The chemical ion equation is a small amount and a large amount problem super simple .

Answer: (1) When CO2 is small: CO2 + Ca2+ +2OH- = CaCO3 + H2O

2) When CO2 is excessive: CO2 + OH- = HCO3- Analysis: (1) When CO2 is small, the chemical equation for the reaction of CO2 and Ca(OH)2 is: CO2 + Ca(OH)2=CaCO3 + H2O

Therefore, its ionic equation is: CO2 + Ca2+ +2OH- = CaCO3 +H2O

2) When there is an excess of CO2, the chemical equation for the reaction of excess CO2 with Ca(OH)2 is:

CO2 + Ca(OH)2=CaCO3 +H2OCAco3+CO2+H2O=Ca(HCO3)2 Therefore, its ionic equation is:

CO2 + Ca2+ +2OH- = CaCO3 +H2OCAco3 + CO2 + H2O = Ca2+ +2HCO3-+ The total ionic equation is: CO2 + OH- = HCO3- Finally, I wish you progress in your studies!!

In small amounts, CO2 + Ca2+ +2OH- = CAC3 + H2O, in excess, CO2 + OH- = HCO3- In fact, in excess, it can be regarded as the product of the first step of reaction, and CaCO3 reacts with CO2 and water.

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

CO2 + Ca(OH)2 = CaCO3 + H2O (when a small amount of CO2 is common) is converted to the ionic equation CO2 + Ca2+ +2OH- = CaCO3 +H2O

CaCO3 + CO2 + H2O = Ca(HCO3)2 (when excess CO2 is introduced) is converted into an ionic equation in which excess CO2 is directly introduced into Ca(OH)2.

2co2 + h2o = 2hco3-

In small amounts, CO2 + 2 OH- = CO32- +H2O

CO2 + OH- = HCO3- in case of overdose

That's probably it, as long as you know that CO2 reacts with water to form carbonic acid, it decomposes into H+ and HCO3- before reacting, and HCO3- will also decompose in small amounts.

CO2 + Ca(OH)2 = CaCO3 + H2O in small amounts

In case of overdose 2CO2 + Ca(OH)2 = CaHCO3 + H2O

Then convert it to ionic formula

Ca(HCO3)2 is generated

ca(oh)2+2co2=(ca2+)+2hco3-

The mobile phone called, it can only be like this, I don't understand and ask again.

The basic principle is that the amount of a substance with a relatively small amount of reactant is set at 1mol, and the amount of another reactant is taken as much as possible.

Add a small amount of NaOH solution to the Ca(HCO3)2 solution.

Add sufficient NaOH solution to Ca(HCO3)2 solution. Please write the ionic equation for the reaction.

Analysis: A small amount of NaOH solution in the reactant is used by the method of "determining one with less", and NaOH is set to 1mol and Ca(HCO3)2 is sufficient.

oh- +hco3- ＝h2o + co32-。

1mol needs 1mol to produce 1mol 1mol.

co32- +ca2+ ＝caco3↓。

1mol takes 1mol to produce 1mol.

Ion equation: oh- +HCO3- +Ca2+ H2O + CaCO3.

A small amount of Ca(HCO3)2 solution is in the reaction, and Ca(HCO3)2 is 1mol, and the amount of NaOH is sufficient.

ca（hco3）2＝ca2+ +2hco3-。

1mol 1mol 2mol。

2hco3- +2oh- ＝2h2o + 2co32-。

2mol takes 2mol to produce 2mol 2mol.

ca2+ +co32- ＝caco3↓。

1mol takes 1mol to produce 1mol. Remaining: 1mol CO32-.

Ion equation: Ca2+ +2HCO3- +2OH- CaCO3 +CO32- +2H2O.

Summary: A small amount of reactant is NaOH solution, which is set at 1mol, and the amount of Ca(HCO3)2 is taken as much as Ca(HCO3)2.

