At school, I had chemistry for show, nothing more. In the 9th grade, this subject was absent for half a year, and the remaining six months were taught by ... a fireman. In grades 10-11, chemistry went like this: I didn’t go to it for half a semester, then I handed in three downloaded presentations, and I was given a proud “five” because I had to go to school 6 days a week for 12 km (I lived in the village, studied in city) was, to put it mildly, laziness.
And in the 11th grade, I decided to take chemistry. My level of knowledge of chemistry was zero. I remember being surprised by the existence of the ammonium ion:
- Tatyana Alexandrovna, what is it? (pointing to NH4+)
– Ammonium ion, formed when ammonia is dissolved in water, similar to potassium ion
- First time I see
Now about Tatyana Alexandrovna. This is my chemistry tutor from October to June 13/14 school year. Until February, I just went to her, sat out my pants, listened to a boring theory on general and inorganic chemistry. Then February came and I realized that the exam was too close... What should I do?! Get ready!
Subscribe to PUtelegram . Only the most important.
Little by little, solving options (at first without organics) I prepared. At the end of March, we finished the study of INORGANICS, there was a sampler, which I wrote for 60 points and for some reason I was very happy. And the goal was powerful, above 90 points (my faculty needed a lot of points). And all knowledge of organics was limited homologous series methane.
For April-May, a difficult task was ahead: to learn all the organics. Well, I sat until 11 at night, until my eyes stuck together, solved tests, filled my hand. I remember that on the last evening before the exam, I analyzed the topic “amines”. In general, time is running out.
How the exam itself went: in the morning I solved one option (to turn on the brain), I came to school. It was the most alert hour of my life. Firstly, chemistry was the most difficult exam for me. Secondly, immediately after chemistry, the results of the Unified State Examination in Russian were to be announced. I barely had enough time on the exam, although I didn’t have enough time to finish task C4. I passed by 86 points, which is not bad for several months of preparation. There were mistakes in part C, one in B (just for amines) and one controversial mistake in A, but A cannot be appealed.
Tatyana Alexandrovna reassured me, saying that it just didn’t fit in my head. But the story doesn't end there...
I did not enter my faculty last year. Therefore, it was decided: the second time it will work!
Started preparing right from the first of September. This time there was no theory, just solving tests, the more and faster the better. In addition, I studied "complex" chemistry for the entrance exam to the university, and for half a year I had a subject called "general and inorganic chemistry", which was led by Olga Valentinovna Arkhangelskaya herself, the organizer All-Russian Olympiad in chemistry. So six months passed. Knowledge of chemistry has grown exponentially. Came home in March, complete isolation. Continued preparation. I was just solving tests! Lot! There are about 100 tests in total, some of them several times. Passed the exam with 97 points in 40 minutes.
1) Be sure to study theory, and not just solve tests. The best textbook I consider "Principles of Chemistry" Eremin and Kuzmenko. If the book seems too big and complicated, then there is a simplified version (which is enough for the exam) - "Chemistry for high school students and university applicants";
2) Separately, pay attention to the topics: production, safety, chemical glassware (no matter how absurd it may sound), aldehydes and ketones, peroxides, d-elements;
3) Having solved the test, be sure to check your mistakes. Do not just count the number of errors, but look at which answer is correct;
4) Use the circular solution method. That is, they solved a collection of 50 tests, solve it again, in a month or two. So you fix the material that is not memorable for you;
5) Cribs - be! Write cheat sheets, always by hand and preferably small. This way you will remember the problem information better. Well, no one forbids using them in the exam (only in the toilet !!!), the main thing is to be careful.
6) Calculate your time along with the clearance. The main problem of the chemistry exam is the lack of time;
7) Make out the tasks (preferably) the way they are made out in the collections. Instead of "nu", write "n", for example.
Egor Sovetnikov told
Part C on the exam in chemistry begins with task C1, which involves compiling a redox reaction (already containing part of the reagents and products). It is worded like this:
C1. Using the electron balance method, write an equation for the reaction. Determine the oxidizing agent and reducing agent.
Often applicants believe that this task does not require special preparation. However, it contains pitfalls that prevent you from getting a full score for it. Let's see what to pay attention to.
Theoretical information.
Potassium permanganate as an oxidizing agent.
+ reducing agents
|
||
| in an acidic environment | in a neutral environment | in an alkaline environment |
| (salt of the acid involved in the reaction) |
Manganate or, - | |
Dichromate and chromate as oxidizing agents.
| (acidic and neutral environment), (alkaline environment) + reducing agents always turns out
|
||
| acidic environment | neutral environment | alkaline environment |
| Salts of those acids that participate in the reaction: | in solution or melt | |
Increasing the oxidation states of chromium and manganese.
| + very strong oxidizing agents (always regardless of the medium!) | ||
| , salts, hydroxo complexes | + very strong oxidizing agents: a), oxygen-containing salts of chlorine (in an alkaline melt) b) (in alkaline solution) |
Alkaline environment: formed chromate |
| , salt | + very strong oxidizing agents in an acidic environment or |
Acid environment: formed dichromate or dichromic acid |
| - oxide, hydroxide, salts | + very strong oxidizing agents: , oxygen-containing salts of chlorine (in melt) |
Alkaline environment: manganate |
| - salt | + very strong oxidizing agents in an acidic environment or |
Acid environment: Permanganate |
Nitric acid with metals.
- no hydrogen is released, nitrogen reduction products are formed.
| The more active the metal and the lower the acid concentration, the further nitrogen is reduced. | ||||
Nonmetals + conc. acid |
Inactive metals (to the right of iron) + dil. acid | Active metals (alkaline, alkaline earth, zinc) + conc. acid | Active metals (alkali, alkaline earth, zinc) + medium dilution acid | Active metals (alkaline, alkaline earth, zinc) + very dil. acid |
| Passivation: do not react with cold concentrated nitric acid: |
||||
| do not react with nitric acid at any concentration: |
||||
Sulfuric acid with metals.
- diluted sulfuric acid reacts like an ordinary mineral acid with metals to the left in a series of voltages, while hydrogen is released;
- when reacting with metals concentrated sulfuric acid no hydrogen is released, sulfur reduction products are formed.
| Inactive metals (to the right of iron) + conc. acid Nonmetals + conc. acid |
Alkaline earth metals + conc. acid | Alkali metals and zinc + concentrated acid. | Dilute sulfuric acid behaves like a normal mineral acid (like hydrochloric acid) | |
| Passivation: do not react with cold concentrated sulfuric acid: |
||||
| do not react with sulfuric acid at any concentration: |
||||
Disproportionation.
Disproportionation reactions are reactions in which the same the element is both an oxidizing agent and a reducing agent, both raising and lowering its oxidation state:
Disproportionation of non-metals - sulfur, phosphorus, halogens (except fluorine).
| Sulfur + alkali 2 salts, metal sulfide and sulfite (reaction occurs during boiling) | and |
| Phosphorus + alkali phosphine and salt hypophosphite(reaction proceeds at boiling) | and |
| Chlorine, bromine, iodine + water (without heating) 2 acids, Chlorine, bromine, iodine + alkali (without heating) 2 salts, and and water |
and |
| Bromine, iodine + water (when heated) 2 acids, Chlorine, bromine, iodine + alkali (when heated) 2 salts, and and water |
and |
Disproportionation of nitric oxide (IV) and salts.
| + water 2 acids, nitric and nitrogenous + alkali 2 salts, nitrate and nitrite |
and |
| and | |
| and |
Activity of metals and non-metals.
To analyze the activity of metals, either electrochemical series stresses of metals, or their position in Periodic table. The more active the metal, the easier it will donate electrons and the better it will be as a reducing agent in redox reactions.
Electrochemical series of voltages of metals.
Features of the behavior of some oxidizing and reducing agents.
a) oxygen-containing salts and acids of chlorine in reactions with reducing agents usually turn into chlorides:
b) if substances participate in the reaction in which the same element has a negative and positive oxidation state, they occur in the zero oxidation state (a simple substance is released).
Required skills.
- Arrangement of oxidation states.
It must be remembered that the degree of oxidation is hypothetical the charge of an atom (i.e. conditional, imaginary), but it must not go beyond common sense. It can be integer, fractional, or zero.Exercise 1: Arrange the oxidation states of the substances:
- Arrangement of oxidation states in organic matter Oh.
