When does aluminum and base react to form metaaluminate

If you use too much NaOH, the latter is used in less NaOH How can a partial aluminum root be formed if you don't use too much NaOH If you don't do it, it will form an AI(OH)3 precipitate, (It seems that when you calculate too much, you often use aluminum to react with four hydroxide radicals) You should not tell you how much NaOH is, so there are two situations. That's why this equation will appear.

First, aluminum and hydrochloric acid react to form aluminum chloride.

Reaction equation: 2Al+6HCl=2ALCl3+3H2 at a small amount of NaOH:

Chemical equation: AlCl3 + 3NaOH = 3NaCl + Al(OH)3 NaOH excess:

The chemical equation: AlCl3+3NaOH=3NaCl+Al(OH)3 Al(OH)3+NaOH=Naalo2+2H2O, at this time, metaaluminate is produced.

Metaaluminic acid: Metaaluminic acid is a weak acid, an acid that is weaker than carbonic acid.

Cannot react with CO2. But if it's a metaaluminate solution, it's fine. Because metaaluminic acid is less acidic than carbonic acid, it is equivalent to the replacement of weak acid with strong acid, (taking sodium metaaluminate as an example) to produce sodium carbonate and aluminum hydroxide.

precipitation, if CO2 is excessive, will produce sodium bicarbonate.

and aluminum hydroxide precipitation.

Metaaluminate is formed when sodium hydroxide is excessive, and aluminum hydroxide is formed when sodium hydroxide is small amount.

Aluminum hydroxide is produced when the ratio of aluminum ions to hydroxide is 1:3.

The reaction formula is: (Al3+)+3OH-=AL(OH)3, when the ratio of aluminum ions to hydroxide is 1; 4 o'clock production metaaluminate.

The reaction is divided into two parts: (1) (Al3+) + 3OH - = AL(OH) 32) AL(OH) 3 + OH-= (ALO2-) + 2H2O The total reaction formula is (AL3+) + 4OH - = ALO2) -2H2O In general, the alkali is added less to produce aluminum hydroxide, and the more is added, metaaluminate Bochang and water are generated.

In fact, there is also a process, but the existence of aluminum hydroxide is very short.

Because alumina is an insoluble solid, when it reacts with a strong base, it becomes aluminum hydroxide, and aluminum hydroxide is an insoluble substance, which will cover the surface of alumina and prevent the reaction from being prepared, but aluminum hydroxide can be dissolved in a strong alkali, so it will be destroyed and reacted quickly, if alumina wants to be completely reacted, then aluminum hydroxide will not exist, so when alumina reacts with a strong alkali, metaaluminate ions are generated.

Soluble aluminum ions are different, even if the resulting aluminum hydroxide is dissolved, as long as the alkali is not excessive, then the metaaluminate ions will react with the aluminum ions and form aluminum hydroxide. Therefore, as long as the strong alkali is not overcovered, aluminum hydroxide will exist.

Aluminium can react with bases to form metaaluminate compounds.

Knowledge point analysis:

A common reaction is the reaction of aluminum with strongly basic sodium hydroxide (NaOH) or potassium hydroxide (KOH) to form metaaluminate ions.

The reaction equation is as follows:

2al + 2naoh + 6h2o 2na[al(oh)4] +3h2

In this reaction, aluminum reacts with hydroxides to form metaaluminate ions (Na[Al(OH)4]) and hydrogen is released at the same time.

Application of knowledge points:

1.Metaaluminate ions are widely used in the production process of Portland cement and glass in the chemical industry as thickeners, preservatives, etc.

2.Metaaluminate ions can also be used in wastewater treatment as a precipitant to remove heavy metal ions from wastewater.

Example Explanation:

Question: When 5g of aluminium is added to an excess sodium hydroxide solution, what is the number of moles of metaaluminate ions generated?

Answer: According to the reaction equation, it can be known that 2 mol of aluminum and 6 mol of sodium hydroxide react to form 2 mol metaaluminate ions.

First, calculate the number of moles of aluminum:

The relative atomic mass of aluminium is 27 g mol, so the molar number of 5 g of aluminium is:

5 g / 27 g/mol ≈ mol

Since sodium hydroxide is excessive, the molar ratio is 1:3, and the number of moles of metaaluminate ions is 1 2 mol of aluminum:

mol × 1/2) = mol

So, the number of moles of metaaluminate ions produced is about mol.