A small amount of Ca(HCO3)2 solution in the reactant is set at 1mol, and the amount of NaOH is taken as much as needed.

Excess vs. small amount of ion equation:

1. Chemical equation: Fe + 4Hno3 (dilute) = Fe (No3) 3 + No + 2H2O - ion equation: Fe + 4H+ +NO3- = Fe3+ +No +2H2O.

2. Chemical equation: 3Fe + 8HNO3 (dilute) = 3Fe (NO3) 2 + 2NO + 4H2O - ion equation: 3Fe + 8H+ +2NO3- = 3Fe2+ +2NO +4H2O.

Error-prone analysis. 1. All oxides and peroxides are written with chemical formulas, and beginners are easy to ignore that only soluble and easily ionized electrolytes are represented by ion symbols, which often disassemble many insoluble strong electrolytes, resulting in errors. It must be clear here that active metal oxides or peroxides such as sodium peroxide and sodium oxide are soluble electrolytes but cannot be dismantled.

2. There are also soluble strong electrolytes like sodium bicarbonate, but sometimes (such as carbon dioxide into saturated sodium carbonate) is also written as a chemical formula, it depends on whether it mainly exists in the form of a solid substance or in the form of ions in solution.

You have to name at least a few chemical elements.

A lot of chemical elements.

For example: 1Excess Ca(HCO3)2+NaOH

There are different equations for different elements.

Let's take the reaction between NaOH and Ca(HCO3)2 as an example to teach you a simple and practical method.

Like the above reaction, when encountering an excessive amount or a small amount, you first treat a small amount of the substance as 1mol

The number of measurements of the substance in the equation is considered to be 1Consider the reaction again.

For example, excess NaOH reacts with Ca(HCO3)2. I assume Ca(HCO3)2 as 1molThen 1mol.

Ca(HCO3)2 has 1mol of Ca2+

and 2mol of HCO3-

For the reaction, 2mol of OH - that is, 2mol of NaOH is requiredi.e.: 2HCO3-

2oh-2h2o

2co32-

The resulting 2mol of CO32- reacts with 1mol of Ca2+ to form 1mol of CaCO3 precipitate.

i.e.: CO32-

ca2+caco3↓

So, the ionic equation for the total reaction is: 2HCO3-

2oh-ca2+

caco3↓+

CO32- In addition, if a small amount of NaOH reacts with a sufficient amount of Ca(HCO3)2, OH- is considered as 1mol

The ionic equation of the reaction: HCO3-

oh-ca2+

caco3↓

Hello landlord, the keyboard is not easy to complete the input equation.

So I'll briefly talk about the method.

Of course, there are many ways to do this, but I recommend the "less is one" method.

Using 1 as an example, let me introduce this approach.

The so-called fixed less than 1 is to write the coefficient of a small amount of matter as 1. This gives the NAHCo3 coefficient to 1. Because there is less NaHCO3, it will be completely reactive.

To react to one HCO3, one OH is required. In this way, there is one OH left in Ba(OH)2, which will combine with the Na ion to form NaOH.

To summarize briefly, after setting the coefficient of a small amount of a substance to 1, let's see how to react it with ions.

Fall. Let's take 2, which is the opposite of 1, for example.

If Ba(OH)2 is small, then two OHs need to be reacted off, and two HCO3s are needed, so the coefficient before NaHCO3 is 2, and two carbonates are generated by the reaction.

One binds to the Zhiwu car BA ion and the other binds to two NA ions.

To summarize again, to compare the difference between the two, we need to react all the small amounts of substances and then make up the other substances.

Note: It is easy for the landlord to understand, the amount of the ant or the substance.

This is actually not difficult to understand. Just remember, which is a small amount, just use this as a quantification, split it into ions according to the proportion of the molecular formula, and write it down first, for example, the first one is written oh-, (sodium ion.

You don't need to write it).The second one is ca2+ 2HCO3-Then write about that excess of the substance's ions.