Remember that we are only interested in the oxidation states of those carbon atoms that change their environment in the redox process, while the total charge of the carbon atom and its non-carbon environment is taken as 0.Task 2: Determine the oxidation state of the carbon atoms circled along with the non-carbon environment:
2-methylbutene-2: - =
acetone:
acetic acid: -
- Do not forget to ask yourself the main question: who donates electrons in this reaction, and who accepts them, and what do they turn into? So that it does not work that electrons arrive from nowhere or fly away to nowhere.
Example:
In this reaction, one must see that potassium iodide can be only reducing agent, so potassium nitrite will accept electrons, lowering its degree of oxidation.
Moreover, under these conditions (dilute solution) nitrogen goes from to the nearest oxidation state. - Drawing up an electronic balance sheet is more difficult if formula unit substance contains several atoms of an oxidizing agent or reducing agent.
In this case, this must be taken into account in the half-reaction by calculating the number of electrons.
The most common problem is with potassium dichromate, when it goes into the role of an oxidizing agent:These deuces cannot be forgotten when calling, because they indicate the number of atoms of a given type in the equation.
Task 3: What coefficient should be put before and before
Task 4: What coefficient in the reaction equation will stand in front of magnesium?
- Determine in which medium (acidic, neutral or alkaline) the reaction takes place.
This can be done either about the products of the reduction of manganese and chromium, or by the type of compounds that were obtained on the right side of the reaction: for example, if in the products we see acid, acid oxide - it means that this is definitely not an alkaline environment, and if metal hydroxide precipitates, it is definitely not acidic. And of course, if on the left side we see metal sulfates, and on the right - nothing like sulfur compounds - apparently, the reaction is carried out in the presence of sulfuric acid.Task 5: Determine the environment and substances in each reaction:
- Remember that water is a free traveler, it can both participate in a reaction and be formed.
Task 6:Which side of the reaction will the water be on? What will the zinc go to?
Task 7: Soft and hard oxidation of alkenes.
Add and equalize the reactions, after placing the oxidation states in organic molecules:(cold solution)
(aqueous solution) - Sometimes a reaction product can be determined only by compiling an electronic balance and understanding which particles we have more:
Task 8:What other products will be available? Add and equalize the reaction:
- What are the reactants in the reaction?
If the schemes we have learned do not give an answer to this question, then we need to analyze which oxidizing agent and reducing agent in the reaction are strong or not?
If the oxidizer is of medium strength, it is unlikely that it can oxidize, for example, sulfur from to, usually oxidation only goes up to.
Conversely, if is a strong reducing agent and can recover sulfur from up to , then only up to .Task 9: What will the sulfur turn into? Add and equalize the reactions:
(conc.)
- Check that there is both an oxidizing agent and a reducing agent in the reaction.
Task 10: How many other products are in this reaction, and which ones?
- If both substances can exhibit the properties of both a reducing agent and an oxidizing agent, it is necessary to consider which of them more active oxidant. Then the second one will be the restorer.
Task 11: Which of these halogens is the oxidizing agent and which is the reducing agent?
- If one of the reactants is a typical oxidizing agent or a reducing agent, then the second one will “do his will”, either by donating electrons to the oxidizing agent, or by accepting them from the reducing agent.
Hydrogen peroxide is a substance with dual nature, in the role of an oxidizing agent (which is more characteristic of it) passes into water, and as a reducing agent - passes into free gaseous oxygen.
Task 12: What role does hydrogen peroxide play in each reaction?
The sequence of arrangement of the coefficients in the equation.
First put down the coefficients obtained from the electronic balance.
Remember that you can double or reduce them only together. If any substance acts both as a medium and as an oxidizing agent (reducing agent), it will have to be equalized later, when almost all the coefficients are arranged.
Hydrogen is equalized penultimately, and we only check for oxygen!
Take your time counting the oxygen atoms! Remember to multiply rather than add indices and coefficients.
The number of oxygen atoms on the left and right sides must converge!
If this does not happen (provided that you count them correctly), then there is a mistake somewhere.
Possible mistakes.
- Arrangement of oxidation states: check each substance carefully.
Often mistaken in the following cases:a) oxidation states in hydrogen compounds of non-metals: phosphine - oxidation state of phosphorus - negative;
b) in organic substances - check again whether the entire environment of the atom is taken into account;
c) ammonia and ammonium salts - they contain nitrogen always has an oxidation state;
d) oxygen salts and acids of chlorine - in them chlorine can have an oxidation state;
e) peroxides and superoxides - in them, oxygen does not have an oxidation state, it happens, and in - even;
f) double oxides: - in them, metals have two different oxidation states, usually only one of them is involved in the transfer of electrons.Task 14: Add and equalize:
Task 15: Add and equalize:
- The choice of products without taking into account the transfer of electrons - that is, for example, in the reaction there is only an oxidizing agent without a reducing agent, or vice versa.
Example: free chlorine is often lost in a reaction. It turns out that electrons came to manganese from outer space...
- Incorrect products from a chemical point of view: a substance that interacts with the environment cannot be obtained!
a) in an acidic environment, metal oxide, base, ammonia cannot be obtained;
b) in an alkaline environment, acid or acid oxide will not be obtained;
c) an oxide, let alone a metal that reacts violently with water, is not formed in an aqueous solution.Task 16: Find in reactions erroneous products, explain why they cannot be obtained under these conditions:
Answers and solutions to tasks with explanations.
Exercise 1:
Task 2:
2-methylbutene-2: - =
acetone:
acetic acid: -
Task 3:
Since there are 2 chromium atoms in the dichromate molecule, they donate 2 times more electrons - i.e. 6.
Task 4:
Since in a molecule two nitrogen atoms, this deuce must be taken into account in the electronic balance - i.e. before magnesium it should be coefficient .
Task 5:
If the environment is alkaline, then phosphorus will exist in the form of salt- potassium phosphate.
If the medium is acidic, then phosphine turns into phosphoric acid.
Task 6:
Since zinc is amphoteric metal, in alkaline solution it forms hydroxocomplex. As a result of the arrangement of the coefficients, it turns out that water must be present on the left side of the reaction:
Task 7:
Electrons give away two atoms in an alkene molecule. Therefore, we must take into account general the number of electrons donated by the whole molecule:
(cold solution)
Please note that out of 10 potassium ions, 9 are distributed between two salts, so alkali will turn out only one molecule.
Task 8:
In the balance sheet process, we see that 2 ions have 3 sulfate ions. This means that in addition to potassium sulfate, another sulfuric acid(2 molecules).
Task 9:
(permanganate is not a very strong oxidizing agent in solution; note that water passes during adjustment to the right!)
(conc.)
(concentrated nitric acid is a very strong oxidizing agent)
Task 10:
Don't forget that manganese accepts electrons, wherein chlorine should give them away.
Chlorine is released in the form of a simple substance.
Task 11:
The higher the non-metal in the subgroup, the more active oxidizing agent, i.e. Chlorine is the oxidizing agent in this reaction. Iodine passes into the most stable positive oxidation state for it, forming iodic acid.
Task 12:
(peroxide is an oxidizing agent, because the reducing agent is )
(peroxide is a reducing agent, because the oxidizing agent is potassium permanganate)
(peroxide is an oxidizing agent, since the role of a reducing agent is more characteristic of potassium nitrite, which tends to turn into nitrate)
The total charge of a particle in potassium superoxide is . Therefore, he can only give.
|
(water solution) (acid environment) |
slide 2
“In order to avoid mistakes, one must gain experience; in order to gain experience, one must make mistakes.”
slide 3
C1. Using the electron balance method, write an equation for the reaction. Determine the oxidizing agent and reducing agent.
slide 4
Required Skills
Arrangement of oxidation states Ask yourself the main question: who donates electrons in this reaction, and who accepts them? Determine in which medium (acidic, neutral or alkaline) the reaction takes place. if we see acid in products, acid oxide means that it is definitely not an alkaline medium, and if metal hydroxide precipitates, it is definitely not acidic. Check that there is both an oxidizing agent and a reducing agent in the reaction. If both substances can exhibit the properties of both a reducing agent and an oxidizing agent, it is necessary to consider which of them is a more active oxidizing agent. Then the second one will be the restorer.
slide 5
The sequence of the coefficients in the equation
First, put down the coefficients obtained from the electronic balance. If any substance acts both as a medium and as an oxidizing agent (reductant), it will need to be equalized later, when almost all the coefficients are placed. The penultimate hydrogen is equalized by oxygen, we only check
slide 6
Possible mistakes
Arrangement of oxidation states: a) oxidation states in hydrogen compounds of non-metals: phosphine РН3 - the oxidation state of phosphorus is negative; b) in organic substances - check again whether the entire environment of the C atom is taken into account c) ammonia and ammonium salts - in them nitrogen always has an oxidation state of −3 c) oxygen salts and chlorine acids - in them chlorine can have an oxidation state of +1, +3, +5, +7; d) double oxides: Fe3O4, Pb3O4 - in them, metals have two different oxidation states, usually only one of them is involved in electron transfer.
