A few questions about the ionic reaction equation

1)al3++

Aluminum hydroxide is insoluble in weak bases.

2.It is incorrect to write NH4OH, NH4+ will not stay in NH4OH after combining with OH, but will be rapidly converted into NH3·H2O, so NH4OH cannot actually exist.

Must be written.

Same as NH4OH, you can write either of the ion reaction formulas.

The ionic reaction equation for the reaction of excess ammonia and aluminum chloride.

al3+ += al(oh)3 + 3nh4+

1)al3++

2) NH4OH is written in a wrong way, and it is occasionally seen in old books and should be written.

The first question has passed a few upstairs, so I won't say much.

What exactly is the difference between NH4OH and NH4OH?

I don't think there's any difference, it's okay, nh4oh is not wrong to write like this. Just like writing the chemical formula of benzene, you can draw 3 lines and a circle in it.

Personally, I think it's better to write ionic reaction formulas for NH4OH.

al3+ += al(oh)3 + 3nh4+

There is no difference between the two, both are written now.

An example of an ionic reaction equation is as follows:

The ionic equation for the reaction of sodium carbonate solution and calcium sulfate solid is given below:

co32- +caso4 = caco3↓ +so42-

2. The ion equation for the reaction between chlorine and lime milk:

cl2 + ca（oh）2 = ca2+ +clo- +cl- +h2o

3. The ionic equation of iron oxide dissolved in dilute hydrochloric acid:

fe2o3 + 6h+ =2fe3+ +3h2o

4. The ionic equation for the reaction of calcium carbonate and acetic acid:

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

5. The ion equation for the reaction of chlorine gas with ferrous chloride solution:

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

6. Ionic reaction equation of copper and concentrated sulfuric acid:

Cu + 2H2SO4 (concentrated) = Cu2+ +SO42- +SO2 + 2H2O (Condition: Heated).

7. The ionic equation for the reaction of alumina and sulfuric acid:

al2o3 + 6h+ =2al3+ +3h2o

8. The ionic equation for the reaction of copper and concentrated nitric acid:

cu + 4h+ +2no3- =cu2+ +2no2↑ +2h2o

9. The ionic equation for the reaction of calcium carbonate and water and carbon dioxide:

caco3 + h2o + co2 = ca2+ +2hco3-

10. The ionic equation for the reaction of ferrous oxide and concentrated nitric acid:

feo + 4h+ +no3- =fe3+ +2h2o + no2↑

1) NaCN is a strong alkali and weak salt, hydrolysis is alkaline, and the ionic equation is: CN - H2 O HCN+OH - so the answer is: alkali; cn - h 2 o⇌hcn+oh -

2) cn - ionizing water cn - h2o hcn + oh - water can ionize: h2o oh - h + so the ion concentration magnitude relationship is: c(na + c(cn - c(oh)> c(h + so the answer is:

c(na + salute c(liangshan comic cn - c(oh)> c(h +

3) The conservation of charge is based on the apparent neutrality of the solution, that is, the concentration of the positive charge of the hydrogen ions in the solution is equal to the concentration of the negative charge of the anions, so it is: c(na + c(h + c(oh - c(cn -

The conservation of materials is based on the column formula of CN- partial hydrolysis to HCN, so there is C(Na + C(HCN) + C(CN -

This reaction cannot be written as an ionic equation because it is not this reaction that occurs in water, Na2O reacts when it meets water: Na2O H2O 2NaOH; The essence of Na2O's reaction with CO2 in water is that NaOH reacts with CO2, so Na2O cannot be written as an ion.

2.In the ionic equation, gas, precipitation, and weak electrolyte cannot be written in the form of ions, and elemental and oxide are generally weak electrolytes or gases and precipitates, so they generally cannot be disassembled.

3.The reaction between solids cannot be written as an ion equation, it means that two or more substances react in a solid state, such as grinding sulfur powder and aluminum powder: 2AL3S AL2S3, the powdered solid is dissolved in water, it is no longer in the form of a solid, but generally in the form of (hydrated) ions, which is recorded as "aq", which is no longer satisfied with the "reaction under the solid state", so this sentence is correct, and the reaction between the solid and the silver segment of the gas Ling finger body cannot be written as an ion equation.