Can't coexist. Metaaluminate separates metal aluminum ions from water, and it will combine with hydroxide to form a precipitate, so it cannot coexist. The anions of weak acid groups cannot exist in strong acid solutions, such as carbonate, acetate, sulfur ions, metaaluminate, hypochlorite can not coexist with hydrogen ions in large quantities.

Can metaaluminate coexist with hydroxideMetaaluminate and hydrogen ions cannot coexist in large quantities because the two react. If H+ is small, Al(OH)3 is generated, and the equation: H++AlO2-+H2O==Al(OH)3 (precipitation); If H+ is excessive, Al3+ is generated, Eq:

4H++alo2-==Al3++2H2O metaaluminate and hydroxide ions can coexist in large quantities.

The ionic equation for the reaction of metaaluminate with hydrogen ions.

Low amount of H+ ions: AlO2-+H++H2O=Al(OH)3 H+ ion excess: AlO2-+4H+=Al3++2H2O

Reason: Metaaluminate is a weak acid group, which combines hydrogen ions ionized from water, destroys the ionization balance of water, and makes the ionization of water move forward, resulting in the concentration of hydroxide ions in the solution greater than the concentration of hydrogen ions, making the aqueous solution alkaline. The specific process is shown in the following figure:

The process of forming a weak electrolyte from the ions of salt and the hydrogen ions ionized by water in solution is called the hydrolysis of salts. Hydrolysis of salts: Salts must be soluble in water, and salts must be able to ionize weak acid ions or weak base cations.

Metaaluminate is a chemical substance with the chemical formula Al(OH) The following are some of the chemical reaction equations that are common in high school for metaaluminate lead in high school:

1.Hydrolysis reaction letter grip burning:

al(oh)₄⁻aq) +h₂o(l) al(oh)₃(s) +oh⁻(aq)

2.Oxidation reaction:

2al(oh)₄⁻aq) al₂o₃(s) +4oh⁻(aq) +h₂o(l)

3.Acid-base neutralization reaction:

al(oh)₄⁻aq) +h⁺(aq) al(oh)₃(s) +h₂o(l)

4.Precipitation reaction:

al(oh)₄⁻aq) +ca²⁺(aq) ca(oh)₂(s) +al(oh)₃(s)

5.Pyrolysis reaction:

2al(oh)₄⁻aq) al₂o₃(s) +6oh⁻(aq) +2h₂o(l)

It is important to note that the reaction equations listed here only represent the involvement of metaaluminate in some common reactions. In practical applications and more complex chemical systems, the chemical properties and reaction behavior of metaaluminate may be more diverse and complex. Detailed understanding and research need to refer to more professional chemistry textbooks and materials.

1. The hydrolysis of aluminum ions presents strong acidity, and the hydrolysis of metaaluminate ions presents strong alkalinity, so the two cannot coexist.

2. The hydrolysis equation of aluminum ions is:

ai3+ +3h₂o=ai(oh)₃↓3h+

Metaaluminate ion hydrolysis equation:

aio2-+2h₂o=ai(oh)₃↓oh-。

3. The two are mixed to produce double hydrolysis, and the reaction equation is: AI3+ +3AIO2- +6H O = 4AI(OH).

The ion reaction is carried out in the direction of the weakening of the ions, and the coexistence of ions essentially refers to whether the ions can have a chemical reaction, and if the ions cannot react with each other, they can coexist. On the other hand, if the ions in the solution can react with each other, they cannot coexist.

There are many factors that need to be considered for the reaction between ions, such as: whether volatile substances are formed between ions in the solution, whether insoluble substances are formed, whether weak electrolytes are formed, etc.

2al+6h2o=2al(oh)3 +3h2 al(oh)3 +naoh=naalo2 +2h2o2al+2naoh+2h2o=2naalo2+3h2 The essence of the reaction between aluminum and alkali:

The essence of the reaction between aluminum and alkali is that aluminum reacts with water first, and the aluminum hydroxide produced by the reaction is soluble in a strong alkali solution. So the chemical reaction proceeds to the right, which is manifested as the reaction of aluminum with a strong alkali solution.

But it can only react with a strong alkali solution) This reaction is a redox reaction, the reducing agent is aluminum, and the oxidizing agent is water. However, since the water is eliminated after the addition of the two reaction equations, the direction of electron transfer in the ion equation is to replenish the water.

2al+6h2o=2al(oh)3 +3h2 ↑al(oh)3 +naoh=naalo2 +2h2o2al+2naoh+2h2o=2naalo2+3h2↑