Slide 7
2. The choice of products without taking into account the transfer of electrons - that is, for example, in the reaction there is only an oxidizing agent without a reducing agent, or vice versa 3. Incorrect products from a chemical point of view: a substance that interacts with the environment cannot be obtained! a) in an acidic environment, metal oxide, base, ammonia cannot be obtained; b) in an alkaline environment, acid or acid oxide will not be obtained; c) an oxide, let alone a metal that reacts violently with water, is not formed in an aqueous solution.
Slide 8
Slide 9
Increasing the oxidation states of manganese
Slide 10
Dichromate and chromate as oxidizing agents.
slide 11
Increasing the oxidation states of chromium
slide 12
Nitric acid with metals. - hydrogen is not released, nitrogen reduction products are formed
slide 13
Disproportionation
Disproportionation reactions are reactions in which the same element is both an oxidizing agent and a reducing agent, simultaneously raising and lowering its oxidation state:
Slide 14
Sulfuric acid with metals
Dilute sulfuric acid reacts like an ordinary mineral acid with metals to the left of H in a series of voltages, while hydrogen is released; - when reacting with concentrated sulfuric acid metals, hydrogen is not released, sulfur reduction products are formed.
slide 15
Disproportionation of nitric oxide (IV) and salts.
slide 16
C 2. Relationship between different classes of inorganic substances
Changes in KIM 2012
Slide 17
Task C2 is offered in two formats. In some versions of the CMM, it will be offered in the old format, and in others in a new one, when the task condition is a description of a specific chemical experiment, the course of which the examinee will have to reflect through the equations of the corresponding reactions.
Slide 18
C2.1. (old format) - 4 points. Substances are given: nitric oxide (IV), copper, potassium hydroxide solution and concentrated sulfuric acid. Write the equations for four possible reactions between all the proposed substances, without repeating the pairs of reactants.
C2.2. (In the new format) - 4 points. The salt obtained by dissolving iron in hot concentrated sulfuric acid was treated with an excess of sodium hydroxide solution. The brown precipitate formed was filtered off and dried. The resulting substance was fused with iron. Write the equations of the described reactions.
Slide 19
1 or 2 reactions usually "lie on the surface", showing either acidic or basic properties of the substance. In a set of four substances, as a rule, typical oxidizing and reducing agents are found. In this case, at least one is an OVR. To write reactions between an oxidizing agent and a reducing agent, it is necessary: 1. to assume to what possible value the degree of oxidation of the reducing atom will increase and in which reaction product it will manifest it; 2. to suggest to what possible value the degree of oxidation of the oxidizing atom will decrease and in what reaction product it will manifest it. Mandatory minimum knowledge
Slide 20
Typical oxidizing and reducing agents in order of decreasing oxidizing and reducing properties
slide 21
Four substances are given: nitric oxide (IV), hydrogen iodide, potassium hydroxide solution, oxygen. 1. acid + alkali a) there are 2 oxidizing agents: NO2 and O2 b) reducing agent: HI 2. 4HI + O2 = 2I2 + 2H2O 3. NO2 + 2HI = NO + I2 + H2O Disproportionation in alkali solutions 4.2NO2 + 2NaOH = NaNO2 + NaNO3 + H2O
slide 22
C 3. Genetic relationship between the main classes of organic substances
slide 23
General properties classes of organic substances General methods for obtaining organic substances Specific properties of some specific substances Required minimum knowledge
slide 24
Most of the transformations of hydrocarbons into oxygen-containing compounds occur through halogen derivatives during the subsequent action of alkalis on them Interconversions of hydrocarbons and oxygen-containing organic substances
Slide 25
Basic transformations of benzene and its derivatives
Note that for benzoic acid and nitrobenzene, substitution reactions occur in the meta positions, while for most other benzene derivatives, in the ortho and para positions.
slide 26
Obtaining nitrogen-containing organic substances
Slide 27
Interconversions of nitrogen-containing compounds
It must be remembered that the interaction of amines with haloalkanes occurs with an increase in the number of radicals at the nitrogen atom. So it is possible to obtain salts of secondary amines from primary amines, and then from them to obtain secondary amines.
Slide 28
Redox properties of oxygen-containing compounds
Oxidizing alcohols are most often copper (II) oxide or potassium permanganate, and oxidizing agents for aldehydes and ketones - copper (II) hydroxide, ammonia solution of silver oxide and other oxidizing agents. The reducing agent is hydrogen
Slide 29
Obtaining derivatives of carboxylic acids
Sector 1 - chemical reactions with the breaking of O-H bonds (obtaining salts) Sector 2 - chemical reactions with the replacement of a hydroxo group by a halogen, an amino group or the production of anhydrides Sector 3 - obtaining nitriles
slide 30
Genetic relationship between carboxylic acid derivatives
Slide 31
Typical mistakes when performing the SZ task: ignorance of the flow conditions chemical reactions, genetic connection classes of organic compounds; ignorance of the mechanisms, nature and conditions of reactions involving organic substances, properties and formulas of organic compounds; inability to predict properties organic compound on the basis of ideas about the mutual influence of atoms in a molecule; ignorance of redox reactions (for example, with potassium permanganate).
slide 32
С 4. Calculations by reaction equations
Slide 33
Task classification
slide 34
Calculations by reaction equations. The gas released during the interaction of 110 ml of a 18% solution of HCl (ρ = 1.1 g / ml) and 50 g of a 1.56% solution of Na2S was passed through 64 g of a 10.5% solution of lead nitrate. Determine the mass of salt precipitated.
Slide 35
II. Tasks for a mixture of substances To neutralize 7.6 g of a mixture of formic and acetic acids, 35 ml of a 20% potassium hydroxide solution (density 1.20 g / ml) was used. calculate the mass of acetic acid and its mass fraction in the initial mixture of acids.
slide 36
III. Determination of the composition of the reaction product (tasks for the "type of salt") Ammonia with a volume of 4.48 l (N.U.) was passed through 200 g of a 4.9% solution of phosphoric acid. Name the salt formed as a result of the reaction, and determine its mass.
Slide 37
IV. Finding the mass fraction of one of the reaction products in solution according to the material balance equation The oxide formed during the combustion of 18.6 g of phosphorus in 44.8 l (N.O.) of oxygen was dissolved in 100 ml of distilled water. Calculate the mass fraction of phosphoric acid in the resulting solution.
Slide 38
Finding the mass of one of the starting substances according to the material balance equation What mass of lithium hydride must be dissolved in 200 ml of water to obtain a solution with mass fraction hydroxide 10%? What color will methyl orange acquire when it is added to the resulting solution? Write down the reaction equation and the results of intermediate calculations.
Municipal budgetary educational institution
"Average comprehensive school № 37
with in-depth study of individual subjects "
Vyborg, Leningrad region
"Solving computational problems of an increased level of complexity"
(materials for preparing for the exam)
chemistry teacher
Podkladova Lyubov Mikhailovna
2015
Statistics conducting the exam shows that about half of the students cope with half of the tasks. Analyzing the results of the check USE results in chemistry students of our school, I came to the conclusion that it is necessary to strengthen the work on solving calculation problems, so I chose methodical theme"Solving problems of increased complexity."
Tasks are a special type of tasks that require students to apply knowledge in compiling reaction equations, sometimes several, compiling a logical chain in carrying out calculations. As a result of the decision, new facts, information, values of quantities should be obtained from a certain set of initial data. If the algorithm for completing a task is known in advance, it turns from a task into an exercise, the purpose of which is to turn skills into skills, bringing them to automatism. Therefore, in the first classes in preparing students for the exam, I remind you of the values \u200b\u200band units of their measurement.
Value
Designation
Units
g, mg, kg, t, ... * (1g \u003d 10 -3 kg)
l, ml, cm 3, m 3, ...
