How to trim NOO2 H2O HNO3?

In this reaction, the valency of nitrogen increases from +2 to +5, increases by 3, and loses 3 electrons.

The oxygen element in oxygen is reduced from 0 valence to -2 valence, and the two oxygen atoms are reduced by a total of 4 valence, giving 4 electrons.

According to the number of electrons gained and lost, the number of electrons should be equal, no should be matched with 4 as coefficients, and oxygen should be matched with 3 coefficients.

That is, 4NO+3O2+H2O=HNO3

According to the conservation of nitrogen atoms, nitric acid should be matched by a coefficient of 4

That is, 4NO+3O2+H2O=4HNO3

According to the conservation of hydrogen atoms, water should be matched with a coefficient of 2

That is, 4NO+3O2+2H2O=4HNO3

2no+o2=2no2

3no2+h2o=2hno3+no.

Your equation doesn't seem to have this reaction!

As can be seen from the reaction, the valency of the element n is increased by 1, while the valency of oxygen is decreased by 2

Therefore, one NO2 increases the first valence, and one O2 decreases the 4 valence.

So the ratio of the number of meters of No2 and O2 is 4:1

Then the measurement number of No2 is 4, and the measurement number of O2 is 1

After the substitute punch cavity into the loose clothes, you can know that the reactive experience is:

4no2+2h2o+o2=4hno3

4NO+2HNO3===3N2O3+H2O2NO(+2 valence) -2E--- Summoning N2O3 (+3 valence) 2HNO3 (+5 valence) +4E---N2O3 (+3 valence) The minimum common sum collapse multiple is 4, so 2NO (+2 valence) front shirt Qing multiplied by 2 The total n atom is 6, so N2O3 is multiplied by 3

Both sides of the hydrogen are 2, and finally check whether the number of oxygen on both sides is balanced.

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Problem description: Is it so difficult to make N2 both an oxidation product and a reduction product?

Analysis: NH4NO3, the front n is -3 valence, and the back n is +5 valence. n is used as both an oxidant and a reducing agent.

The 3-valent n is from -3 to 0, and the +5-valent n is reduced from 5 to 0, so the ratio of the oxidation leakage defibrillant and the original agent is 5:3, so the coefficient x of n2 should ensure that 2*x is a multiple of 8, and the celebration is matched with n2 with 4, and the rest of the coefficients are matched according to the coefficient of n2.

5nh4no3=4n2+2hno3+9h2o

n : 3 liters to socks call 0 3*2=6*2

o: 0 to -2 price 2*2=4*3

4NH3+3O2——2N2+6H2O,2,4NH3+3O2=2N2+6H2O,2,4NH3+3O2=2N2+6H2O 【Reason】n in NH3 is -3 valence, and n of N2 is 0 valence, so it is 3*2=6

In O2 o is 0 valence, H2O o is -2 valence, and the electron transfer is 2*2=4 According to the conservation of charge, the charge of the oxidation reaction and the reduction counter-accuser are equal, so the upper edge is 6*2=12, and the lower edge is 4*3=12

After this trimming, it is 4NH3+3O2=2N2+6H2O,2,4NH3+3O2=2N2+6H2O,1,4NH3+3O2=2N2+6H2O,1,4NH3+3O2——2N2+6H2O,0,

5nh4no3 ==4n2 + 2hno3 + 9h2o

2fes2 + 6h2so4 ==fe2[so4]3 + 4s + 3so2 + 6h2o

The second equation upstairs was written incorrectly, and there was no S generation.

The i of the i2 is 0 valence.

If you want to hno3, then beam n is +5 valence.

The i of the HIO3 is +5 valence.

The n of no is +2 valence.

So, n(i):n(n)=3:5

i.e. n(i2):n(hno3)=3:10

The following follows the principle of quality and spring collapse and balance, and the result is:

3i2+10hno3=6hio3+10no+2h2o