Very difficult tasks 3 chemistry exam. Tasks C3 Unified State Exam in Chemistry

We discussed the general algorithm for solving problem No. 35 (C5). It's time to look at specific examples and offer you a selection of problems to solve on your own.

Example 2. The complete hydrogenation of 5.4 g of some alkyne requires 4.48 liters of hydrogen (n.s.). Determine the molecular formula of this alkyne.

Solution. We will act in accordance with the general plan. Let a molecule of an unknown alkyne contain n carbon atoms. General formula of the homologous series C n H 2n-2. Hydrogenation of alkynes proceeds according to the equation:

C n H 2n-2 + 2H 2 = C n H 2n+2.

The amount of hydrogen that reacted can be found using the formula n = V/Vm. In this case, n = 4.48/22.4 = 0.2 mol.

The equation shows that 1 mole of alkyne adds 2 moles of hydrogen (recall that in the problem statement we are talking about complete hydrogenation), therefore, n(C n H 2n-2) = 0.1 mol.

Based on the mass and amount of the alkyne, we find its molar mass: M(C n H 2n-2) = m(mass)/n(amount) = 5.4/0.1 = 54 (g/mol).

The relative molecular weight of an alkyne is the sum of n atomic masses of carbon and 2n-2 atomic masses of hydrogen. We get the equation:

12n + 2n - 2 = 54.

We solve the linear equation, we get: n = 4. Alkyne formula: C 4 H 6.

Answer: C 4 H 6 .

I would like to draw attention to one significant point: the molecular formula C 4 H 6 corresponds to several isomers, including two alkynes (butyn-1 and butyn-2). Based on these problems, we will not be able to unambiguously establish the structural formula of the substance under study. However, in this case this is not required!

Example 3. When 112 liters (n.a.) of an unknown cycloalkane are burned in excess oxygen, 336 liters of CO 2 are formed. Establish the structural formula of the cycloalkane.

Solution. The general formula of the homologous series of cycloalkanes: C n H 2n. With complete combustion of cycloalkanes, as with the combustion of any hydrocarbons, carbon dioxide and water are formed:

C n H 2n + 1.5n O 2 = n CO 2 + n H 2 O.

Please note: the coefficients in the reaction equation in this case depend on n!

During the reaction, 336/22.4 = 15 moles of carbon dioxide were formed. 112/22.4 = 5 moles of hydrocarbon entered the reaction.

Further reasoning is obvious: if 15 moles of CO 2 are formed per 5 moles of cycloalkane, then 15 molecules of carbon dioxide are formed per 5 molecules of hydrocarbon, i.e., one cycloalkane molecule produces 3 CO 2 molecules. Since each molecule of carbon monoxide (IV) contains one carbon atom, we can conclude: one cycloalkane molecule contains 3 carbon atoms.

Conclusion: n = 3, cycloalkane formula - C 3 H 6.

As you can see, the solution to this problem does not “fit” into the general algorithm. We did not look for the molar mass of the compound here, nor did we create any equation. According to formal criteria, this example is not similar to the standard problem C5. But I already emphasized above that it is important not to memorize the algorithm, but to understand the MEANING of the actions being performed. If you understand the meaning, you yourself will be able to make changes to the general scheme at the Unified State Exam and choose the most rational solution.

There is one more “oddity” in this example: it is necessary to find not only the molecular, but also the structural formula of the compound. In the previous task we were not able to do this, but in this example - please! The fact is that the formula C 3 H 6 corresponds to only one isomer - cyclopropane.

Answer: cyclopropane.

Example 4. 116 g of some saturated aldehyde were heated for a long time with an ammonia solution of silver oxide. The reaction produced 432 g of metallic silver. Determine the molecular formula of the aldehyde.

Solution. The general formula of the homologous series of saturated aldehydes is: C n H 2n+1 COH. Aldehydes are easily oxidized to carboxylic acids, in particular, under the action of an ammonia solution of silver oxide:

C n H 2n+1 COH + Ag 2 O = C n H 2n+1 COOH + 2 Ag.

Note. In reality, the reaction is described by a more complex equation. When Ag 2 O is added to an aqueous ammonia solution, a complex compound OH is formed - diammine silver hydroxide. It is this compound that acts as an oxidizing agent. During the reaction, an ammonium salt of a carboxylic acid is formed:

C n H 2n+1 COH + 2OH = C n H 2n+1 COONH 4 + 2Ag + 3NH 3 + H 2 O.

Another important point! The oxidation of formaldehyde (HCOH) is not described by the given equation. When HCOH reacts with an ammonia solution of silver oxide, 4 moles of Ag per 1 mole of aldehyde are released:

НCOH + 2Ag2O = CO2 + H2O + 4Ag.

Be careful when solving problems involving the oxidation of carbonyl compounds!