*(1ml \u003d 1cm 3, 1 m 3 \u003d 1000l)
Density
g/ml, kg/l, g/l,…
Relative atomic mass
Relative molecular weight
g/mol, …
Molar volume
Vm or Vm
l / mol, ... (at n.o. - 22.4 l / mol)
Amount of substance
mole, kmol, mlmol
Relative density of one gas over another
Mass fraction of a substance in a mixture or solution
Volume fraction of a substance in a mixture or solution
Molar concentration
mol/l
Product output from theoretically possible
Avogadro constant
N A
6.02 10 23 mol -1
Temperature
t0 or
Celsius
on the Kelvin scale
Pressure
Pa, kPa, atm., mm. rt. Art.
Universal gas constant
8.31 J/mol∙K
Normal conditions
t 0 \u003d 0 0 C or T \u003d 273K
P \u003d 101.3 kPa \u003d 1 atm \u003d 760 mm. rt. Art.
Then I propose an algorithm for solving problems, which I have been using for several years in my work.
"An algorithm for solving computational problems".
V(r-ra)V(r-ra)
↓ρ ∙ Vm/ ρ
m(r-ra)m(r-ra)
↓m∙ ω m/ ω
m(in-va)m(in-va)
↓ m/ MM∙ n
n 1 (in-va)-- by ur. districts.→ n 2 (in-va)→
V(gas) / V M ↓ n∙ V M
V 1 (gas)V 2 (gas)
Formulas used to solve problems.
n = m / Mn(gas) = V(gas) / V M n = N / N A
ρ = m / V
D = M 1(gas) / M 2(gas)
D(H 2 ) = M(gas) / 2 D(air) = M(gas) / 29
(M (H 2) \u003d 2 g / mol; M (air.) \u003d 29 g / mol)
ω = m(in-va) / m(mixtures or solutions) = V(in-va) / V(mixtures or solutions)
= m(pract.) / m(theor.) = n(pract.) / n(theor.) = V(pract.) / V(theor.)
C = n / V
M (gas mixtures) = V 1 (gas) ∙ M 1(gas) + V 2 (gas) ∙ M 2(gas) / V(gas mixtures)
The Mendeleev-Clapeyron equation:
P∙ V = n∙ R∙ T
For passing the exam, where the types of tasks are quite standard (No. 24, 25, 26), the student must first of all show knowledge of standard calculation algorithms, and only in task No. 39 can he meet a task with an undefined algorithm for him.
Classification chemical problems increased complexity is hampered by the fact that most of them are combined tasks. I divided the calculation tasks into two groups.
1. Tasks without using reaction equations. Some state of matter or a complex system is described. Knowing some characteristics of this state, it is necessary to find others. An example would be tasks:
1.1 Calculations according to the formula of the substance, the characteristics of the portion of the substance
1.2 Calculations according to the characteristics of the composition of the mixture, solution.
Tasks are found in the Unified State Examination - No. 24. For students, the solution of such problems does not cause difficulties.
2. Tasks using one or more reaction equations. To solve them, in addition to the characteristics of substances, it is necessary to use the characteristics of processes. In the tasks of this group, the following types of tasks of increased complexity can be distinguished:
2.1 Formation of solutions.
1) What mass of sodium oxide must be dissolved in 33.8 ml of water to obtain a 4% sodium hydroxide solution.
To find:
m (Na 2 O)
Given:
V (H 2 O) = 33.8 ml
ω(NaOH) = 4%
ρ (H 2 O) \u003d 1 g / ml
M (NaOH) \u003d 40 g / mol
m (H 2 O) = 33.8 g
Na 2 O + H 2 O \u003d 2 NaOH
1 mol 2mol
Let the mass of Na 2 O = x.
n (Na 2 O) \u003d x / 62
n(NaOH) = x/31
m(NaOH) = 40x /31
m (solution) = 33.8 + x
0.04 = 40x /31 ∙ (33.8+x)
x \u003d 1.08, m (Na 2 O) \u003d 1.08 g
Answer: m (Na 2 O) \u003d 1.08 g
2) To 200 ml of sodium hydroxide solution (ρ \u003d 1.2 g / ml) with a mass fraction of alkali of 20% was added metallic sodium weighing 69 g.
What is the mass fraction of the substance in the resulting solution?
To find:
ω 2 (NaOH)
Given:
V (NaO H) solution = 200 ml
ρ (solution) = 1.2 g/ml
ω 1 (NaOH) \u003d 20%
m (Na) \u003d 69 g
M (Na) \u003d 23 g / mol
Metallic sodium interacts with water in an alkali solution.
2Na + 2H 2 O \u003d 2 NaOH + H 2
1 mol 2mol
m 1 (p-ra) = 200 ∙ 1.2 = 240 (g)
m 1 (NaOH) in-va \u003d 240 ∙ 0.2 = 48 (g)
n (Na) \u003d 69/23 \u003d 3 (mol)
n 2 (NaOH) \u003d 3 (mol)
m 2 (NaOH) \u003d 3 ∙ 40 = 120 (g)
m total (NaOH) \u003d 120 + 48 \u003d 168 (g)
n (H 2) \u003d 1.5 mol
m (H 2) \u003d 3 g
m (p-ra after p-tion) \u003d 240 + 69 - 3 \u003d 306 (g)
ω 2 (NaOH) \u003d 168 / 306 \u003d 0.55 (55%)
Answer: ω 2 (NaOH) \u003d 55%
3) What is the mass of selenium oxide (VI) should be added to 100 g of a 15% solution of selenic acid to double its mass fraction?
To find:
m (SeO 3)
Given:
m 1 (H 2 SeO 4) solution = 100 g
ω 1 (H 2 SeO 4) = 15%
ω 2 (H 2 SeO 4) = 30%
M (SeO 3) \u003d 127 g / mol
M (H 2 SeO 4) \u003d 145 g / mol
m 1 (H 2 SeO 4 ) = 15 g
SeO 3 + H 2 O \u003d H 2 SeO 4
1 mol 1mol
Let m (SeO 3) = x
n(SeO 3 ) = x/127 = 0.0079x
n 2 (H 2 SeO 4 ) = 0.0079x
m 2 (H 2 SeO 4 ) = 145 ∙ 0.079x = 1.1455x
m total . (H 2 SeO 4 ) = 1.1455x + 15
m 2 (r-ra) \u003d 100 + x
ω (NaOH) \u003d m (NaOH) / m (solution)
0.3 = (1.1455x + 1) / 100 + x
x = 17.8, m (SeO 3 ) = 17.8 g
Answer: m (SeO 3) = 17.8 g
2.2 Calculation by reaction equations when one of the substances is in excess /
1) To a solution containing 9.84 g of calcium nitrate was added a solution containing 9.84 g of sodium orthophosphate. The precipitate formed was filtered off and the filtrate was evaporated. Determine the masses of the reaction products and the composition of the dry residue in mass fractions after evaporation of the filtrate, assuming that anhydrous salts are formed.
To find:
ω (NaNO3)
ω (Na 3 PO 4)
Given:
m (Ca (NO 3) 2) \u003d 9.84 g
m (Na 3 PO 4) \u003d 9.84 g
M (Na 3 PO 4) = 164 g / mol
M (Ca (NO 3) 2) \u003d 164 g / mol
M (NaNO 3) \u003d 85 g / mol
M (Ca 3 (PO 4) 2) = 310 g / mol
2Na 3 PO 4 + 3 Сa (NO 3) 2 \u003d 6NaNO 3 + Ca 3 (PO 4) 2 ↓
2 mole 3 mole 6 mole 1 mole
n (Сa(NO 3 ) 2 ) total = n (Na 3 PO 4 ) total. = 9.84/164 =
Ca (NO 3) 2 0.06 / 3< 0,06/2 Na 3 PO 4
Na 3 PO 4 is taken in excess,
we carry out calculations for n (Сa (NO 3) 2).
n (Ca 3 (PO 4) 2) = 0.02 mol
m (Ca 3 (PO 4) 2) \u003d 310 ∙ 0.02 \u003d 6.2 (g)
n (NaNO 3) \u003d 0.12 mol
m (NaNO 3) \u003d 85 ∙ 0.12 \u003d 10.2 (g)
The composition of the filtrate includes a solution of NaNO 3 and
solution of excess Na 3 PO 4.
n proreact. (Na 3 PO 4) \u003d 0.04 mol
n rest. (Na 3 PO 4) \u003d 0.06 - 0.04 \u003d 0.02 (mol)
m rest. (Na 3 PO 4) \u003d 164 ∙ 0.02 \u003d 3.28 (g)
The dry residue contains a mixture of NaNO 3 and Na 3 PO 4 salts.
m (dry rest.) \u003d 3.28 + 10.2 \u003d 13.48 (g)
ω (NaNO 3) \u003d 10.2 / 13.48 \u003d 0.76 (76%)
ω (Na 3 PO 4) \u003d 24%
Answer: ω (NaNO 3) = 76%, ω (Na 3 PO 4) = 24%
2) How many liters of chlorine will be released if 200 ml of 35% of hydrochloric acid
(ρ \u003d 1.17 g / ml) add 26.1 g of manganese oxide (IV) ? How many grams of sodium hydroxide in a cold solution will react with this amount of chlorine?
