What are the ions that actually participate in the reaction in the ion equation

Ionic equations.

The ions written in are the ions that actually participate in the reaction.

For example, the NaCl+AgNO3 reaction.

In NaCl is Na+ Cl-

In agno3 is ag+ no3-

When putting the cations and cations on top together, if it is a precipitate, gas, or water. Then react.

So the above 4 ions together will form AgCl precipitate, while Na+ NO3- has no reaction.

The ionic equation ag++cl-=agcl (precipitation) Of course, in addition to the formation of precipitation, there are many more gases and waters. For example, redox reactions.

Thorough double hydrolysis, complexation reaction. Wait.

For example, when a solution of Na2CO3 and NaHSO4 is mixed, the ions that actually participate in the reaction are.

In high school, we still think that bisulfate ions can be completely ionized, so the answer to this question in high school is: hydrogen ions and carbonate ions; In excess of sodium bisulfate. co3

2h+h2o

CO2 (rising); When the amount of sodium bisulfate is insufficient. CO3H+HCO3- (rise);

However, I hope the following will not mislead you, the university believes that the first step of sulfuric acid ionization is complete, and the second step of ionization is incomplete, HSO4-

The H+SO4 ionization equilibrium constant is only 10-2 orders of magnitude "moderately strong acid" (this does not mean that sulfuric acid is not a strong acid: the first ionization is still complete). Therefore, at the university level, this question should be carbonate and bisulfate ions.

Ionic equation: The formula that represents an ionic reaction with the symbol of the ions that actually participate in the reaction.

Ions whose concentration changes before and after the reaction.

A Febr 3 solution does not contain ferrous ions, only bromine ions are dry as reacted with chlorine water, the correct ion equation is: 2br -+Cl 2=br 2+2Cl -, so A is wrong;

B Acetic acid is added to the baking soda solution, baking soda is sodium bicarbonate, and the correct ion equation is: HCO3 -+CH 3COOH=CO2 +H2O+CH3COO -, so B is wrong;

c Add a small amount of chlorine water to the sodium carbonate solution to generate bicarbonate ions, the correct ion equation is: CO3 2-+H+=HCO3-, so C is wrong;

d The sodium carbonate solution absorbs a small amount of sulfur dioxide, and the reaction generates bicarbonate ions, and the ion equation of the reaction is: 2CO3 2-+SO2+H2O=2HCO3 -+SO3 2-, so D is correct;

Therefore, d ,8 is selected, and the ionic equation that can correctly represent the following reactions is ( ).

A Febr3 solution with excess chlorine water 2Fe 2++4BR -+3Cl 2=2BR 2+6Cl -+2Fe 3+

b Add acetic acid CO3 2-+2CH 3COOH=CO2 +H2O+2CH3COO -to the baking soda solution

c Add a small amount of chlorine water CO3 2-+2H +=CO3 2-+H2O to the sodium carbonate solution

d Absorb a small amount of dioxane hail sulfur 2CO3 2-+SO2+H2O=2HCO3-+SO32- with sodium carbonate solution

A is false, nitric acid has strong oxidizing property regardless of whether it is concentrated or dilute. fes+4h++no3-==fe3++s+2h2o+no↑。

b error, a small amount of sulfur dioxide, the formation of sulfate ions. 2c6h5o-+so2+h2o==2c6h5oh+so32-。

c False, under acidic conditions, there is H+ in the reactant difference, and there will be no OH- in the product, but.

d correct. For the problem of small amount and excess, the principle of less and more variable can be used, that is, a small amount of substance is set at 1mol to see how much excess can be consumed.

E error, HCO is a weak acid, to write the molecular formula, and is a reversible reaction. cl2+h2o↔h++cl-+hclo。

f False, chlorine is excessive, it will oxidize sulfite ions, it should be the formation of sulfate ions. s2o32-+4cl2+5h2o=2so42-+8cl-+10h+。

g false, hypochlorous acid has strong oxidizing properties and can oxidize sulfur dioxide. Ca2++2ClO-+2SO2+2H2O==CaSO4 +SO42-+2Cl-+4H+, which is a reaction of excess SO2.

h error, the generated copper should have a precipitation symbol.

iTrue, because the sodium carbonate solution is saturated and the solubility of sodium bicarbonate is less, so the sodium bicarbonate produced is mostly precipitated.

j False, ethyl bromide does not contain bromide ions, and silver bromide precipitates cannot be generated. The correct way to do this is to add a slightly excess sodium hydroxide solution to the ethyl bromide and heat it, then add an excess of nitric acid to the solution, and finally add silver nitrate.

