Ask a high school chemistry question about nitrogen oxide calculations .

NO2 may be 30ml or.

Analysis: There are two scenarios. The remaining gas may be O2, and the reaction is carried out according to 2H2O+4NO+3O2=4HNO3 and 2H2O+4NO2+O2=4HNO3, and all NO2 and NO2 are reacted.

It is also possible to have no left, because 3NO2+H2O=2Hno3+NOnO, we can assume that NO2 is consumed first, that is, 2H2O+4NO2+O2=4HNO3 occurs, and then the reaction 2H2O+4NO+3O2=4HNO3 occurs, and finally part of NOno is left.

1) If the remaining gas is O2. Then the reaction consumes O2 volume of 20-5=15ml

By the reaction of 2H2O + 4O + 3O2 = 4Hno3, it is known that the oxidation of 40mL NO requires O2 30mL; By the reaction of 2H2O + 4 NO2 + O2 = 4Hno3, it is known that the oxidation of 40ml of No2 requires O2 10ml. The actual oxidation of 40ml of mixed gas consumes O2 15ml. According to the criss-cross method, the formula is as follows:

no 30/40 15/40-10/40

no2 10/40 30/40-15/40

Therefore, the volume ratio of the original mixture NO and NO2 is 1 3

Therefore, the volume of NO2 in the original mixture is 40 * (3 4) = 30ml.

2) If the remaining gas is の. Then the volume of O2 consumed by the reaction is 20ml, and the mixture consumes 40-5=35ml.

By the reaction of 2H2O + 4O + 3O2 = 4Hno3, it is known that the oxidation of 40mL NO requires O2 30mL; By the reaction of 2H2O + 4 NO2 + O2 = 4Hno3, it is known that the oxidation of 40ml of No2 requires O2 10ml. The actual oxidation of 35ml of mixed gas, consumption of O2 20ml. According to the criss-cross method, the formula is as follows:

no 30/40 20/35-10/40

no2 10/40 30/40-20/35

So, the volume ratio of NO and NO2 in the consumed mixture is 9 5

Therefore, the volume of NO2 in the original mixture is 35*(5 14)=.

Because 3NO2+H2O=2HNO3+NOn, 2NO+O2=2NO2, the two equations are added to obtain equations such as (1)4NO2+O2+2H2O=4HNO3 (2)4NO+3O2+2H2O=4HNO3 (3)NO2+NO+O2+H2O=2HNO3, and then calculate.

It's empty to say too much, let's take a look at a few example questions!

Q: A total of 9ml of mixed gas filled with NO2 and O2 is inverted in a tank filled with water, and there is no remaining gas after full reaction, what is the volume of NO2 and O2 in the mixed gas?

In this question, 3NO2+H2O=2HNO3+NO

2NO+O2=2NO2.

Answer: After the merger, it is 4NO2+O2+2H2O==4HNO3

It can be seen that in order for the gas to react exactly completely, then Wu Duqing no2:o2=4:1

So no2 volume = 9 (4+1) *4 =

O2 volume = 4NO2 + O2 + 2H2O = = 4Hno3 is best remembered.

Question: 2. When the mixed gas of NO2 and O2 is dissolved in water, according to 4NO2 + O2 + 2H2O = 4HNO3, it can be seen that: V(NO2) V(O2)= 4 1, there is a complete reaction before the cavity, and no gas remains; When v(no2) v(o2)>4 1, NO2 is excessive, and the remaining gas is NO (calculated according to 3NO2+H2O=2HNO3+NO); When v(no2) v(o2)4:

1 hour. Then the amount of the substance of NO2 participating in the reaction is equal to 4 times the amount of the substance of O2. The amount of No2 is greater than 4 times that of O2, so No2 will remain, that is, there is an excess of No2, and then No2 can react with water, 3NO2 H2O 2Hno3 NoTherefore, the final residual gas is no.

Do you understand? Question: GFe is put into hot concentrated nitric acid to produce reddish-brown gas A, the resulting solution is evaporated under reduced pressure to obtain a mixture of 20g Fe(NO3)2 and Fe(NO3)3, the solid is heated at high temperature to obtain reddish-brown Fe2O3 and mixed gases B, A, B gas are mixed into sufficient water, and the volume of remaining gas under standard conditions is (

a．2240ml b．4480ml c．3360ml d．1120ml

Answer: Conservation law.

g fe = +3 valence fe --- transfer electrons.