To find:
V(Cl2)
m (NaO H)
Given:
m (MnO 2) = 26.1 g
ρ (HCl solution) = 1.17 g/ml
ω(HCl) = 35%
V (HCl) solution) = 200 ml.
M (MnO 2) \u003d 87 g / mol
M (HCl) \u003d 36.5 g / mol
M (NaOH) \u003d 40 g / mol
V (Cl 2) = 6.72 (l)
m (NaOH) = 24 (g)
MnO 2 + 4 HCl \u003d MnCl 2 + Cl 2 + 2 H 2 O
1 mol 4 mol 1 mol
2 NaO H + Cl 2 = Na Cl + Na ClO + H 2 O
2 mol 1 mol
n (MnO 2) \u003d 26.1 / 87 \u003d 0.3 (mol)
m solution (НCl) = 200 ∙ 1.17 = 234 (g)
m total (НCl) = 234 ∙ 0.35 = 81.9 (g)
n (НCl) \u003d 81.9 / 36.5 \u003d 2.24 (mol)
0,3 < 2.24 /4
HCl - in excess, calculations for n (MnO 2)
n (MnO 2) \u003d n (Cl 2) \u003d 0.3 mol
V (Cl 2) \u003d 0.3 ∙ 22.4 = 6.72 (l)
n(NaOH) = 0.6 mol
m(NaOH) = 0.6 ∙ 40 = 24 (d)
2.3 Composition of the solution obtained during the reaction.
1) In 25 ml of 25% sodium hydroxide solution (ρ \u003d 1.28 g / ml) phosphorus oxide is dissolved (V) obtained by the oxidation of 6.2 g of phosphorus. What is the composition of the salt and what is its mass fraction in solution?
To find:
ω (salts)
Given:
V (NaOH) solution = 25 ml
ω(NaOH) = 25%
m (P) = 6.2 g
ρ (NaOH) solution = 1.28 g / ml
M (NaOH) \u003d 40 g / mol
M (P) \u003d 31 g / mol
M (P 2 O 5) \u003d 142 g / mol
M (NaH 2 PO 4) \u003d 120 g / mol
4P + 5O 2 \u003d 2 P 2 O 5
4mol 2mol
6 NaO H + P 2 O 5 \u003d 2 Na 3 RO 4 + 3 H 2 O
4 NaO H + P 2 O 5 \u003d 2 Na 2 H PO 4 + H 2 O
n (P) \u003d 6.2 / 31 \u003d 0.2 (mol)
n (P 2 O 5) = 0.1 mol
m (P 2 O 5) \u003d 0.1 ∙ 142 = 14.2 (g)
m (NaO H) solution = 25 ∙ 1.28 = 32 (g)
m (NaO H) in-va \u003d 0.25 ∙ 32 = 8 (g)
n (NaO H) in-va \u003d 8/40 \u003d 0.2 (mol)
According to the quantitative ratio of NaO H and P 2 O 5
it can be concluded that the acid salt NaH 2 PO 4 is formed.
2 NaO H + P 2 O 5 + H 2 O \u003d 2 NaH 2 PO 4
2mol 1mol 2mol
0.2mol 0.1mol 0.2mol
n (NaH 2 PO 4) = 0.2 mol
m (NaH 2 PO 4) \u003d 0.2 ∙ 120 = 24 (g)
m (p-ra after p-tion) \u003d 32 + 14.2 \u003d 46.2 (g)
ω (NaH 2 PO 4) \u003d 24 / 46.2 \u003d 0 52 (52%)
Answer: ω (NaH 2 PO 4) = 52%
2) At electrolysis 2 l aqueous solution sodium sulfate with a mass fraction of salt 4%
(ρ = 1.025 g/ml) 448 l of gas (n.o.) were released on the insoluble anode. Determine the mass fraction of sodium sulfate in the solution after electrolysis.
To find:
m (Na 2 O)
Given:
V (r-ra Na 2 SO 4) \u003d 2l \u003d 2000 ml
ω (Na 2 SO 4 ) = 4%
ρ (r-ra Na 2 SO 4) \u003d 1 g / ml
M (H 2 O) \u003d 18 g / mol
V (O 2) \u003d 448 l
V M \u003d 22.4 l / mol
During the electrolysis of sodium sulfate, water decomposes, oxygen gas is released at the anode.
2 H 2 O \u003d 2 H 2 + O 2
2 mol 1mol
n (O 2) \u003d 448 / 22.4 \u003d 20 (mol)
n (H 2 O) \u003d 40 mol
m (H 2 O ) decomp. = 40 ∙ 18 = 720 (g)
m (r-ra to el-za) = 2000 ∙ 1.025 = 2050 (g)
m (Na 2 SO 4) in-va \u003d 2050 ∙ 0.04 = 82 (g)
m (solution after el-za) \u003d 2050 - 720 \u003d 1330 (g)
ω (Na 2 SO 4 ) \u003d 82 / 1330 \u003d 0.062 (6.2%)
Answer: ω (Na 2 SO 4 ) = 0.062 (6.2%)
2.4 A mixture of a known composition enters into the reaction; it is necessary to find portions of spent reagents and / or products obtained.
1) Determine the volume of the sulfur oxide gas mixture (IV) and nitrogen, which contains 20% sulfur dioxide by mass, which must be passed through 1000 g of a 4% sodium hydroxide solution so that the mass fractions of salts formed in the solution become the same.
To find:
V (gases)
Given:
m(NaOH) = 1000 g
ω(NaOH) = 4%
m (medium salt) =
m (acid salt)
M (NaOH) \u003d 40 g / mol
Answer: V (gases) = 156.8
NaO H + SO 2 = NaHSO 3 (1)
1 mole 1 mole
2NaO H + SO 2 = Na 2 SO 3 + H 2 O (2)
2 mol 1mol
m (NaOH) in-va \u003d 1000 ∙ 0.04 = 40 (g)
n(NaOH) = 40/40 = 1 (mol)
Let n 1 (NaOH) \u003d x, then n 2 (NaOH) \u003d 1 - x
n 1 (SO 2) \u003d n (NaHSO 3) \u003d x
M (NaHSO 3) \u003d 104 x n 2 (SO 2) \u003d (1 - x) / 2 \u003d 0.5 ∙ (1-x)
m (Na 2 SO 3) \u003d 0.5 ∙ (1-x) ∙ 126 \u003d 63 (1 - x)
104 x \u003d 63 (1 - x)
x = 0.38 mol
n 1 (SO 2) \u003d 0.38 mol
n 2 (SO 2 ) = 0.31 mol
n total (SO 2 ) = 0.69 mol
m total (SO 2) \u003d 0.69 ∙ 64 \u003d 44.16 (g) - this is 20% of the mass of the gas mixture. The mass of nitrogen gas is 80%.
m (N 2) \u003d 176.6 g, n 1 (N 2) \u003d 176.6 / 28 \u003d 6.31 mol
n total (gases) \u003d 0.69 + 6.31 \u003d 7 mol
V (gases) = 7 ∙ 22.4 = 156.8 (l)
2) When dissolving 2.22 g of a mixture of iron and aluminum filings in an 18.25% hydrochloric acid solution (ρ = 1.09 g/ml) 1344 ml of hydrogen (n.o.) were released. Find the percentage of each metal in the mixture and determine the volume of hydrochloric acid required to dissolve 2.22 g of the mixture.
To find:
ω(Fe)
ω(Al)
V (HCl) solution
Given:
m (mixtures) = 2.22 g
ρ (HCl solution) = 1.09 g/ml
ω(HCl) = 18.25%
M (Fe) \u003d 56 g / mol
M (Al) \u003d 27 g / mol
M (HCl) \u003d 36.5 g / mol
Answer: ω (Fe) = 75.7%,
ω(Al) = 24.3%,
V (HCl) solution) = 22 ml.