Because it is excessive, it is certain that the void wisdom will consume all the ions of the small amount of substances written above, so the number of excess ions should correspond to the previous one, for example, the first one, in order to consume an oh-, it is necessary to use a hco3-, so write a hco3-, and the reaction between the two will produce a carbonate.

will and calcium ions.

A precipitate is generated, so a calcium ion needs to be added, so that the left side of the equation is written, and the right side only needs to be written as a generated water and a calcium carbonate.

Precipitation is fine. The first two panicles are the same, first to consume two HCO3-, then, you need two OH-, obviously to write bucket answer 2oh- (finish writing on the left side of the equation), so that two carbonate ions will be generated, but before there is only one calcium ion, then, only one calcium carbonate precipitate can be generated, so on the right side of the equation write a calcium carbonate precipitate, two generated water, and one carbonate ion.

It's very detailed, if you don't know, you can continue to discuss it with me.

High pressure ==2no

2no+o2==2no2

Wild reform. 3no2+h2o=2hno3+no

Or: 4NO2 + O2 + 2H2O = 4Hno3 concentrated) == Heating = = MnCl2 + Cl2 +2H2OMN (+4) Mn (+2 is reduced) (2E-) Cl(-1) Cl2 (0 is oxidized) (loses 2E-) 3A small amount of chlorine gas is passed into the ferrous bromide solution, and because the ferrous ions are more reducible than the bromine ions, only the ferrous ions are oxidized.

The chemical equation is: 3Cl2+6Febr2===4Febr3+2FeCl3 is replaced by an ionic equation: Cl2

2fe2+==2fe3+

2cl- sufficient chlorine gas is passed into the ferrous bromide solution, and in addition to the oxidation of ferrous ions, the bromide ions are also oxidized

The chemical equation is 3Cl2

2febr2===2fecl3

The equation for 2br2 to ion is: 3cl2+2fe2+4br-==2fe3+

2br26cl-

1.Baking soda (sodium bicarbonate).

If Ca(OH)2 is insufficient.

ca(oh)2

2nahco3===caco3↓

na2co3

2H2O ion equation: 2oh-

ca2+2hco3-

caco3↓+2h2o

If Ca(OH)2 is excessive.

ca(oh)2

nahco3===caco3↓

naohh2o

Ion equation: Ca2+HCO3-

caco3↓+oh-h2o

2nahso4

baohbaso4↓+

H2ONa2SO4, which is in case of sodium bisulfate excess.

2nahso4

2baoh2baso4

H2O2NaOH, which is in the case of excess barium hydroxide ridge sliding.

Ionic equation: (BA2+)+OH-)+H+)+SO42-)=BASO4 +H2O

Write the chemical equation first and then split it into ionic equations.

ca（hco3）2

2naohcaco3↓

na2co3

2H2O split, remove 2Na+ on 2 sides to obtain.

ca2+2hco3-

2oh-caco3↓

co32-2h2o

Considering the case of excess NaOH, 1 part Ca(HCO3)2 in the chemical equation above

and 2 parts of NaOH just react completely, while if there is an excess of NaOH will not react with the three products CaCO3 and Na2CO3

H2O any kind of reaction, so"Overdose"It does not affect the writing of equations.

Finally, let's see why ca(oh)2 is not generated, because the reason mentioned upstairs is that caco3 is more insoluble than ca(oh)2.

Specific analysis: Assuming that Ca(OH)2 is generated, the remaining components in the solution must have NaHCO3, and these two cannot coexist.

ca(oh)2

2nahco3

caco3↓+

na2co3

2H2O calcium hydroxide in small amounts).

ca(oh)2

nahco3

caco3↓+

naohh2o

Excess calcium hydroxide, excess calcium hydroxide will produce Ca(OH)2 with the sodium carbonate produced above

na2co3

caco3↓+

2naoh, which is obtained after stacking).