High one chemical magnesium aluminum alloy problem, high one chemical problem, sodium magnesium alum

It's how many grams of magnesium it contains. It should be grams. The volume of sodium hydroxide solution required to reach the maximum precipitated value is. (After adding hydrochloric acid, it should be indicated that it is exactly a complete reaction, otherwise this question cannot be done).

The input of HCL converts all Mg, Al into Mg2+, Al3+. Then the following reaction occurs:

Mg2+ +2OH- = Mg(OH)2 (precipitate) Al3+ +3OH- = Al(OH)3 (precipitate).

When the maximum value of precipitation is reached, the above two reactions are complete. When sodium hydroxide is added again, the following reaction occurs: Al(OH)3 + OH- = ALO2- +2H2O

Only mg(OH)2 remains, i.e., the reaction is complete. This reaction adds NaOH, indicating that there is Al3+, i.e. by the conservation of Al atoms, known to have elemental matter, i.e. Therefore mg has.

When the maximum value of precipitation is reached, Al consumption, Mg consumption, and total consumption are reached. So consume 2mol l of sodium hydroxide.

Magnesium and aluminum alloys first react with hydrochloric acid, and the product reacts with sodium hydroxide, and when the amount of precipitation is the largest, there is a relation: N(HCl)=N(NaOH).

Here's how: the hydrogen gas generated is obtained by replacing hydrogen ions in hydrochloric acid. Therefore, sodium, magnesium, and aluminum react with hydrochloric acid to become sodium salts, magnesium salts, and aluminum salts, and the number of electron transfers is also equal.

2na---h2

mg---h2

2/3al---h2

Therefore, the ratio of the amount of dust to the substances that require sodium, magnesium, and aluminum is: 2:1:

2 3 6:3:4 mass ratio is 6*23:

18 I hope it can help you, in fact, just write out their chemical reaction equations.

Any remorse has a metal one.

Molar electron mass me, that is.

Metal is released.

1mole-

The quality consumed.

Known metal, me=

Molar quality and filial piety.

Valency. And the average.

ME = alloy quality

The amount of matter of the electron.

Title. Average.

me=10g

10gmol

The ME of zinc, iron, aluminum, magnesium, and four reed metals are.

Only. Less than.

Must contain al

A: Select AAnalysis: Because mg does not react with Naoh, while Al reacts with Naoh:

2al+2naoh+2h2o=2naalo2+3h2, so the residual solid is mg

As can be seen from the title, the mass (Ag) of the reaction of Mg with O2 to form MGO is equal to the mass of the mixture of magnesium and aluminium (Ag), so the mass of aluminium in the mixture is equal to the mass of O2 participating in the reaction.

Let the mass of aluminum in the mixture be x, then the mass of magnesium is ag-x, and the mass of O2 participating in the reaction is also x, 2mgo2 ignition=

2mgoag-x~~~x

The column proportional solution yields: x=

Therefore, the mass fraction of aluminum in the magnesium-aluminum mixture is: (

This is mainly about the level of understanding of chemical reactions.

mg+cl2===mgcl2

mg+br2===mgbr2

2mg+o2===2mgo

mg+s===mgs

mg+2h2o===mg（oh）2+h2

2mg+TIC4(melted)===Ti+2mgCl2mg+2RBCl===MgCl2+2rb

2mg+co2===2mgo+c

2mg+sio2===2mgo+si

mg+h2s===mgs+h2

mg+h2so4===mgso4+h2

2al+3cl2===2alcl3

4al+3o2===2al2o3 (passivation).

4al(hg)+3o2+2xh2o===2(4al+3mno2===2al2o3+3mn2al+cr2o3===al2o3+2cr2al+fe2o3===al2o3+2fe2al+3feo===al2o3+3fe

2al+6HCl===2ALCL3+3H22AL+3H2SO4===AL2(SO4)3+3H22AL+6H2SO4 (concentrated)===AL2(SO4)3+3SO2+6H2O

Al,Fe passivated in cold, concentrated H2SO4,Hno3) AL+4HNO(dilute)===AL(NO3)3+NO+2H2O2AL+2NaOH+2H2O====2Naalo2+3H22Fe+3BR2===2Febr3

fe+i2===fei2

fe+s===fes

3fe+4h2o(g)===fe3o4+4h2fe+2hcl===fecl2+h2

fe+cucl2===fecl2+cu

fe+sncl4===fecl2+sncl2

The final BG is Al2O3, the mass is 102, where the mass of aluminum is b 102*54, and the mass fraction should be the mass of aluminum divided by the sample amount and then multiplied by 100%, so the answer is b divided by 102 and then multiplied by 54 and divided by a, and finally multiplied by 100%, [b 102)*54] a}*100%=(9b 17a)*100%.

b g alumina, alumina mass 102

where the mass number of aluminum is 54

b 102*54 is the quality of aluminum.

Divide by a for the answer.

Sodium hydroxide was added, and the Al(OH)3 precipitate was dissolved

al(oh)3+naoh=naalo2+2h2ox...90-80)*

x = moles. So the quantity of the substance of al=, the mass =

The second question: the amount of sodium hydroxide substance consumed by Wang Cheng Al(OH)3 = , volume = 30ml;

If the mixture is all AL, then the maximum value of a = 80-30 = 50,;

aMinimum=0

Question 3: If A takes the minimum value = 0, the maximum Mg(OH)2 is generated, then 50ml of sodium hydroxide reacts with Mg ions to form a precipitate;

Then the amount of the substance of mg = 50*, mass =

So n(mg) n(al).

The maximum value is =