Fe + 2HCl \u003d 2 FeCl 2 + H 2
1 mol 2 mol 1 mol
2Al + 6HCl \u003d 2 AlCl 3 + 3H 2
2 mol 6 mol 3mol
n (H 2) \u003d 1.344 / 22.4 \u003d 0.06 (mol)
Let m (Al) \u003d x, then m (Fe) \u003d 2.22 - x;
n 1 (H 2) \u003d n (Fe) \u003d (2.22 - x) / 56
n (Al) \u003d x / 27
n 2 (H 2) \u003d 3x / 27 ∙ 2 = x / 18
x / 18 + (2.22 - x) / 56 \u003d 0.06
x \u003d 0.54, m (Al) \u003d 0.54 g
ω (Al) = 0.54 / 2.22 = 0.243 (24.3%)
ω(Fe) = 75.7%
n (Al) = 0.54 / 27 = 0.02 (mol)
m (Fe) \u003d 2.22 - 0.54 \u003d 1.68 (g)
n (Fe) \u003d 1.68 / 56 \u003d 0.03 (mol)
n 1 (НCl) = 0.06 mol
n(NaOH) = 0.05 mol
m solution (NaOH) = 0.05 ∙ 40/0.4 = 5 (g)
V (HCl) solution = 24 / 1.09 = 22 (ml)
3) The gas obtained by dissolving 9.6 g of copper in concentrated sulfuric acid was passed through 200 ml of potassium hydroxide solution (ρ =1 g/ml, ω (TO Oh) = 2.8%. What is the composition of the salt? Determine its mass.
To find:
m (salts)
Given:
m(Cu) = 9.6 g
V (KO H) solution = 200 ml
ω (KOH) \u003d 2.8%
ρ (H 2 O) \u003d 1 g / ml
M (Cu) \u003d 64 g / mol
M (KOH) \u003d 56 g / mol
M (KHSO 3) \u003d 120 g / mol
Answer: m (KHSO 3) = 12 g
Cu + 2H 2 SO 4 \u003d CuSO 4 + SO 2 + 2H 2 O
1 mole 1 mole
KO H + SO 2 \u003d KHSO 3
1 mole 1 mole
2 KO H + SO 2 \u003d K 2 SO 3 + H 2 O
2 mol 1mol
n (SO 2) \u003d n (Cu) \u003d 6.4 / 64 \u003d 0.1 (mol)
m (KO H) solution = 200 g
m (KO H) in-va \u003d 200 g ∙ 0.028 = 5.6 g
n (KO H) \u003d 5.6 / 56 \u003d 0.1 (mol)
According to the quantitative ratio of SO 2 and KOH, it can be concluded that the acid salt KHSO 3 is formed.
KO H + SO 2 \u003d KHSO 3
1 mol 1 mol
n (KHSO 3) = 0.1 mol
m (KHSO 3) = 0.1 ∙ 120 = 12 g
4) After 100 ml of a 12.33% solution of ferric chloride (II) (ρ =1.03g/ml) passed chlorine until the concentration of ferric chloride (III) in the solution did not become equal to the concentration of ferric chloride (II). Determine the volume of absorbed chlorine (N.O.)
To find:
V(Cl2)
Given:
V (FeCl 2) = 100 ml
ω (FeCl 2) = 12.33%
ρ (r-ra FeCl 2) \u003d 1.03 g / ml
M (FeCl 2) \u003d 127 g / mol
M (FeCl 3) \u003d 162.5 g / mol
V M \u003d 22.4 l / mol
m (FeCl 2) solution = 1.03 ∙ 100 = 103 (g)
m (FeCl 2) p-in-va \u003d 103 ∙ 0.1233 = 12.7 (g)
2FeCl 2 + Cl 2 = 2 FeCl 3
2 mol 1 mol 2 mol
Let n (FeCl 2) proreact. \u003d x, then n (FeCl 3) arr. = x;
m (FeCl 2) proreact. = 127x
m (FeCl 2) rest. = 12.7 - 127x
m (FeCl 3) arr. = 162.5x
According to the condition of the problem m (FeCl 2) rest. \u003d m (FeCl 3)
12.7 - 127x = 162.5x
x \u003d 0.044, n (FeCl 2) proreact. = 0.044 mol
n (Cl 2) \u003d 0.022 mol
V (Cl 2) \u003d 0.022 ∙ 22.4 = 0.5 (l)
Answer: V (Cl 2) \u003d 0.5 (l)
5) After calcining a mixture of magnesium and calcium carbonates, the mass of the released gas turned out to be equal to the mass of the solid residue. Determine the mass fractions of substances in the initial mixture. What volume of carbon dioxide (N.O.) can be absorbed by 40 g of this mixture, which is in the form of a suspension.
To find:
ω (MgCO 3)
ω (CaCO 3)
Given:
m (solid product) \u003d m (gas)
m ( mixtures of carbonates)=40g
M (MgO) \u003d 40 g / mol
M CaO = 56 g/mol
M (CO 2) \u003d 44 g / mol
M (MgCO 3) \u003d 84 g / mol
M (CaCO 3) \u003d 100 g / mol
1) We will carry out calculations using 1 mol of a mixture of carbonates.
MgCO 3 \u003d MgO + CO 2
1mol 1mol 1mol
CaCO 3 \u003d CaO + CO 2
1 mole 1 mole 1 mole
Let n (MgCO 3) \u003d x, then n (CaCO 3) \u003d 1 - x.
n (MgO) = x, n (CaO) = 1 - x
m(MgO) = 40x
m (СаO) = 56 ∙ (1 - x) \u003d 56 - 56x
From a mixture taken in an amount of 1 mol, carbon dioxide is formed in an amount of 1 mol.
m (CO 2) = 44.g
m (tv.prod.) = 40x + 56 - 56x = 56 - 16x
56 - 16x = 44
x = 0.75,
n (MgCO 3) = 0.75 mol
n (CaCO 3) = 0.25 mol
m (MgCO 3) \u003d 63 g
m (CaCO 3) = 25 g
m (mixtures of carbonates) = 88 g
ω (MgCO 3) \u003d 63/88 \u003d 0.716 (71.6%)
ω (CaCO 3) = 28.4%
2) A suspension of a mixture of carbonates, when carbon dioxide is passed through, turns into a mixture of hydrocarbons.
MgCO 3 + CO 2 + H 2 O \u003d Mg (HCO 3) 2 (1)
1 mole 1 mole
CaCO 3 + CO 2 + H 2 O \u003d Ca (HCO 3) 2 (2)
1 mol 1mol
m (MgCO 3) \u003d 40 ∙ 0.75 = 28.64(g)
n 1 (CO 2) \u003d n (MgCO 3) \u003d 28.64 / 84 \u003d 0.341 (mol)
m (CaCO 3) = 11.36 g
n 2 (CO 2) \u003d n (CaCO 3) \u003d 11.36 / 100 \u003d 0.1136 mol
n total (CO 2) \u003d 0.4546 mol
V (CO 2) = n total (CO2) ∙ V M = 0.4546 ∙ 22.4 = 10.18 (l)
Answer: ω (MgCO 3) = 71.6%, ω (CaCO 3) = 28.4%,
V (CO 2 ) \u003d 10.18 liters.
6) A mixture of powders of aluminum and copper weighing 2.46 g was heated in a stream of oxygen. The resulting solid was dissolved in 15 ml of a sulfuric acid solution (acid mass fraction 39.2%, density 1.33 g/ml). The mixture completely dissolved without evolution of gas. To neutralize the excess acid, 21 ml of sodium bicarbonate solution with a concentration of 1.9 mol/l was required. Calculate the mass fractions of metals in the mixture and the volume of oxygen (N.O.) that reacted.
