About ionic reactions, what an overdose or something

I have my own way of writing this kind of reaction equation, which is actually a sentence: whoever has less will take it out, and use a few more to add a few. For example, let's use the one you said.

A small amount of NaHCO3 is added to BA(OH)2, and there is less sodium bicarbonate in this reaction, so take out the bicarbonate ions of sodium bicarbonate (all reactions are calculated according to 1mol, and 1mol of sodium bicarbonate contains 1mol bicarbonate ions, so take out one) and then barium hydroxide is excessive, add a few with a few, 1mol of bicarbonate ions need to react with 1mol of hydroxide ions, 1mol of Ba2+ to generate barium carbonate and water. So the reaction equation is.

hco3-+ba2++oh-=baco3+h2o

If there is a small amount of barium hydroxide, sodium bicarbonate is excessive. So according to my method, 1mol of barium hydroxide in 1mol of ba2+ and 2moloh- are all taken out, and 2moloh- needs to react with 2mol of bicarbonate ions, so 2mol of bicarbonate ions are added, and the equation is: ba2++2oh-+2HCO3-=BAC3+2H2O+CO32-

I've been writing these equations this way, and I've never made a mistake so far, so please forgive me if I'm not clear about my statement. Note that the rule only applies to acid salt and alkali reactions, and the others do not apply.

That's a small amount, and when you write the equation, you can just write the exit equation that completely reflects 1 mole of the substance.

The ionic equation for dropping NaHCO3 into the Ba(OH)2 solution is:

hco3-)+ba2+)+oh-)= (baco3)↓+h2o)

You can look at the equation, because there is only one NaHCO3, and there is only one HCO3-, which consumes an OH-, which also generates a CO3-, which is combined with BA2+ to form BACO3

But if it is Ba(OH)2 drops into NaHCO3.

2(hco3-)+ba2+)+2(oh-)= (baco3)↓+h2o)+co3-

Similarly, Ba(OH)2 is a small amount, so the ion equation.

That is, the equation that consumes a BA(OH)2.

If you have any chemistry problems in the future, you can talk to me, because I am in my third year of high school, and many knowledge points have been picked up, and some rules are too useful for your first and second years of high school.

Ionic reactions, that is, reactions between ions, will be expressed by different ionic reaction equations due to the different amounts of reactive ions. The general idea of the question of who is in excess is to set the coefficient of the small amount of the ion to 1, and the excess ion changes with the coefficient of the small number of ions. If you keep this in mind, you can write the correct corresponding ionic reaction equation in the ionic reaction equation that has a question about quantity.

For example, an excess of Ca(OH)2 reacts with NaHCO3.

If the coefficient of NaHCO3 is set to 1, the coefficient of Ca(OH)2 changes with the coefficient of NaHCO3. It is written as the change of NaHCO3+1 2Ca(OH)2. Then the ionic reaction equation is expressed as:

hco3-+ca2++oh-=caco3+h2o

For example, a small amount of NaHCO3 is added to BA(OH)2, because NaHCO3 is a small amount, so a small amount is taken as the standard, and only one H+ is needed to neutralize HCO3-, so the chemical equation is NaHCO3 + BA(OH)2===BaCO3 precipitation + H2O + NaOH

If NaHCO3 is excessive, it means that Ba(OH)2 is a small amount, taking a small amount as the standard, there are two OH- in Ba(OH)2, so two HCO3- are needed to react with it, and the chemical equation is 2NaHCO3 + BA(OH)2===BAC3 precipitation + Na2CO3 + 2H2O

Remember, a small amount is the standard.

I can only give you some examples, if you are good at chemistry, you should be able to understand it yourself

1 HCO3- with OH-

It's not that it's too much, it's just that we look at the valence ratio of the previous ion (na ca ba). For example, NAHCo3 is added to Ba(OH)2

Ca(HCO3)2 with NAOH is the proportional requirement for equation writing.

2 naoh+co2

2:1 (C02 minor) CO32-

1:1 (C02 overdose) HCO3-

The reason is CO2 + H2O + CO32- 2HCO3- (SO2 H2S reacts on the same principle as OH-).

3. Aluminum triangle conversion.

al3+ +3oh- →al（oh)3

al（oh)3+oh- →alo2-

Yield: Al3+ + strong base (a small amount) Al(OH)3 precipitate.

Al3+ + strong base (excess) AlO2- (precipitate appears and disappears).

Similarly, inversely: alo2- +h+ al(oh)3

al（oh)3 +3h+ →al3+ +6h2o

Yields: AlO2- + strong acid (a small amount) Al(OH)3 precipitate.

AlO2- + strong acid (excess) Al3+ (precipitate appears and disappears).

Note: Al3+ and weak alkali ALO2- and weak acid can only form Al(OH)3 precipitate if it is too much.

4 Fe with dilute nitric acid (the concentrated one is passivated).

Fe (in small amounts) is oxidized by dilute nitric acid to form Fe3+ and No, but (in excess) Fe is oxidized by dilute nitric acid to produce Fe2+ and No

The reason is Fe+2Fe3+ 3Fe2+

There are examples, but they require your usual attention and accumulation, I hope you can continue to love chemistry, I wish you all the best in your studies!

I have a method that works. When writing an ion reaction, only a small amount of substance is used, and the ratio of each ion in the reaction formula is in accordance with its chemical formula. And the one that overdoses doesn't matter.

Then you can figure out what happens to the individual ions, and add them up in proportion. Your example is too simple to be representative.

HCO3-)+BA2+)+OH-)= (BAC3) +H2O) Ba2+ and OH- ratios do not have to be 1:2

If it is a small amount of barium hydroxide, 2HCO3-)+BA2+)+2OH-)=(BAC3) +CO32-)+2H2O)).