Let's return to our example. Based on the mass of released silver, you can find the amount of this metal: n(Ag) = m/M = 432/108 = 4 (mol). According to the equation, 2 moles of silver are formed per 1 mole of aldehyde, therefore, n(aldehyde) = 0.5n(Ag) = 0.5*4 = 2 moles.

Molar mass of aldehyde = 116/2 = 58 g/mol. Try to do the next steps yourself: you need to create an equation, solve it and draw conclusions.

Answer: C 2 H 5 COH.

Example 5. When 3.1 g of a certain primary amine reacts with a sufficient amount of HBr, 11.2 g of salt is formed. Determine the formula of the amine.

Solution. Primary amines (C n H 2n + 1 NH 2) when interacting with acids form alkylammonium salts:

С n H 2n+1 NH 2 + HBr = [С n H 2n+1 NH 3 ] + Br - .

Unfortunately, based on the mass of the amine and the salt formed, we will not be able to find their quantities (since the molar masses are unknown). Let's take a different path. Let us remember the law of conservation of mass: m(amine) + m(HBr) = m(salt), therefore, m(HBr) = m(salt) - m(amine) = 11.2 - 3.1 = 8.1.

Pay attention to this technique, which is very often used when solving C 5. Even if the mass of the reagent is not given explicitly in the problem statement, you can try to find it from the masses of other compounds.

So, we are back on track with the standard algorithm. Based on the mass of hydrogen bromide, we find the amount, n(HBr) = n(amine), M(amine) = 31 g/mol.

Answer: CH 3 NH 2 .

Example 6. A certain amount of alkene X, when reacting with an excess of chlorine, forms 11.3 g of dichloride, and when reacting with an excess of bromine, 20.2 g of dibromide. Determine the molecular formula of X.

Solution. Alkenes add chlorine and bromine to form dihalogen derivatives:

C n H 2n + Cl 2 = C n H 2n Cl 2,

C n H 2n + Br 2 = C n H 2n Br 2.

In this problem it is pointless to try to find the amount of dichloride or dibromide (their molar masses are unknown) or the amount of chlorine or bromine (their masses are unknown).

We use one non-standard technique. The molar mass of C n H 2n Cl 2 is 12n + 2n + 71 = 14n + 71. M(C n H 2n Br 2) = 14n + 160.

The masses of dihalides are also known. You can find the amounts of substances obtained: n(C n H 2n Cl 2) = m/M = 11.3/(14n + 71). n(C n H 2n Br 2) = 20.2/(14n + 160).

By convention, the amount of dichloride is equal to the amount of dibromide. This fact allows us to create the equation: 11.3/(14n + 71) = 20.2/(14n + 160).

This equation has a unique solution: n = 3.

Answer: C 3 H 6

In the final part, I offer you a selection of C5 type problems of varying difficulty. Try to solve them yourself - it will be excellent training before taking the Unified State Exam in Chemistry!

To practice, take the VPR test 2017 chemistry grade 11 Option 3

Instructions for performing the work

The test includes 15 tasks. 1 hour 30 minutes (90 minutes) is allotted to complete the chemistry work.

Formulate your answers in the text of the work according to the instructions for the assignments. If you write down an incorrect answer, cross it out and write a new one next to it.

When performing work, you are allowed to use the following additional materials:

– Periodic table of chemical elements D.I. Mendeleev;

– table of solubility of salts, acids and bases in water;

– electrochemical series of metal voltages;

– non-programmable calculator.

When completing assignments, you can use a draft. Entries in draft will not be reviewed or graded.

We advise you to complete the tasks in the order in which they are given. To save time, skip a task that you cannot complete immediately and move on to the next one. If you have time left after completing all the work, you can return to the missed tasks.

The points you receive for completed tasks are summed up. Try to complete as many tasks as possible and score the most points.

We wish you success!

Name

№1 From your chemistry course, you know the following methods for separating mixtures: sedimentation, filtration, distillation (distillation), magnetic action, evaporation, crystallization. In Fig. 1 and 2 show two of these methods.
From the mixtures listed below, select those that can be separated using the methods shown in the pictures.

A) river sand and sawdust.

B) water and olive oil.

B) granulated sugar and starch.

D) chalk and water

Write down the names of the selected mixtures, the number of the corresponding figure and the name of the method of separating the mixture.

Water and olive oil
Figure number

Mixture separation method

Mixture

Chalk and water

Figure number

Mixture separation method

№2 The figure shows a diagram of the distribution of electrons across energy levels in an atom of a certain chemical element.
Based on the proposed scheme, complete the following tasks:

1) write down the symbol of the chemical element to which this atomic structure diagram corresponds;

2) write down the period number and group number in the Periodic Table of Chemical Elements D.I. Mendeleev, in which this element is located;
Enter the period number in your answer.

Enter the group number in your answer.

