Ionic equations. Add AlCl3 solution dropwise to the NaOH solution dropwise and shake while dropping

1: At the beginning, OH-excess: AL3+ +4OH- = ALO2- +2H2O.

Continue to add Al3+:3ALO2- +AL3+ +6H2O = 4AL(OH)3 (precipitate).

2: At the beginning, NaHSO4 is in excess and Ba(OH)2 is insufficient, so BA(OH)2 reacts completely, i.e., the ratio of BA2+ to OH- is 1:2 BA2+ +2OH- +2H+ +SO42- = BASO4 + 2H2O

solution to alkaline, at this time H+ is insufficient, the ratio of H+ to SO42- is 1:1 H+ + SO42- +OH- +BA2+ = BASO4 + H2O

I only know 56 two, accelerants: nano aluminum and nickel, etc.

Co-solvent: 4 carbon chloride, 1,1 2 chloroethane....

1: Because it is added drop by drop, it is necessary to write 2 ion equations.

1) Al3+ +3OH- = AL(OH)32)AL(OH)3+OH- = ALO2- +2H2O2: Because it is added drop-by-drop, it is OH- insufficient in the early stage.

2H+ +SO4 2- +Ba(OH)2= BaSO4+2H2O, etc., is H+ insufficient.

ba(oh)2 + h+ +so4 2- =oh- +baso4 +h2o

At first, there was an excess of NaOH, so there was no white precipitate of the spine, and the phenomenon was that there was no obvious change in the equation AlCl3 + 4NaOH==Naalo2 + 2H2O, and after the NaOH was consumed by droplets, the white group field precipitation equation was 3Naalo2 + AlCl3 + 6H2O ==4Al(OH)3 + 3NaCl

Phenomenon: There is a white precipitate first, and then it is dissolved as a precipitate.

Ion equation: Al3+ +3OH-= AL(OH)3 (Shen Zao Chun Qidian) Mori Ant.

al（oh）3+ oh- =alo2- +2h2o

AlCl3 solution was added dropwise to Naalo 2 solution, and aluminum ions and metaaluminate ions were decomposed in water to form aluminum hydroxide precipitate, and the ion equation of the reaction was lead: 3alo2 - al3+ +6H2 o=4al(OH) 3;

So the answer is: 3alo 2 - al 3+ +6h 2 o=4al(oh) 3 ;

Solution: 1. AlCl was added to the NaOH solution at the beginning, and the NaOH was excessive. At this time, NaOH has a chemical reaction with AlCl without obvious reaction phenomenon.

The main phenomenon is the generation of heat during the reaction. The chemical reaction that occurs is as follows.

4NaOH+AlCl = Naalo +2HO+3NaCl In this reaction, AlCl is used as an acid, and NaOH is neutralized to form Naalo salt and water.

2. When NaOH is completely reacted and AlCl is added, the Naalo generated earlier will continue to react with AlCl, and the reaction phenomenon is that there is a white precipitate. The chemical reaction that occurs is as follows.

3naalo₂+alcl₃+6h₂o=4al(oh)₃+3nacl

At first, there was an excess of NaOH, so there was no white precipitate, and there was no significant change.

Equation: ALCL + 4NAoh ==NAALO + 2H O After the NaOH is consumed by another drop, a white precipitate will appear.

The equation is: 3Naalo +AlCl +6H O ==4Al(OH) 3NaCl

Aluminum chloride is a colorless transparent crystal or a crystalline powder with white and microstriped light yellow. The vapor of aluminum chloride exists as a covalent dimeric molecule when dissolved in a non-polar solvent or in a molten state.

Soluble in water and many organic solvents. The aqueous solution is acidic. In the presence of aromatic hydrocarbons, aluminum chloride can be mixed with aluminum to synthesize dimetallic complexes.

During the process of slowly dropping, a precipitate will be produced first. When naoh:alcl3=3:

At 1, Cl- is just completely precipitated, and the chemical equation at this time is 3NaOH+AlCl3=3NaCl+Al(OH)3 (remember to write the precipitation number!). Otherwise, it's written in vain... Same upstairs), continue to add NaOH dropwise, and the precipitate will slowly dissolve, because Al(OH)3 is an amphoteric substance, right?

It can react with both acids and bases. So, when naoh:alcl=4:

At 1, the reaction formula: Al(OH)3+NaOH=Na, the precipitate disappears. Understand, right?

The total equation is the addition of two things: 4NaOH+ALCl3=3NaCl+Na

I was wrong, I just reversed the topic. I thought it was adding NaOH drops to ALCL3. However, you can also collect it, so I won't delete it. It will be useful later.

Correct answer:

At the beginning, there was an excess of NaOH, so the reaction formula should be: 4NaOH + AlCl3 = 3NaCl + Na, no precipitate appeared, and later, AlCl3 became more, and the reaction: AlCl3 + 3Na = 4Al(OH)3 (precipitation number!!

3nacl+ won't match anymore. Wait a minute.

I'm so depressed.,It's already ready.,But,I'm struggling.,In the end, ** is wrong.。。。 Alas...

Upstairs, it seems to be the old version. We don't need those.,I don't know which version the landlord learned?