The ionic reactions that occur are as follows: C2H5BR+OH-= heating = C2H5OH+BR-, H++OH-===H2O (adding nitric acid is to neutralize the excess sodium hydroxide), BR-+AG+==AGBR.

A False, nitric acid has free cracking oxidation, which will produce S precipitation, and dimethyl iron will be oxidized.

B False, a small amount of SO32- is generated, and an excess of HSO3-C is generated, how can acidic lead be produced to OH-, which is a reaction formula under alkaline conditions.

d False, excess barium hydroxide is, 2HCO3-+Ba2++2OH-===BaCO3 +CO32-+H2O

e Wrong, did you write it wrong, or? CL is not the same left and right.

f False, SO32- will be oxidized to SO42-

g False, CLO - has strong oxidizing properties, oxidizing SO32-H is correct. i Correct (Looks like you're typing the wrong one again...)

j False, the bromine in ethyl bromide is covalently bonded to the molecule, and there is no br-

A nitric acid can oxidize ferric iron, so wrong.

b should be the generation of sulfite, liquid accompaniment instead of bisulfite loss root, sulfur dioxide is not empty stupid as carbon dioxide.

In the case of c, it is impossible to have a large amount of hydroxide under acidic conditions, wrong.

D right. e hypochlorous acid is a weak acid and cannot be disassembled.

f Chlorine gas has strong oxidizing properties, oxidizing sulfite to sulfate, so wrong.

g Hypochlorous acid has strong oxidizing properties and can oxidize calcium sulfite into calcium sulfate.

The h-point solution should produce copper and hydrogen.

Bromine in i-j bromoethane is non-ionizing, so it cannot be combined with silver ions.

So right things d and i

I'm so tired, give me these 10 fortunes.

(1) Iron powder reacts with water vapor 3Fe+4H2O=Fe3O4+4H2 (2) Aluminum reacts with hydrochloric acid 2Al+6HCl=2AlCl3+3H2 2Al+6H+=2Al3++3H2

3) Reaction of aluminum with sodium hydroxide solution 2Al+2NaOH+2H2O=2Naalo2+3H2

2al+2oh-+2h2o=2alo2-+3h2 (4) Zinc particles react with copper chloride solution Zn+CuCl2=ZnCl2+Cu Zn+Cu2+=Zn2++Cu

1. Iron powder and water vapor reaction: 3Fe + 4H2O (gas) = = Fe3O4 + 4H2 (gas);

Ion equation: 3Fe + 4H2O (gas) = = Fe3O4 + 4H2 (gas);

2. Reaction of aluminum with hydrochloric acid: 2Al+6HCl==2AlCl3+3H2 (gas) ion equation: 2Al+6H+==2Al3+

3h2 (gas).

3. Reaction of aluminum with sodium hydroxide solution: 2Al + 2NaOH + 2H2O = = 2Naalo2 + 3H2 (gas).

Ion equation: 2al+2oh-

2h2o==2alo2-+

3H2 (gas) (metaaluminate is -1 valence, the negative sign is on it) 4 zinc particles react with copper chloride solution: Zn+CuCl2==ZnCl2+Cu ion equation: Zn+Cu2+==Zn2++Cu

1) Look: whether the reaction conditions are reactions between free ions. The ionic reactions carried out in non-solution cannot be represented by ionic equations.

2) See: whether the reaction can occur. For example, AgCl+Hno3 ≠, because AgCl is insoluble in acid.

3) Look: whether it conforms to the law of reaction. For example, 2Fe + 6H + = 2Fe3 + + 3H2 (false) Fe can only be oxidized to Fe3+ when it encounters a strong oxidant

4) Look: whether it conforms to the rules for writing ion equations. Soluble strong electrolytes are written as ionic symbols, and insoluble substances, gases, and weak electrolytes are written as molecular formulas.