The Ba2+ and OH- quantity ratio must be 1:2

1 HCO3- with OH-

It's not that it's too much, it's just that we look at the valence ratio of the previous ion (na ca ba). For example, NAHCo3 is added to Ba(OH)2

Ca(HCO3)2 with NAOH is the proportional requirement for equation writing.

2 naoh+co2

2:1 (C02 minor) CO32-

1:1 (C02 overdose) HCO3-

The reason is CO2 + H2O + CO32- 2HCO3- (SO2 H2S reacts on the same principle as OH-).

3. Aluminum triangle conversion.

al3+ +3oh- →al（oh)3

al（oh)3+oh- →alo2-

Yield: Al3+ + strong base (a small amount) Al(OH)3 precipitate.

Al3+ + strong base (excess) AlO2- (precipitate appears and disappears).

Similarly, inversely: alo2- +h+ al(oh)3

al（oh)3 +3h+ →al3+ +6h2o

Yields: AlO2- + strong acid (a small amount) Al(OH)3 precipitate.

AlO2- + strong acid (excess) Al3+ (precipitate appears and disappears).

Note: Al3+ and weak alkali ALO2- and weak acid can only form Al(OH)3 precipitate if it is too much.

4 Fe with dilute nitric acid (the concentrated one is passivated).

Fe (in small amounts) is oxidized by dilute nitric acid to form Fe3+ and No, but (in excess) Fe is oxidized by dilute nitric acid to produce Fe2+ and No

The reason is Fe+2Fe3+ 3Fe2+

There are a total of 3 aspects of the reactive formula, just write it down, and any problem can be solved.

An excess is the number of ions that make it up, which can be seen as an infinite number relative to the substances that react with it.

As you gave in the example.

The ionic equation for NaHCO3 (less) + Ba(OH)2 (more) is.

The ionic equation for HCO3-)+BA2+)+OH-)= (BAC3) +H2O)NaHCO3 (more) + BA(OH)2 (less) is.

2hco3-)+ba2+)+2oh-)=(baco3)↓+co32-)+2h2o)

PS: The parentheses don't make any sense, just to distinguish matter.

Hope you are satisfied.

Mainly from the essence of the ionic reaction, for example, a small amount of NaHCO3 and Ba(OH)2 reaction, a NaHCO3 ionizes a HCO3-, so he can only react with one OH-, while a BA(OH)2 can ionize two OH-, so there must be one Oh- left in the solution, so the reaction formula should be.

NaHCO3 + BA(OH)2 = BAC3 + NAOH+H2O, but if the formula NaHCO3 is excessive, all the OH- ionized by BA(OH)2 can be consumed, and there is no OH- in the solution, and the reaction is.

2nahco3+ba(oh)2=baco3+naco3+2h2o

Theoretically accepted?

This question is a bit harder.

Bicarbonate ions are acidic, and of course react with alkaline hydroxide ions, which react with hydroxide ions to form water. Excess is the reaction of acid gas and lye, which means that when there is an excess of acid gas, acid salts are formed, and when there is an excess of alkaline salts, such as NaHCO3 can be regarded as a mixed crystal of H2CO3 and Na2CO3.

It's up to us to understand it slowly.

Well, maybe it's because you're good at chemistry that your teacher didn't help you talk about it, but in fact, the formula for chemistry isn't just a simple thing.

Some equations are a combination of multiple equations and they have to be exactly proportional, but there are different phenomena in experiments that are not proportional (and it's very common!). Just as inequalities are always more than equations)

At this time

The equations that are combined are split (of course, the teacher will tell you) and then readjusted according to their respective proportions

That's why there are experimental methods such as titration

So you don't have to rush to ask your teacher to help you find a time to talk about it

The main thing in the writing of the equation is to clarify who is overdosed and who is in excess of what is the difference between the products produced when a small amount is excessive.

The key is to take a small amount of that reactant as a benchmark, and then react it completely to see what the rest of the part can compose, and then write it down, the key is that the product is sometimes different, for example, the salt may be a normal salt or an acid salt in different environments.

Let's start with the simplest chemistry

For example, h2 + 2o2 = 2h20

If there is a known excess of H2 or O2, the amount of water produced can be calculated based on the excess!

It's the same with ionic reactions, but it's not a few words that can be said clearly!

This needs to be analyzed on a case-by-case basis, you have to see what the excess is, judge whether the solution is acidic or alkaline after the reaction, and then judge the possible substances after the reaction to determine the acidity and alkalinity.

Increasing the concentration of ions that do not participate in the reaction has an effect on the rate of the reaction, and this effect is chemically called"Salt effect".

For example, ferric chloride.

The reaction with the KSCN, the ionic equation of its reaction.

Fe3+ +SCN- =Fe(SCN)2+ [The process of this reaction is relatively complex, this is one of the ion equations, and some materials also write other reactions, such as: Fe3+ +6SCN- =Fe(SCN)63-, etc.].

The phenomenon that can be seen is that after dropping KSCN into the ferric chloride solution, the solution turns from yellow to blood red.

If KCl or Na2SO4 is added to the solution (it must be a strong electrolyte.

, you can see that the blood red color of the solution will become lighter or even disappear.

This phenomenon is called"Salt effect".

The chemical explanation of the salt effect is as follows: the ions added though do not participate in the ionic reaction.

However, the addition of strong electrolytes increases the concentration of total ions in the solution, which will make the ions participating in the reaction"Collisions"to the point of affecting the rate of the reaction.

You can look it up online"Salt effect"related theories. There will be further understanding!