Phenomenon: No precipitate at the beginning (no precipitate in the whole solution, during the dripping process, you may see a small amount of precipitate formed but it disappears after oscillation) and then a large amount of precipitate appears.

The chemical equation is: 4NaOH + AlCl3 = Naalo2 + 2H2O

3 naalo2 + alcl3 + 6h2o = 4ai(oh)3 ↓ 3naci

Because AlCl3 is added dropwise to NaOH to an excessive amount, no precipitate is generated, and a white precipitate is produced. In the reaction process, there is always an excess of NaOH first, and Al(OH)3 formed by the reaction of AlCl3 reacts with an excess of NaOH to form NaALO2. So the precipitation is not visible.

After an excess of AlCl3, Naalo2 reacts with AlCl3 and H2O to produce NaCl and Al(OH)3

So after the precipitation is produced.

Equation: 4NaOH + ALC3 = Naalo2 + 2H2O + 3NaCl3 Naalo2 + ALCL3 + 6H2O = 3NaCl + 4AL(OH)3 Ionization equation: 4OH - +AL3 + = ALO2-+2H2O3ALO2-+(AL3+) + 6H2O = 4AL(OH)3

A should be selected, the image of water should rise slowly first and then rise quickly, because there is water in the NaOH solution, there is no water added to the reactant, and there will be no horizontal straight line, 3NaOH + AlCl3 = Al(OH)3 + 3NaCl reaction is anhydrous, and Al(OH)3 + NaOH = Naalo2 + 2H2O is generated by water, so Figure A should be divided into two segments of slow and fast first.

I. AlCl3 was added to Al(OH)3, AL3++3OH-AL(OH)3 occurred, AL(OH)3 was amphoteric, and AL(OH)3+OH-ALO2 was generated when NaOH solution was added

2H2O, the dissolution of the precipitate can be observed, 1) add NaOH solution dropwise to the AlCl3 solution to an excessive amount: a white precipitate is produced first, and then the precipitate is gradually dissolved until it disappears completely;

Therefore, the answer is: white precipitate is produced first, and then the precipitate is gradually dissolved until it disappears completely;

2) Add NaOH solution dropwise to the ALCL3 solution to the excess, and the reactions occur: Al3++3OH-AL(OH)3, AL(OH)3+OH-ALO2

2h2o；So the answer is: al3++3oh- al(oh)3, al(oh)3+oh- alo2

ii. Add AlCl3 solution dropwise to the NaOH solution to an excessive amount, at the beginning NaOH is excessive, the generated Al(OH)3 is dissolved in the NaOH solution to generate NaALO2, and then with the increase of the amount of AlCl3, when the sodium hydroxide reaction is completed, the alCl3 and Naalo2 droplet into the droplet occur to generate Al(OH)3 precipitate, 1) add the AlCl3 solution dropwise to the NaOH solution to the excess, The experimental phenomenon is as follows: at the beginning, there is a local white precipitate, and the precipitate disappears after oscillation, and then there is a white precipitate, which does not disappear after oscillation;

Therefore, the answer is: at the beginning, there is a local white precipitate, and the precipitate disappears after oscillation, and then there is a white precipitate, which does not disappear after oscillation;

2) AlCl3 solution was added dropwise to the NaOH solution to the excess, and the reaction occurred: Al3++4OH-=ALO2 successively

2h2o；3alo2

al3++6h2o=4al（oh）3↓；

So the answer is: al3++4oh-=alo2

2h2o；3alo2

al3++6h2o=4al（oh）3↓；

3) AlCl3 solution was added dropwise to the NaOH solution to the excess, and the reaction occurred successively: Al3++4OH-=ALO2

2h2o；3alo2

(1) Sodium hydroxide preferentially reacts with hydrochloric acid to form sodium chloride and water, and the ion equation of the reaction is: H++OH- H2O, so the answer is: H++OH- H2O;

2) After the reaction of sodium hydroxide is complete, sodium carbonate reacts with hydrochloric acid to form sodium carbonate and sodium chloride, and the ion equation of the reaction is: H++CO3

2-=HCO3, so the answer is: H++CO3

2-=hco3

(3) Finally, sodium bicarbonate reacts with hydrochloric acid to produce sodium chloride, carbon dioxide and water, and the ion equation of the reaction is: HCO3

h+=CO2 +H2O, so the answer is: HCO3

h+=co2↑+h2o．

al 4oh=alo2 2h2o, the mobile phone can't play the ion symbol, you make up for it yourself, give a good review to make yourself happier.

Add AlCl3 solution dropwise to the NaOH solution until the excess is NaOH at the beginning, so there is no white precipitate, and the phenomenon is no obvious change Equation AlCl3 + 4NaOH==Naalo2 + 2H2OAfter the NaOH is consumed by droplet, a white precipitate will appear The equation is 3Naalo2 + AlCl3 + 6H2O ==4Al(OH)3 + 3NaCl

If the AlCl3 solution is dropped into the NaOH solution to an excess, the ionic reaction is written step by step

1）al3++4oh-==alo2-+2h2o （2）3alo2-+al3++6h2o==4al(oh)3↓