The remaining gas cannot be O2, it can only be No

1mol of electrons is produced and 3mol of electrons is transferred, so the generated no is 2240ml

G Fe eventually all become +3 valence, and Hno3 eventually shifts to NO, and then uses the conservation of electrons).

Sum up! This kind of problem is generally solved by the conservation method.

"These gases are mixed with the standard condition and then passed into the water", so the liter gas here should be oxygen, right?

Analysis: The reaction of copper with concentrated nitric acid and the reaction of mixed gas with water are all redox reactions. Therefore, the law of conservation of charge should be observed during the reaction.

In the reaction between copper and concentrated nitric acid, the number of electrons lost by copper is equal to the number of electrons gained by nitrogen (nitrogen gets electrons to form nitrogen dioxide, nitrogen tetroxide, and nitric oxide). Nitrogen dioxide, nitrogen tetroxide, nitric oxide mixed with oxygen In the reaction with water, the number of electrons obtained by oxygen is equal to the number of electrons lost by nitrogen (in nitrogen dioxide, nitrogen tetroxide, nitric oxide). So the number of electrons of the oxygen element is equal to the number of electrons lost by the copper element.

Amount of oxygen = Amount of substance

During the reaction, an oxygen atom needs to get 2e, and oxygen needs to get electrons.

Then, the amount of matter in which copper loses electrons is also. During the reaction, a copper atom loses 2e, so the amount of copper participating in the reaction is .

According to: cu(2

2oh(-)=cu(oh)2↓

Volume of sodium hydroxide consumed =

Chemical reaction.

The total formula is 4NO

3o2+2h2o=4

Hno3No is insoluble in water, so the remaining gas is all.

no, then the reaction.

The volume of の is 8 ml

The volume of の and.

The volume ratio of the O2 reaction is:

4:3, so the volume of oxygen introduced is 6ml

The most basic are these two equations.

3no2+h2o==2hno3+no

1)2no+o2==2no2

2) (1) *2 + (2) can be obtained.

4no2+o2+2h2o==4hno3。

So at this time v(no2) v(o2)=

4 1, when v(no2) v(o2) > 4:1, that is, no2 excess, so (1) type excess then there will naturally be no surplus. In the same way, when v(no2) v(o2) <

4 1, i.e. O2 excess, there must be oxygen left.

1. A is no and its volume is 6 ml2, calculate the range of values of v.

6<=v<=18 process:

1.According to the equation.

3no2h2o

2hno3+no

It can be seen that the generated gas is no (no does not react with water), so the remaining gas is no, and then according to the equation.

2no+o2=

2NO2 shows that the volume ratio of NO to O2 participating in the reaction is 2:1

If the volume of O2 is known to be equal to the volume of gas A, i.e., NO, then if O2 is excessive, there are both O2 and NO. Let the volume of O2 participating in the reaction be A, then the volume of the colorless gas A, i.e., NO, is 2A

2no+o2=

2no23no2

h2o2hno3+no

12aa2a

2A2 3A Remaining O2 Volume + No Volume Generated After Water Again = 5ml = A + 2 3A

Get a = 3ml

2a=6ml

So, the volume of colorless gas A is 6ml

2.Use the difference method and the extreme value method to solve the problem.

It is known that the volume of gas A is 6ml, that is, there is 6ml of gas remaining after the reaction, so the volume of gas reduced by the reaction is V-6ml

3no2h2o

2hno3+no

Difference 313-1

xv-6ml

The solution yields x = 3 2 (v-6 ml).

The solution volume is the volume of NO2 in the gas mixture, and according to the extreme value method, the volume range of NO2 in the gas mixture is 0<=x<=V

So 0<=3 2(v-6ml)<=v

Solution 6<=

v<=18

4no2 + o2 + 2h2o=4hno3 4no + 3o2 + 2h2o=4hno3

10ml 10ml(10-2)ml 6ml

Because NO2 is soluble in water, the residual gas can only be NO or O2, see the above formula, if NO2 is excessive, then O2 is consumed, if the excess is oxygen, then the amount of oxygen consumed is.