To find:
ω(Al); ω(Cu)
V(O2)
Given:
m (mixes) = 2.46 g
V (NaHCO 3 ) = 21 ml =
0.021 l
V (H 2 SO 4 ) = 15 ml
ω(H 2 SO 4 ) = 39.2%
ρ (H 2 SO 4 ) \u003d 1.33 g / ml
C (NaHCO 3) \u003d 1.9 mol / l
M (Al) \u003d 27 g / mol
М(Cu)=64 g/mol
M (H 2 SO 4) \u003d 98 g / mol
V m \u003d 22.4 l / mol
Answer: ω (Al ) = 21.95%;
ω ( Cu) = 78.05%;
V (O 2) = 0,672
4Al + 3O 2 = 2Al 2 O 3
4mol 3mol 2mol
2Cu + O 2 = 2CuO
2mol 1mol 2mol
Al 2 O 3 + 3H 2 SO 4 = Al 2 (SO 4 ) 3 + 3H 2 O(1)
1 mole 3 mole
CuO + H 2 SO 4 = CuSO 4 + H 2 O(2)
1 mole 1 mole
2 NaHCO 3 + H 2 SO 4 = Na 2 SO 4 + 2H 2 O+ SO 2 (3)
2 mol 1 mol
m (H 2 SO 4) solution = 15 ∙ 1.33 = 19.95 (g)
m (H 2 SO 4) in-va = 19.95 ∙ 0.393 = 7.8204 (g)
n ( H 2 SO 4) total = 7.8204/98 = 0.0798 (mol)
n (NaHCO 3) = 1,9 ∙ 0.021 = 0.0399 (mol)
n 3 (H 2 SO 4 ) = 0,01995 ( mole )
n 1+2 (H 2 SO 4 ) =0,0798 – 0,01995 = 0,05985 ( mole )
4) Let be n (Al) = x, . m(Al) = 27x
n (Cu) = y, m (Cu) = 64y
27x + 64y = 2.46
n(Al 2 O 3 ) = 1.5x
n(CuO) = y
1.5x + y = 0.0585
x = 0.02; n(Al) = 0.02 mole
27x + 64y = 2.46
y=0.03; n(Cu)=0.03 mole
m(Al) = 0.02∙ 27 = 0,54
ω (Al) = 0.54 / 2.46 = 0.2195 (21.95%)
ω (Cu) = 78.05%
n 1 (O 2 ) = 0.015 mole
n 2 (O 2 ) = 0.015 mole
n common . (O 2 ) = 0.03 mole
V(O 2 ) = 22,4 ∙ 0 03 = 0,672 ( l )
7) When dissolving 15.4 g of an alloy of potassium with sodium in water, 6.72 liters of hydrogen (n.o.) were released. Determine the molar ratio of metals in the alloy.
To find:
n (K) : n( Na)
m (Na 2 O)
Given:
m(alloy) = 15.4 g
V (H 2) = 6.72 l
M ( Na) =23 g/mol
M (K) \u003d 39 g/mol
n (K) : n ( Na) = 1: 5
2K + 2 H 2 O= 2 K Oh+ H 2
2 mol 1 mol
2Na + 2H 2 O = 2 NaOH+ H 2
2 mol 1 mol
Let n(K) = x, n ( Na) = y, then
n 1 (H 2) = 0.5 x; n 2 (H 2) \u003d 0.5y
n (H 2) \u003d 6.72 / 22.4 \u003d 0.3 (mol)
m(K) = 39 x; m (Na) = 23 y
39x + 23y = 15.4
x = 0.1, n(K) = 0.1 mol;
0.5x + 0.5y = 0.3
y = 0.5, n ( Na) = 0.5 mol
8) When processing 9 g of a mixture of aluminum with aluminum oxide with a 40% sodium hydroxide solution (ρ \u003d 1.4 g / ml) 3.36 l of gas (n.o.) were released. Determine the mass fractions of substances in the initial mixture and the volume of the alkali solution that entered into the reaction.
To find:
ω (Al)
ω (Al 2 O 3)
V r-ra ( NaOH)
Given:
M(see) = 9 g
V(H 2) = 33.8ml
ω (NaOH) = 40%
M( Al) = 27 g/mol
M( Al 2 O 3) = 102 g/mol
M( NaOH) = 40 g/mol
2Al + 2NaOH + 6H 2 O = 2Na + 3H 2
2 mole 2 mole 3 mole
Al 2 O 3 + 2NaOH + 3H 2 O = 2 Na
1mol 2mol
n ( H 2) \u003d 3.36 / 22.4 \u003d 0.15 (mol)
n ( Al) = 0.1 mol m (Al) = 2.7 g
ω (Al) = 2.7 / 9 = 0.3 (30%)
ω(Al 2 O 3 ) = 70%
m (Al 2 O 3 ) = 9 – 2.7 = 6.3 ( G )
n(Al 2 O 3 ) = 6,3 / 102 = 0,06 ( mole )
n 1 (NaOH) = 0.1 mole
n 2 (NaOH) = 0.12 mole
n common . (NaOH) = 0.22 mole
m R - ra (NaOH) = 0.22∙ 40 /0.4 = 22 ( G )
V R - ra (NaOH) = 22 / 1.4 = 16 ( ml )
Answer : ω(Al) = 30%, ω(Al 2 O 3 ) = 70%, V R - ra (NaOH) = 16 ml
9) An alloy of aluminum and copper weighing 2 g was treated with a solution of sodium hydroxide, with a mass fraction of alkali 40% (ρ =1.4 g/ml). The undissolved precipitate was filtered off, washed, and treated with nitric acid solution. The resulting mixture was evaporated to dryness, the residue was calcined. The mass of the resulting product was 0.8 g. Determine the mass fraction of metals in the alloy and the volume of the spent sodium hydroxide solution.
To find:
ω (Cu); ω (Al)
V r-ra ( NaOH)
Given:
m(mixture)=2 g
ω (NaOH)=40%
M( Al)=27 g/mol
M( Cu)=64 g/mol
M( NaOH)=40 g/mol
Alkali dissolves only aluminum.
2Al + 2NaOH + 6H 2 O = 2 Na + 3 H 2
2mol 2mol 3mol
Copper is an undissolved residue.
3Cu + 8HNO 3 = 3Cu(NO 3 ) 2 +4H 2 O + 2 NO
3 mole 3 mole
2Cu(NO 3 ) 2 = 2 CuO + 4NO 2 + O 2
2mol 2mol
n (CuO) = 0.8 / 80 = 0.01 (mol)
n (CuO) = n (Cu(NO 3 ) 2 ) = n(Cu) = 0.1 mole
m(Cu) = 0.64 G
ω (Cu) = 0.64 / 2 = 0.32 (32%)
ω(Al) = 68%
m(Al) = 9 - 0.64 = 1.36(g)
n ( Al) = 1.36 / 27 = 0.05 (mol)
n ( NaOH) = 0.05 mol
m r-ra ( NaOH) = 0,05 ∙ 40 / 0.4 = 5 (g)
V r-ra ( NaOH) = 5 / 1.43 = 3.5 (ml)
Answer: ω (Cu) = 32%, ω (Al) = 68%, V r-ra ( NaOH) = 3.5 ml
10) A mixture of potassium, copper and silver nitrates was calcined, weighing 18.36 g. The volume of released gases was 4.32 l (n.o.). The solid residue was treated with water, after which its mass decreased by 3.4 g. Find the mass fractions of nitrates in the initial mixture.
To find:
ω (KNO 3 )
ω (Cu(NO 3 ) 2 )
ω (AgNO 3)
Given:
m(blends) = 18.36 g
∆m(hard. rest.)=3.4 g
V (CO 2) = 4.32 l
M(K NO 2) \u003d 85 g / mol
M(K NO 3) =101 g/mol
2 K NO 3 = 2 K NO 2 + O 2 (1)
2 mol 2 mol 1mol
2 Cu(NO 3 ) 2 = 2 CuO + 4 NO 2 + O 2 (2)
2 mol 2 mol 4 mol 1 mol
2 AgNO 3 = 2 Ag + 2 NO 2 + O 2 (3)
2 mol 2 mol 2 mol 1 mol
CuO + 2H 2 O= interaction not possible
Ag+ 2H 2 O= interaction not possible
To NO 2 + 2H 2 O= salt dissolution
The change in the mass of the solid residue occurred due to the dissolution of the salt, therefore:
m(TO NO 2) = 3.4 g
n(K NO 2) = 3.4 / 85 = 0.04 (mol)
n(K NO 3) = 0.04 (mol)
m(TO NO 3) = 0,04∙ 101 = 4.04 (g)
ω (KNO 3) = 4,04 / 18,36 = 0,22 (22%)
n 1 (O 2) = 0.02 (mol)
n total (gases) = 4.32 / 22.4 = 0.19 (mol)
n 2+3 (gases) = 0.17 (mol)
m(mixtures without K NO 3) \u003d 18.36 - 4.04 \u003d 14.32 (g)
Let be m (Cu(NO 3 ) 2 ) = x, then m (AgNO 3 ) = 14.32 – x.
n (Cu(NO 3 ) 2 ) = x / 188,
n (AgNO 3) = (14,32 – x) / 170
n 2 (gases) = 2.5x / 188,
n 3 (gases) = 1.5 ∙ (14.32 - x) / 170,
2.5x/188 + 1.5 ∙ (14.32 - x) / 170 \u003d 0.17
X = 9.75, m (Cu(NO 3 ) 2 ) = 9,75 G
ω (Cu(NO 3 ) 2 ) = 9,75 / 18,36 = 0,531 (53,1%)
ω (AgNO 3 ) = 24,09%
Answer : ω (KNO 3 ) = 22%, ω (Cu(NO 3 ) 2 ) = 53.1%, ω (AgNO 3 ) = 24,09%.
