A high school chemistry calculation problem about nitric acid

Analysis: First of all, Cu is oxidized by nitric acid and loses electrons, nitric acid gets electrons and converts into the three gases, and all the three gases are oxidized by O2, that is, the valence of the three gases rises and is oxidized, and all the electrons just obtained from Cu are lost to element O.

That is: there are electrons lost by Cu = electrons gained by O2.

From this, the number of electrons lost by Cu can be calculated.

The number of electrons lost in copper is equal to the number of electrons gained in oxygen.

Then *4=n(cu)*2 is solved to get n(cu)=1mol copper ions corresponding to 2mol hydroxide, so v(naoh)=2n divided by c= divided by 5=

I'm sorry, it's too late to go to bed, I can't take your question seriously, so I made this decision in order to get full of today's score. But I still hope you and I, newbie, will give me some encouragement!

1) m (mixed gas) =

g mol is determined by the law of conservation of mass, n (mixed gas) m (mixed gas) = n (a) m (a).

So m(a)=58

G mol infers that the chemical formula of A is C4H10

ch3ch2ch2ch3, butane.

CH3CH2(CH3)2,2-propane.

2)①b：ch4，c：c3h6

b：c2h6，c：c2h4

3) When B is methane, CH4+2O2 CO2+2H2O satisfies the condition.

b：ch4，c：c3h6

When B is ethane, C2H6+7 2O2 2CO2+3H2O does not meet the condition.

So B can only be methane, and C can only be ethylene.

abcd

In general, the reaction between metals and acids will have the formation of acid anions, a

Concentrated nitric acid has strong oxidation and acidity, and reacts with active metals to have NO2,3 valent Febc, I think it should be nitric acid thinning, and Fe reaction to form Rat

n（cu）=

n(h+)=

n(no3-)=

3cu+8h+

2no3-→3cu2+

2no↑+4h2o

Because n(h+) is insufficient, it is calculated according to it, so n(cu2+)=, c(cu2+)=

If you think this kind of problem is best done with ionic equations, chemical equations can't be done.

Simple, I'll teach you.

This one has 2 reactions, so it can't be done in a normal way. It should be conserved with elements.

Cu + 4 Hno3 = Cu (No3) 2 + 2 O + 2H2O3 Cu8 Hno3 = 3 Cu (No3) 2 + No + H2O This requires the conservation of elements, n (Hno3) = 1 2n (Cu) + 1 2n (gas).

HNO3 has three destinations.

Cu(NO3)2 is conserved by copper.

No and no2 are altogether, i.e., no matter what ratio n of the two is.

So a total of nitric acid is consumed.

The amount of the substance of copper is.

There is xmolno ymolno2

Then: x+y=

3x+y=2*conservation according to gains and losses electrons).

Solve the system of equations, get.

x= volume of y=no.

The volume of no2.

Common calculation methods for solving the reaction between metals and nitric acid:

1) Conservation of atoms: When nitric acid reacts with metals, part of it exists in the form of NO3- and part of it is converted into a reduction product (i.e., no or NO2), and the amount of N in these two parts is equal to the amount of N in Hno3 consumed by the reaction.

2) Conservation of electrons: The reaction between nitric acid and metals is a redox reaction, and the number of electrons obtained by n atoms is equal to the number of electrons lost by metal atoms.

3) Use the ion equation to calculate instead of the chemical equation, because there is NO3- after the reaction of Hno3, if the solution contains H+, the reaction can continue, so the answer calculated by the chemical equation is usually wrong.

1 charge conservation n 5 valence down to 3 4 valence find out the number of electrons transferred (to be accurate).

2 Conservation of Elements There are several n to 3 Valence 4 Valence to find out (to be accurate) 3 Find out the reducing agent.

Moles are used as the base unit.

Hello, this problem is a typical application of conservation method calculations.

Suppose that the volume of the gas (nitric oxide) collected under standard conditions is XL at the end, and only look at the electron transfer at the beginning and end of the reaction.