3) determine whether the simple substance that forms this element is a metal or non-metal.

Write your answer as metal or non-metal.

№3 Periodic table of chemical elements D.I. Mendeleev is a rich repository of information about chemical elements, their properties and the properties of their compounds. For example, it is known that with an increase in the atomic number of a chemical element, the radii of atoms in periods decrease and in groups increase. Considering these patterns, arrange the following elements in order of decreasing atomic radius: B, Al, C, N. Write the symbols of the elements in the desired sequence.

For example Be; Li; Mg; Na

№4 The table below shows some characteristics of covalent and ionic types of chemical bonds.

Using this information, determine the type of chemical bond in the molecules:

1) sulfur(IV) oxide (SO2);

2) hydrogen sulfide (H2S).

1) In the sulfur oxide(IV) molecule_____________________

2) In the hydrogen sulfide molecule______________________

№5 Complex inorganic substances can be conditionally distributed, that is, classified, into four classes, as shown in the diagram. In this diagram, enter the missing names of two classes and two formulas of substances that are representatives of the corresponding classes.

The assignment of this part is checked separately by experts. The answers may be different, for example Acids: HNO3

Fill in the name of the missing group for P2O5

The assignment of this part is checked separately by experts. The answers may be different, for example Salts: Na3PO4

In response, fill in the missing group name for KOH

To complete tasks 6–8, use the information contained in this text.

Aluminum is the third most abundant element in the earth's crust. Aluminum-based alloys are produced that have high strength and are relatively cheap to produce. Pots, pans, baking sheets, ladles and other household utensils are made from these alloys. Aluminum cookware conducts heat well, heats up very quickly, and is easy to clean. Meat is baked in the oven and pies are baked on aluminum foil; Butter and margarine, cheese, chocolate and candy are packaged in aluminum foil. Metallic aluminum is a reactive metal, but is resistant to corrosion, since when interacting with oxygen in the air, a thin layer of aluminum oxide (Al2O3) is formed on its surface, which has greater strength. If the oxide film is removed, then aluminum easily enters into chemical reactions with non-metals, for example with halogens.
The most common natural compounds of aluminum are its oxide and hydroxide. These compounds have amphoteric properties, that is, they can exhibit both basic and acidic properties, depending on the nature of the substance that reacts with them. Due to its ability to neutralize acid, aluminum hydroxide (Al(OH)3) is used in medicine in the manufacture of medicines for ulcers and heartburn. In the laboratory, aluminum hydroxide can be obtained by the action of alkalis (without excess) on solutions of aluminum salts.
№6

1) Write an equation for the reaction of aluminum with chlorine.

2) What determines the corrosion resistance of aluminum metal?

№7

1) Write a molecular equation for the reaction between aluminum hydroxide and nitric acid (HNO3).

2) Indicate what properties (basic or acidic) aluminum hydroxide exhibits in this reaction.

Write down your answer, and after testing, look at the answers with a detailed solution.

№8

1) Write a shortened ionic equation for the reaction between solutions of aluminum chloride (AlCl3) and potassium hydroxide (without excess).
2) Explain why there should not be an excess of alkali in this reaction to obtain aluminum hydroxide.

Write down your answer, and after testing, look at the answers with a detailed solution.

№9 A scheme of the redox reaction is given.

H2S + Cl2 + H2O → H2SO4 +HCl

1. Make an electronic balance for this reaction.

2. Identify the oxidizing agent and reducing agent.

3. Arrange the coefficients in the reaction equation.
Write down your answer, and after testing, look at the answers with a detailed solution.

№10 The transformation scheme is given:

CuO → CuCl2 → Cu(NO3)2 → Cu(OH)2

Write molecular reaction equations that can be used to carry out these transformations.

Write down your answer, and after testing, look at the answers with a detailed solution.

№11 Establish a correspondence between the formula of an organic substance and the class/group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number.

FORMULA OF THE SUBSTANCE

CLASS/GROUP

1) carboxylic acids

2) hydrocarbons

3) aldehydes

4) alcohols

Write down the selected numbers under the corresponding letters.

Tasks C3 are devoted to reactions that confirm the relationship between various classes of hydrocarbons and oxygen-containing organic compounds. They represent a chain of five stages of transformation of organic substances, and are assessed at 5 primary points. Let's look at examples of the most difficult chains of 2004-2009 (in brackets - success rate in percentage for students in the Tyumen region, first wave)

C3 (2004, 11%)

Acetaldehyde ® potassium acetate ® ethanoic acid ® ethyl acetate ® calcium acetate ® acetone

The fact that this chain contains not formulas, but names of substances, also probably led to the fact that it turned out to be the most difficult for students. Let's rewrite:

CH 3 CHO ® CH 3 COOK ® CH 3 COOH ® CH 3 COOC 2 H 5 ® (CH 3 COO) 2 Ca ® (CH 3) 2 CO

The type of reaction can be suggested by a comparison of the composition of the starting and resulting substances. So, for the first transformation it is clear that it is necessary to oxidize the aldehyde in an alkaline medium, for example:

CH 3 CHO + 2KMnO 4 + 3KOH ® CH 3 COOK + 2K 2 MnO 4 + 2H 2 O

Half-reaction equations for arranging coefficients:

CH 3 CHO + 3OH – – 2ē = CH 3 COO – + 2H 2 O |1

MnO 4 – + ē = MnO 4 2– |2

The following two reactions should not be difficult:

CH 3 COOK + HCl = CH 3 COOH + KCl

CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O

To obtain acetate from ether, it is necessary to hydrolyze it in an alkaline medium, using calcium hydroxide as the alkali:

2CH 3 COOC 2 H 5 + Ca(OH) 2 (CH 3 COO) 2 Ca + 2C 2 H 5 OH

The latter transformation could be particularly difficult, since methods for producing ketones are not usually covered in a basic chemistry course. To carry it out, pyrolysis (thermal decomposition) of calcium acetate is carried out:

(CH 3 COO) 2 Ca (CH 3) 2 CO + CaCO 3

The most difficult of the tasks 2005 chains turned out to include electrolysis of salt solutions, for example:

C3 (2005, 8%) Give reaction equations that can be used to carry out the following transformations

Potassium acetate X 1 X 2 X3®

X 4 X 5

Electrolysis of potassium acetate solution:

K(-) (K +) – not reduced, alkali metal

2H 2 O + 2ē = H 2 + 2OH – | 2

A(+) 2CH 3 COO – –2ē = CH 3 -CH 3 + 2CO 2 | 2

Summary equation:

2CH 3 COO – + 2H 2 O = H 2 + 2OH – + CH 3 -CH 3 + 2CO 2

Or 2CH 3 COOK + 2H 2 O = H 2 + 2KOH + CH 3 -CH 3 + 2CO 2

When ethane is heated in the presence of a Ni, Pt catalyst, dehydrogenation occurs, X 2 - ethene: CH 3 -CH 3 ® CH 2 =CH 2 + H 2

The next stage is ethene hydration:

CH 2 =CH 2 + H 2 O ® CH 3 -CH 2 OH; X 3 – ethanol

Potassium permanganate in an acidic environment is a strong oxidizing agent and oxidizes alcohols to carboxylic acids, X 4 - acetic acid:

5C 2 H 5 OH + 4KMnO 4 + 6H 2 SO 4 = 5CH 3 COOH + 4MnSO 4 + 2K 2 SO 4 + 11H 2 O

Finally, the interaction of acetic acid (X 4) and alcohol (X 3) will lead to the formation of an ester, X 5 - ethyl acetate:

CH 3 COOH + C 2 H 5 OH = CH 3 COOC 2 H 5 + H 2 O

The complexity of this chain also lies in the fact that if you do not know the first reaction, it is impossible to understand what substances we are talking about in the rest of it.

Let us consider a number of other transformations that caused difficulties for schoolchildren during the 2005 exam.

Decomposition of oxalic and formic acid under the action of concentrated sulfuric acid:

H2C2O4 H 2 O + CO 2 + CO

HCOOH H2O+CO

Aldehyde oxidation:

CH 3 CHO X

Here we must recall the material of inorganic chemistry, the oxidizing properties of bromine. The aldehyde is oxidized to a carboxylic acid, and since the reaction occurs in the presence of NaOH, the reaction product will be a salt:

CH 3 CHO + Br 2 + 3NaOH ® CH 3 COONa + 2NaBr + 2H 2 O

Oxidation of aldehydes with an ammonia solution of silver oxide.

HCHO X

Textbooks usually write that it leads to the formation of carboxylic acids. In fact, since the reaction occurs in the presence of excess ammonia, the corresponding ammonium salts are formed. In this case, it is necessary to take into account that formic acid and its salts can oxidize further, to carbonic acid salts:

HCHO + 2Ag 2 O + 2NH 3 ® (NH 4) 2 CO 3 + 4Ag, or more precisely:

HCHO + 4OH ® (NH 4) 2 CO 3 + 4Ag + 2H 2 O + 6NH 3

For independent consideration, chains of transformations that caused the greatest difficulties in the exam are offered. Give reaction equations that can be used to carry out the following transformations:

1. potassium methoxide X 1 ® bromomethane X 2 X 3 ethanal
Here we need to figure out what “potassium methylate” is, but the last stage turned out to be the most difficult, since such a reaction is not discussed in most school textbooks.

2. CH 3 CHO X 1 X 2 ® ethylene ® CH 3 CHO X 3

3. potassium ® potassium ethylate X 1 CH 2 =CH 2 X 2 X 3