11) A mixture of barium hydroxide, calcium and magnesium carbonates weighing 3.05 g was calcined to remove volatile substances. The mass of the solid residue was 2.21 g. Volatile products were brought to normal conditions and the gas was passed through a solution of potassium hydroxide, the mass of which increased by 0.66 g. Find the mass fractions of substances in the initial mixture.
ω (AT a(O H) 2)
ω (WITH a With O 3)
ω (mg With O 3)
m(mixture) = 3.05 g
m(solid rest) = 2.21 g
∆ m(KOH) = 0.66 g
M ( H 2 O) =18 g/mol
M (CO 2) \u003d 44 g / mol
M (B a(O H) 2) \u003d 171 g / mol
M (CaCO 2) \u003d 100 g / mol
M ( mg CO 2) \u003d 84 g / mol
AT a(O H) 2 = H 2 O+ V aO
1 mol 1mol
With a With O 3 \u003d CO 2 + C aO
1 mol 1mol
mg With O 3 \u003d CO 2 + MgO
1 mol 1mol
The mass of KOH increased due to the mass of absorbed CO 2
KOH + CO 2 →…
According to the law of conservation of mass of substances
m (H 2 O) \u003d 3.05 - 2.21 - 0.66 \u003d 0.18 g
n ( H 2 O) = 0.01 mol
n (B a(O H) 2) = 0.01 mol
m(AT a(O H) 2) = 1.71 g
ω (AT a(O H) 2) = 1.71 / 3.05 = 0.56 (56%)
m(carbonates) = 3.05 - 1.71 = 1.34 g
Let be m(WITH a With O 3) = x, then m(WITH a With O 3) = 1,34 – x
n 1 (C O 2) = n (C a With O 3) = x /100
n 2 (C O 2) = n ( mg With O 3) = (1,34 - x)/84
x /100 + (1,34 - x)/84 = 0,015
x = 0,05, m(WITH a With O 3) = 0.05 g
ω (WITH a With O 3) = 0,05/3,05 = 0,16 (16%)
ω (mg With O 3) =28%
Answer: ω (AT a(O H) 2) = 56%, ω (WITH a With O 3) = 16%, ω (mg With O 3) =28%
2.5 An unknown substance enters the reaction o / is formed during the reaction.
1) When a hydrogen compound of a monovalent metal interacted with 100 g of water, a solution was obtained with a mass fraction of a substance of 2.38%. The mass of the solution turned out to be 0.2 g less than the sum of the masses of water and the initial hydrogen compound. Determine which connection was taken.
To find:
Given:
m (H 2 O) = 100 g
ω (Me Oh) = 2,38%
∆ m(solution) = 0.2 g
M ( H 2 O) = 18 g/mol
Men + H 2 O= Me Oh+ H 2
1 mole 1 mole 1 mole
0.1 mol 0.1 mol 0.1 mol
The mass of the final solution decreased by the mass of hydrogen gas.
n (H 2) \u003d 0.2 / 2 \u003d 0.1 (mol)
n ( H 2 O) proreact. = 0.1 mol
m (H 2 O) proreag = 1.8 g
m (H 2 O in solution) = 100 - 1.8 = 98.2 (g)
ω (Me Oh) = m(Me Oh) / m(r-ra g/mol
Let be m(Me Oh) = x
0.0238 = x / (98.2 + x)
x = 2,4, m(Me O H) = 2.4 g
n(Me O H) = 0.1 mol
M (Me O H) \u003d 2.4 / 0.1 \u003d 24 (g / mol)
M (Me) = 7 g/mol
Me - Li
Answer: Li N.
2) When dissolving 260 g of an unknown metal in highly dilute nitric acid two salts are formed: Me(NO 3 ) 2 andX. When heatedXwith calcium hydroxide, gas is released, which with phosphoric acid forms 66 g of ammonium hydroorthophosphate. Determine the metal and salt formulaX.
To find:
Given:
m(Me) = 260 g
m ((NH 4) 2 HPO 4) = 66 g
M (( NH 4) 2 HPO 4) =132 g/mol
Answer: Zn, salt - NH 4 NO 3.
4Me + 10HNO 3 = 4Me(NO 3 ) 2 +NH 4 NO 3 + 3H 2 O
4 mole 1 mole
2NH 4 NO 3 +Ca(OH) 2 = Ca(NO 3 ) 2 +2NH 3 + 2H 2 O
2 mole 2 mole
2NH 3 + H 3 PO 4 = (NH 4 ) 2 HPO 4
2 mol 1mol
n ((NH 4) 2 HPO 4) = 66/132 = 0.5 (mol)
n (N H 3) = n (NH 4 NO 3) = 1 mol
n (Me) = 4mol
M (Me) = 260/4 = 65 g/mol
Me - Zn
3) In 198.2 ml of aluminum sulfate solution (ρ = 1 g/ml) lowered a plate of an unknown divalent metal. After some time, the mass of the plate decreased by 1.8 g, and the concentration of the formed salt was 18%. Define metal.
To find:
ω 2 (NaOH)
Given:
V solution = 198.2 ml
ρ (solution) = 1 g/ml
ω 1 (salt) = 18%
∆m(p-ra) \u003d 1.8 g
M ( Al) =27 g/mol
Al 2 (SO 4 ) 3 + 3Me = 2Al+ 3MeSO 4
3 mole 2 mole 3 mole
m(r-ra to r-tion) = 198.2 (g)
m(p-ra after p-tion) \u003d 198.2 + 1.8 \u003d 200 (g)
m (MeSO 4) in-va \u003d 200 ∙ 0.18 = 36 (g)
Let M (Me) = x, then M ( MeSO 4) = x + 96
n ( MeSO 4) = 36 / (x + 96)
n (Me) \u003d 36 / (x + 96)
m(Me) = 36 x/ (x + 96)
n ( Al) = 24 / (x + 96),
m (Al) = 24 ∙ 27/(x+96)
m(Me) ─ m (Al) = ∆m(r-ra)
36x/ (x + 96) ─ 24 ∙ 27 / (x + 96) = 1.8
x \u003d 24, M (Me) \u003d 24 g / mol
Metal - mg
Answer: mg.
4) During thermal decomposition of 6.4 g of salt in a vessel with a capacity of 1 l at 300.3 0 With a pressure of 1430 kPa. Determine the formula of salt if, during its decomposition, water and a gas poorly soluble in it are formed.
To find:
salt formula
Given:
m(salt) = 6.4 g
V(vessel) = 1 l
P = 1430 kPa
t=300.3 0 C
R= 8.31J/mol ∙ To
n (gas) = PV/RT = 1430∙1 / 8,31∙ 573.3 = 0.3 (mol)
The condition of the problem corresponds to two equations:
NH 4 NO 2 = N 2 + 2 H 2 O ( gas)
1 mol 3 mol
NH 4 NO 3 = N 2 O + 2 H 2 O (gas)
1 mol 3 mol
n (salts) = 0.1 mol
M (salt) \u003d 6.4 / 0.1 \u003d 64 g / mol ( NH 4 NO 2)
Answer: NH 4 N
Literature.
1. N.E. Kuzmenko, V.V. Eremin, A.V. Popkov "Chemistry for high school students and university applicants", Moscow, "Drofa" 1999
2. G.P. Khomchenko, I.G. Khomchenko "Collection of problems in chemistry", Moscow "New Wave * Onyx" 2000
3. K.N. Zelenin, V.P. Sergutina, O.V., O.V. Solod "Manual in chemistry for those entering the Military - medical academy and other higher medical educational establishments»,
St. Petersburg, 1999
4. A guide for applicants to medical institutes "Problems in chemistry with solutions",
St. Petersburg Medical Institute named after I.P. Pavlov
5. FIPI "USE CHEMISTRY" 2009 - 2015