The answer to this question lies in whether O2 is caused by an excess.

1. Excess O2, that is, the last 2ml of O2 remains

4no+3o2+2h2o=4hno3

4no2+o2+2h2o=4hno3

In this way, no and no2 exactly react exactly and o2 requires o2=

If there is an excess of 2 ml, there is 12 ml

2. O2 is not excessive, at this time there is only 2ml of NO, that is, 8ml of NO participates in the reaction and the same equation as above to obtain NO2 consumption O2=; の consumption o2 = 6 ml of one plus.

Iron oxide is, so the consumption of hydrogen ions should be, trivalent iron ions, and the reaction with iron is consumed to form divalent iron. Iron is removed from the reaction with ferric iron, and the reaction with hydrogen ions is left to consume hydrogen ions. Co-consumption of hydrogen ions.

So the total is sulfuric acid, and the amount and concentration of sulfuric acid substances should be.

This equation is obtained by combining the two equations 3NO2+H2O=2Hno3+NO2 and 2NO+O2=2NO2. The specific process is as follows: the original gas is NO2, NO2 is dissolved in water to obtain nitric acid and NO, and NO2 is combined with O2 to produce NO2, and the cycle repeats, and finally all NO2 is converted into HNO3

That is to say, there is no excess NO remaining, at this time, it is necessary to expand the first equation by 2 times as a whole, because the coefficient of NO in the second equation is 2, in order to convert all the generated NO, so the first equation is expanded by 2 times, and the equation 6NO2+2H2O=4HNO3+2NO2 is added, and this equation is added with 2NO+O2=2NO2 (left and left, right and right are added) to obtain 6NO2 + 2H2O + 2NO+O2 = 4HNO3 + 2NO+2NO2, merge both sides of the equation in the same way to get 4NO2+O2+2H2O=4HNO3.

For the volume ratio, this is the coefficient ratio of the equation that we get. : v(no2) v(o2) = 4 1, exactly complete reaction, no gas remaining; This is easy to understand, that is, the coincidence coefficient ratio happens to be a complete response.

When V(NO2) V(O2)>4 1, NO2 is excessive, and the remaining gas is NO; The same can be found in this equation that we get. Greater than 4:1, we can see that O2 is 1, NO2 > 4, when O2 is 1, NO2 is exactly needed to be 4, at this time NO2 > 4, so NO2 remains, and NO2 can react with water, so the final remaining gas is NO

When V(NO2) V(O2)<4 1, O2 is excessive, and the remaining gas is O2In the same way, if we think of no2 as 4, then o2 > 1Because when No2 is 4, O2 happens to be 1, and O2 > 1, so O2 is excessive.

The most basic are these two equations.

3NO2+H2O==2HNO3+NO (1)2NO+O2==2NO2 (2)(1)*2+(2) gives 4NO2+O2+2H2O==4HNO3.

So at this time, v(no2) v(o2) = 4 1, when v(no2) v(o2) > 4:1, that is, no2 excess, so (1) excess then there will naturally be no remainder. In the same way, when v(no2) v(o2)< 4 1, i.e., there is an excess of o2, there must be oxygen left.

According to the chemical equation 4NO2 + O2 + 2H2O = 4HNO3, at room temperature: because nitrogen dioxide and oxygen are both gases, the coefficient ratio can be simply regarded as the volume ratio, so when V(NO2) V(O2) = 4 1, it happens to be completely reacted, and there is no gas left. When v(no2) v(o2) > 4 1, the assumption is 5:

1, that is to say, nitrogen dioxide is 5 volumes, oxygen is 1 volume, 1 volume of oxygen only needs 4 volumes of nitrogen dioxide is enough, the extra is nitrogen dioxide, but the extra nitrogen dioxide will also react with water to form nitric oxide, so the remaining gas is NO, when V(NO2) V(O2)<4 1, assuming 2:1, then nitrogen dioxide is 2 volumes, only the volume of oxygen is enough, now there is 1 volume of oxygen, so it is said that there is too much oxygen, oxygen is not easy to dissolve, and there is no reaction, so the remaining gas is O2