Copper (0 - >2), nitrogen (+5 in nitrate - > +2 in nitric oxide)2*

The chemical reaction that occurs in this question is: Cu + 4Hno3 = Cu(No3)2 + 2NO2 2H2O

As the nitric acid concentration thins out, the reaction is: 3Cu + 8Hno3 = 3Cu(NO3)2 + 2NO2H2O

When the gas mixture passes through the water, a reaction occurs: 3NO2 + H2O = 2HNO3 + NO

The total amount of matter for the No and No2 gases is: i.e., the nitrogen atom n(n)= with the change in valency

Through process analysis, we know that the final change process is that the electrons lost by copper, the nitrogen atoms in nitric acid get electrons, and finally all become no, we regard all the substances in the whole process as a whole, and the electron gain and loss are also equal, so:

The amount of total electrons lost by Cu becoming Cu2+ (each Cu loses 2 electrons): n loss = 64 * 2 = mol

Although in the first two steps of the reaction, Hno3 is converted into NO2 and NO2 respectively, but due to the third step of the reaction, NO2 is converted into NO, and the final change of the whole process is that all the N atoms in HNO3 with valency changes become NO (note that NO2 in the third step of the reaction is also converted to HNO3, and the final effect is that N in the original HNO3 is converted to NO).

Change from Hno3 to No (3 electrons for every n Let no be x mol then the total amount of electrons obtained: n = 3x mol

So there is the equation: 3x = mol x =

v(no)= * v (no)=

This is a multiple-choice question, and it is better to give options if it is to be discussed on a case-by-case basis.

Due to the possibility of the existence of the reaction Fe+2Fe3+ === 3Fe2+, there are three cases implied in the question, let Hno3 be Xg, and the reduced HNo3 be Y mol:

1) Fe happens to react with Hno3 to form Fe(No3)3Fe+ 4Hno3=== Fe(No3)3+ No + 2H2O56g 252g 1mol40g xg ymolx = 180g; y=

Reduced nitric acid:

2) Fe happens to react with Hno3 to form Fe(No3)23Fe+ 8Hno3=== 3Fe(No3)2+ 2O +4H2O168g 504g 2mol40g xg ymolx = 120g; y=

Nitric acid mass fraction: 20%.

Reduced nitric acid: (3) Fe reacts completely with Hno3 to form Fe(No3)2 and Fe(No3)3

The nitric acid mass fraction is between 20% and 30%, and the reduced nitric acid is between :.

Add KNO3, in which the combination of nitrate and H+ in the original solution is equivalent to the reaction of copper and nitric acid) The gas produced by the oxidation of copper by nitric acid is oxidized by O2 and finally becomes nitric acid, and the whole process is equivalent to the oxidation of copper by O2, and the oxidized copper is calculated by the amount of O2).

After adding hydrochloric acid, the combination of H+ and nitrate in the original solution can be re-reacted with the metal) The element may be a +2 valence metal or +4 carbon, which can be judged by the conservation of gain and loss electrons) is calculated by the ionic equation of the reaction between copper and dilute nitric acid, and the ratio of the measurement number of each substance according to the equation is Cu: H+: NO3- =3:

8:2, and the ratio of the three to the data is 3:6:

4. It can be seen that H+ is insufficient, and the amount of Cu2+ in the solution should be calculated by the mass of H+, and then converted into concentration).

The principle is the same as that of question 4).

10.(1) V(NO) is about V(NO2) is about 2) [(mol l.]]

Analysis: After the release of gas, the solution may contain Hno3 in addition to Cu(NO3)2. If no HNO3 is present, the added NaOH reacts directly with Cu(NO3)2 to form Cu(OH)2;If HNO3 is present, the added NaOH reacts with Hno3 and then Cu(NO3)2 to form Cu(OH)2.

In either case, the amount of Na+ and No3- is equal, so the amount of NaOH can be found to the amount of No3- in the solution, and the amount of nitric acid that is reduced to gas is the total amount of primary nitric acid, and then the amount of the original nitric acid can be calculated.

1.If the metal is monovalent, electron transfer is conserved, and the nitride valency decreases by 2 to 3, which is 3, and there is no such option; If the metal is bivalent and nitrogen is monovalent, it is nitrous oxide.

C+4Hno3===CO2+4NO2+2H2O, so it is positive bivalent or positive tetravalent.

Calculate according to the atomic mass you think.