CaO and H2O to form Ca(Oh)2, in which Ca(Oh)2 can be completely ionized, but because the solubility of Ca(Oh)2 is not very large, when the title says that clarified lime water is generated, Ca(Oh)2 can be split into Ca2+ and Oh- ions, and the ion reaction equation is CaO+2H2O==Ca2+ +2Oh-; When lime milk is generated, Ca(OH)2 is considered to be a microsoluble, so it is written in molecular form, so there is no ion equation, and it is written as CaO+2H2O==Ca2(OH)2.

For ionic reactions and ionic reaction equations, it is necessary to grasp the concept of electrolyte so that you can judge when a molecule can be split into ions.

CO2 and H2O react to form H2CO3 The ionic equation for this is CO2 + H2O = 2H+ +CO3 2—

This is not right.

Should be. co2+h2o=h2co3

Weak electrolytes are also not dismantling, such as, acetic acid, carbonic acid, etc., if ions are generated, the estimated concentration is very low, and the degree of ionization is relatively large.

cao+h2o==ca2+ +2oh-

Only a single substance can be ionized, and you are confusing the concept of ionization.

For example, sulfuric acid can ionize H2SO4==2H+ +SO42-ionization The concept is to divide a substance into two ions to make it conductive.

Only ionic reactions need to write ionic equations, so that the essence of the reaction can be seen. But not all reactions are ionic reactions, and not all reactions have to write ionic equations. Seek truth from facts!

（1）2h+ +so42- +ba2+ +2hco3- -baso4 + 2co2 + 2h2o

2）h+ +so42- +ba2+ +hco3- -baso4 + co2 + h2o

3）2oh- +ca2+ +2hco3- -caco3 + co32- +2h2o

4）oh- +ca2+ +hco3- -caco3 + h2o

The excess is no longer reacted with the product.

1 2hpo4- +2oh- +3ca2+ === ca3(po4)2 + 2h2o

2 caco3 + 2h+ === ca2+ +h2o +co23 hco3- +h2o ==== h2co3 + oh-4 s2- +h2o == hs- +oh-hs- +h2o === h2s + oh-5 al(oh)3 + oh- === 2h2o + alo2 -6 2na2o2 + alcl3 ==== 4na+ +3cl- +alo2-

7 2fe3+ +cu == 2fe2+ +cu2+'

Do you still have to write chemical equations.

It's better to put it to me, I'm in pursuit of the rate, and it's all 80% now, thank you.

1 2hpo4- +2oh- +3ca2+ = ca3(po4)2 + 2h2o

2 caco3 + 2ch3cooh = ca2+ +2ch3coo- +h2o + co2 ↑

3 HCO3- +H2O = reversible number = H2CO3 + OH-4 S2- +H2O = reversible sign = HS- +OH- main.

HS- +H2O = reversible number = H2S + OH- times)5 AI2O3 + 2OH- = 2AIo2 - H2O6 2NA2O2 + 2H2O = 4NA+ +4OH- +O2 AI3+ +4OH- = ALO2- +2H2O7 2Fe3+ +CU = 2FE2+ +Cu2+

1. An ionic reaction of adding excess clarified lime water to the disodium hydrogen phosphate solution.

2. Use vinegar to remove the limescale on the inner wall of the thermos gallbladder.

3. At room temperature, NaHCO3 solution is alkaline.

4. The aqueous solution of sodium sulfide is alkaline.

5. Remove the oxide film on the aluminum surface with sodium hydroxide solution.

6. Put 2mol of sodium peroxide into a solution containing 1mol of AlCl3.

7. Ferric chloride solution corrodes copper to make printed circuit boards.

A correct. aSodium bicarbonate has low solubility and can be precipitated.

B error, excess SO2 generates CAHSO3, not precipitate.

c False, the O in hydrogen peroxide is oxidized by permanganate, and hydrogen peroxide itself does not show oxidation. So it should be: 2mNO4 +5H2O2 + 6H+ =2mn2++ 5O2 + 8H2O (the original equation is equivalent to adding a hydrogen peroxide decomposition on the basis of this equation 2H2O2 ==O2 + 2H2O, where H2O2 is both an oxidant and a reducing agent.)

But when it reacts with potassium permanganate, it only acts as a reducing agent, so the decomposition reaction does not occur).

