7 8 grams of sodium peroxide reacts with a sufficient amount of water, and the number of electrons t

grams of sodium peroxide.

Reacting with a sufficient amount of water, the number of electrons transferred is moles.

The calculation process is as follows:

The molecular weight of sodium peroxide is 78, and the gram of sodium peroxide = mol.

2na₂o₂+2h₂o==4naoh+o₂↑

x2:1=：x

Solution x = moles.

o Changes in the valence state of the element. In sodium peroxide is -1, the product is sodium hydroxide.

-2 in the medium and 0 valence in oxygen.

The oxygen in the sodium peroxide is all electron transferred.

Oxygen is a diatomic molecule, so the transfer electrons are 4x=mol.

2Na2O2+2H2O——O2+4Naohna is +1 valence before and after.

o Changes in the valence state of the element. -1 in sodium peroxide, -2 in sodium hydroxide, and 0 in oxygen.

So the number of electrons transferred in the original chemical equation is 2mol

Sodium peroxide is.

Therefore, when there is only sodium peroxide, the coefficient of the equation sodium peroxide is that the number of transferred electrons changes from 2mol.

So the answer is.

2Na2O2+2H2O==4NaOH+O2, the valency of oxygen in sodium peroxide is -1 valence, so 2mol of electrons are transferred for every 1mol of O2 generated.

The reaction equation is:

2na2o2 + 2h2o 4naoh + o2

According to the equation, it can be seen that for every 2 molecules of sodium peroxide, 1 molecule of oxygen is released and 4 electrons are transferred at the same time. Thus, reacting one sodium peroxide molecule transfers 2 electrons.

According to the principle of chemical number accompaniment metrology, when a sufficient amount of sodium peroxide is reacted in liters of water, the amount of the substance of sodium peroxide should be twice the amount of the substance of water, therefore, the amount of the substance of the reacted sodium peroxide is:

n(na2o2) =2 × n(h2o) =2 × l) ÷g/mol) ≈mol

Therefore, the number of electrons transferred by the reaction is:

2 × n(na2o2) ×2 = 2 × mol × 2 = mol e-

Because of this, under standard conditions, a liter of water reacts with a sufficient amount of sodium peroxide to transfer the molar electric chaos ruler.

The number of transferred electrons is na.

The equation for the reaction of sodium peroxide with water is:

First, calculate the total number of electrons transferred in the whole equation, only count the number of electrons obtained by the oxidant or only the number of electrons lost by the reducing agent, and then divide the number of transferred electrons by the coefficient of each substance, calculate how many electrons are transferred for each reaction of a substance, and finally calculate the number of transferred electrons with the amount of matter given in the question as a multiple.

1 mol The number of electrons transferred in the reaction of sodium peroxide with water is 1 mol, because it is a disproportionation reaction, it is the transfer between the oxygen elements in sodium peroxide, one of which loses an electron, from -1 valence, to 0 valence, and one gets an electron, which becomes -2 valence.

The final result of the reaction of 2mol of sodium peroxide and 2mol of water is 4mol of sodium hydroxide and 1mol of oxygen, and a total of 1mol of electrons is transferred in the whole process. It can be roughly considered that the oxygen sensible -1 valence in sodium peroxide can be calculated by using the double-line bridge formula.

Sodium peroxide reacts with water without transferring electrons. The remaining hydrogen peroxide decomposition has electron transfer. Eventually transfer 1 molar electron.

Although it has been 2 years, in order not to mislead other students, I will explain, it should be.

2na2o2+2h2o=4naoh+o2↑

Only Na2O2 participates in the reaction, and the oxygen 2 decreases from -1 valence to -2 valence (Na2O2), and the other 2 increase from -1 valence to 0 valence (O2), so there are only 2 electrons in the formation of oxygen, and the other 2 are on the Na2O2 side, so the number of electrons transferred to 1mol of oxygen is that is, H2O does not participate in the reaction.

In the reaction of sodium peroxide with water, for each oxygen generated, the number of electrons transferred is .

The reaction belongs to the redox reaction, the sodium peroxide period oxidant, one of the oxygen-1 valence loses electrons and becomes 0 valence oxygen; The other one gets electrons that become -2 valence oxygen.

2na2o2+ 2h2o = 4naoh +o2↑

Electron transfer, as shown below.

The essence of this reaction:

Step 1: Metathesis reaction, the valency does not change. 2na2o2+ 2h2o= 4naoh+h2o2

Step 2: H2O2 decomposition reaction, i.e., disproportionation reaction. 2h2o2=2h2o+o2(^)

The sum of the previous two reactions is the total reaction: 2Na2O2+ 2H2O= 4NaOH+O2( ).

As can be seen from the above reaction essence, it is the 2-1 valence O in Na2O2 that has been disproportionated, so.

1 mol The number of electrons transferred in the reaction of sodium peroxide with water is 1 mol, because it is a disproportionation reaction, it is the transfer between the oxygen elements in sodium peroxide, one of which loses an electron, from -1 valence, to 0 valence, and one gets an electron, which becomes -2 valence.

If you don't understand, please ask

The number of electrons transferred is.

2Na2O2 + 2H2O ==4NaOH + O2 (gas) sodium peroxide. The reaction with water is a disproportionation reaction.

O2]2- medium-1 valence O, one loses electrons and one gains electrons.

-2 valence oh- and 0 valence O2 are generated, respectively

That is, only electrons are transferred.

The link is full of holes, and it is disturbing

2Na2O2+2H2O==4NaOH+O2, the valency of oxygen in sodium peroxide is -1 valence, so 2mol of electrons are transferred for every 1mol of O2 generated.

The answer to this question is 0, one becomes 0 valence O, in each Na2O2, 2H2O2 2H2O O2, actually written as 2Na2O2 4H2O 4NaOH 2H2O2, one electron becomes -2 valence O, so each molar Na2O2 transfers 1 lead electron annihilation segment. 8 grams of Na2O2 is 0