5) Look: whether there are microsoluble substances in the reactants and products. (6) Look:

Whether the two conservation of the ion equation is achieved or not, i.e., the conservation of mass and the conservation of charge. (7) Look: whether the ion ratio is correct.

A few questions that are easy to ignore: 1 Is it possible to write an ionic equation for all ionic reactions? Not necessarily.

The ionic equation is written only in the ionic reaction that takes place in solution or in the molten state. For example, the reaction between solid ammonium chloride and slaked lime in the laboratory is essentially an ionic reaction, but it is not a reaction in the form of freely moving ions, so it can only be expressed by molecular formulas, and cannot be written with ionic equations.

2 Can the ion equation represent all the same type of ionic reactions? "The ionic equation represents not only a certain reaction between a given substance, but all ionic reactions of the same type".Is this true for all, without exception?

As it turns out, this is just a general rule. For example, H++OH-=H2O is the ionic equation for the reaction between a strong acid and a strong base, but not all reactions between a strong acid and a strong base can be expressed by this ionic equation.

The ionic equation for the reaction of sulfuric acid with barium hydroxide solution is: that is, this ionic equation does not represent a class of reactions. The reason for this particular is that in these reactions, all the ions in the reactants are involved in the reaction, and there can be no second substance to replace them.

This determines the 1-to-1 correspondence between these ionic equations and chemical equations. 3 Is the ionic equation necessarily represented by the symbols of the ions that actually participate in the reaction? "The formula that represents an ionic reaction by the symbol of the ions that actually participate in the reaction is called the ionic equation.

However, this does not mean that all ionic equations must be expressed in this way, for example, when the refractory ionization substances, insoluble substances, and concentrated sulfuric acid in the reaction participate in the reaction, they must be expressed by molecular formula or chemical formula in accordance with the regulations, and cannot be disassembled into ion symbols.

1.How to solve the ionic reaction equation for a substance in small quantities and in sufficient quantities.

The empirical method is to set the coefficient before a small amount of substances to 1, and then determine the coefficient of excess (the coefficient does not appear as a fraction), for example, the more difficult in middle school is 1 NaHCO3 + Ca(OH)2 (excess) = CaCO3 + NaOH + H2O, 1 Ca(OH)2 + 2NaHCO3 (excess) = CaCO3 + Na2CO3 + 2H2O

2.How to solve the ionic reaction equation when the quantity ratio of the substance is different, such as ferrous bromide and chlorine.

The reaction of Fe2+ with chlorine is written first, and then the reaction of bromine ions with chlorine is written.

3.How to solve the equation of the ionic reaction when the sodium bisulfate solution is dropped into the barium hydroxide solution to neutral and to the sulfate ion is just completely precipitated.

Focus on the builds.

Neutral, then no alkali or acid is required in the product: 2NaHSO4 + BA(OH)2=Na2SO4 + BASO4 + 2H2O

Sulfate precipitation is complete: NaHSO4 + BA(OH)2 = NaOH + BASO4 + H2O

1: A small amount or an excess, this kind of question requires you to pay more attention to the reaction of the substance, different amounts will cause different products, such as iron (Fe), which reacts with Hno3 and other problems.

2: This kind of problem needs to be calculated, and the calculation is to see which substance has more content, and there is more Cl, and the reaction products are only FeCl3 and Br2A small amount, an extra product, FeCl2

After that, according to the chemical reaction equation, it is simple to write the ion equation. Just pay attention to the product.

3: You have to ask this kind of question one by one, and what products are there. The product of this problem is the production of water and barium sulfate precipitates. It's simple.

From the above analysis, it is not difficult to see that this kind of problem can write ionic equations as long as you pay attention to the amount of the product and the amount of substances involved in the reflection to determine the product! Hope you understand.

The amount should be specific, for example, the reaction of barium hydroxide and sodium bisulfate, when the reaction solution of the two is neutral, it is: BA(OH)2+2NaHSO4==BASO4(precipitation)+Na2SO4+2H2O

When the barium ion is completely precipitated, it is: Ba(OH)2+2NaHSO4==BASO4(precipitation)+2NaOH+H2O