D: False. A small amount of naoh only provides one oh-, not two. In the reaction equation is 2 oh- participated, so it is wrong.

a, the problem of solubility, can produce precipitation.

In b, hypochlorous acid has strong oxidation and can generate sulfate.

c, the gain and loss electrons are not conserved.

d, a small amount of sodium hydroxide is added, no ammonia monohydrate is produced, and only reacts with acid acid.

A is not the H+ and NO3- in hi to form Hno3! Hno3 reacts first than Fe3+, but Hi is excessive, and it is all reduced in the end.

fe3+ +3no3- +10 i- +12h+==5i2 +fe2+ 3no +6h2o

b Oxidation of NO2- to KMNO4

C is not correct for excess Ba(OH)2, which means that Ba2+ is sufficient, and there will be no CO32- ions, and CO32- will generate BaCO3 precipitation.

hco3- +ba2+ +oh- ==baco3 ↓ h2o

d is not trimmed 2fe3+ +cu = 2fe2+ +cu2+

Pick B

The second answer, the first divalent iron ion is wrong, it should be trivalent, the third, it should be a hydroxide and a bicarbonate heel, and the fourth is not a matched ......

B pair, the oxidation of I2 in A is stronger than that of Fe3+, and the reaction cannot be carried out. If there is an excess of Ba(OH)2 in C, the product should not be unbalanced and the charge is not conserved.

A. False, Hi is a strong acid, excessive Hi, indicating that there is a large amount of H+ in the solution, then according to the existence of nitrate ions, it is equivalent to having a large amount of nitric acid, so there can be no ferrous ions.

b. Correct. c. Error, barium hydroxide is excessive, the product cannot have carbonate D, error, no trim, the correct should be 2Fe3+ +Cu = 2Fe2+ +Cu2+

1) There is H+ in the solution, so NO3- is combined into HNO3, and Fe2+ is oxidized to Fe3+

2) If there is enough hi, then a is also right.

B is true and c is false, and there is an excess of barium ions.

D is wrong No trim 2Fe3+ +Cu = 2Fe2+ +Cu2+ choose B,

It's been a long time since I've done a chemistry problem.

The basic principle of balancing is that the amount of which substance is less, the coefficient of this substance is 1, and the amount of other substances is less according to the needs. Let's take a look at these examples:

1) Adequate amount of nahso

Solution with BA (HCO

The solution is mixed. Analysis: Two solutions are mixed, including two ion reactions. Namely: HCOHCO

ho and. ba

sobaso

And 1 BA (HCO

Contains 2 HCOs

and 1 バ so nahso

The amount is determined by HCO

Decision. i.e. BA (HCO

and nahso

The coefficients are 1 and 2, respectively.

The chemical equation is:

ba(hco

2nahso

basonaso

The 2CO2ho ion equation is. ba

2hco2hso

baso2co

2ho2) a sufficient amount of BA (HCO

solution with nahso

The solution is mixed. Analysis: Two solutions are mixed, including two ion reactions. Namely: HCOHCO

ho and. ba

sobaso

And 1 nahso

Contains H and 1 SO

So BA (HCO

The amount is determined by h and so

Decision. i.e. BA (HCO

and nahso

The coefficients are 1 and 1, respectively.

The chemical square chain stupid closed program is:

ba(hco

nahsobasonahcocoh

oThe ionic equation is Ba

hcohso

basocoh

O3) Add a sufficient amount of NaOH solution to the CA (HCO solution). Analysis: Two solutions are mixed, including two ion reactions. Namely: HCOOHCO

ho and. ca

cocaco

And 1 CA (HCO

Contains 2 HCOs

and 1 CA so the amount of NAOH is determined by HCO

and CA decisions. i.e. CA (HCO

and NaOH have coefficients of 1 and 2, respectively.

The chemical equation is:

ca(hco

2naohcaco

naco2h

oThe ionic equation is ca

2hco2ohcaco

COho4) will be a sufficient amount of CA (HCO

The solution is added to the NaOH solution.

Resolving shed cleavage: two solutions are mixed, including two ion reactions. Namely: HCOOHCO

ho and. ca

cocaco

Whereas, 1 NAOH contains 1 OH

So CA (HCO

The amount is determined by OH. i.e. CA (HCO

and NaOH have coefficients of 1 and 1, respectively.

The chemical equation is:

ca(hco

naohcaco

The Nahcoho ion equation is Ca

hcoohcacoho