Before writing the ion equation, it is necessary to see the state of the reaction.

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

the principle of less is more; There is also to see whether the formed substance can exist stably in the solution after the reaction, and it is not difficult to follow these two points!

It's all a matter of quantity, and you can control what you want, for example, if you have an excess, you write it down, and then add the excess substance to see if it can react, until the product and the excess substance do not react, and then there is the neutralization one, which is actually to control the water, so that H and Oh react completely to produce water, and then balance it with the water.

Aluminum and hydrochloric acid react 2AL+ 6H+ =2AL3+ +3H2

The reaction of alumina and hydrochloric acid Al2O3 + 6H+ =2Al3+ +3H2O

The reaction of aluminium and oxygen 4Al + 3O2 ==2Al2O3

The reaction of aluminum hydroxide and hydrochloric acid Al(OH)3 + 3H+ = Al3+ +3H2O

Reaction of aluminium chloride and sodium hydroxide Al3+ +3OH- =Al(OH)3 (sodium hydroxide insufficient), Al3+ +4OH- =AlO2- +2H2O (sodium hydroxide excess).

Reaction of aluminium hydroxide and sodium hydroxide Al(OH)3 + OH- =ALO2- +2H2O

Alumina and sodium hydroxide Al2O3 + 2OH- =2alo2- +H2O

Reaction of aluminium and sodium hydroxide with water 2AL + 2OH - +2H2O ==2alo2- +3H2

Magnesium sulfate and barium hydroxide soluble strong electrolyte, to write the ionic form, the reaction of Baso4, Mg(OH)2 is the precipitation, can not be written ionic form, potassium carbonate is a soluble strong electrolyte, so:

mg2+ +so42- +ba2+ +2oh- ＝baso4↓ +mg（oh）2↓

Potassium carbonate and the resulting potassium acetate are soluble strong electrolytes, to write the ionic form, acetic acid is a weak electrolyte, the generated H2O is a weak electrolyte, and CO2 is a gas, and the ionic form cannot be written, so:

co32- +2ch3cooh ＝ 2ch3coo- +h2o + co2↑

No, you misunderstood, soluble strong electrolytes can be disassembled, Na2CO3 is salt, and salt is to be disassembled.

When writing the ion equation, you need to pay attention to:

1.Elemental, gas, and precipitate require the complete chemical formula to be written.

2.Refractory electrolytes or non-electrolytes also need to be written with the full chemical formula.

3.If it changes from ionic to non-ionic before and after the reaction, it should be retained.

If these three are ensured, it should be possible to write the ion equation directly without the need to write it through the reaction formula. In order to write ion equations quickly, you need to practice a lot and be familiar with the properties of matter and the reaction equations.

Hope, thank you.

Electrolytes can generally be divided into strong electrolytes and weak electrolytes, and the conductivity of the two varies greatly. Strong electrolytes can be considered to exist entirely in the form of ions in solution, i.e. there are no "molecules" of the electrolyte (at least in the range of dilute solutions).

Strong electrolytes: all salts (high school stage), three strong acids (high school stage inorganic strong acids are only required for these, organic strong acids are only understood, there are trinitrophenol (picric acid), etc.), and alkali metals, alkaline earth metals formed by alkali metals (lithium hydroxide, magnesium hydroxide are medium and strong bases, except for beryllium hydroxide is amphoteric acid).

The reason why sodium carbonate needs to be dismantled is because carbonate is not easy to seize hydrogen ions in water to form bicarbonate. Therefore, it basically exists as carbonate ions in water.

Na2CO3 is soluble and exists in water as sodium ions and carbonate particles and a small amount of bicarbonate particles, and of course it can be disassembled.

Na2CO3, as a salt, is ionizable and is a strong electrolyte that can be completely ionized.

na2co3 = 2na+ +co3^2-

CO3 2- is a weak acid and ion, hydrolyzed, and does not ionize.

Salt is all detachable1! Weak acid and weak alkali are hungry 1 that cannot be dismantled!! Sodium carbonate is a strong alkali and weak salt!! As long as it's salt, you can dismantle it!!