Olympiad tasks in chemistry, municipal stage. Municipal stage of the All-Russian Chemistry Olympiad for schoolchildren (3) - Document

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1 1 ALL-RUSSIAN OLYMPIAD FOR SCHOOLCHILDREN IN CHEMISTRY MUNICIPAL STAGE. 9 CLASS Solutions and criteria for evaluating Olympiad tasks The five solved ones, for which the participant scored the most points, are counted in the final grade of the six proposed problems. One of the problems with the lowest score is not taken into account. 1. Properties of acidic oxides Write reaction equations in which an acidic oxide reacts with another substance in a molar ratio: a) 1: 1, b) 1: 2, c) 2: 1, d) 1: 3, e) 1: 6 (For each reaction you can choose your own acid oxide.) a) SO 3 + H 2 O = H 2 SO 4 b) CO 2 + 2KOH = K 2 CO 3 + H 2 O c) 2SO 2 + O 2 = 2SO 3 d) P 2 O 5 + 3H 2 O = 2H 3 PO 4 e) P 2 O 5 + 6KOH = 2K 3 PO 4 + 3H 2 O 2 points for each equation. In each case, several options are possible, any reasonable answers that meet the conditions are accepted. 2. Reactions with volume reduction There is a mixture of hydrogen and oxygen. The mixture is ignited with an electric spark. At what volumetric ratio of gases will the volume of the mixture after passing a spark decrease by exactly one quarter (at constant temperature and pressure)? The volume of water formed can be neglected. Give two answer options and support them with calculations. 2H 2 (g) + O 2 (g) = 2H 2 O (l) According to the reaction equation, 2 moles of water are formed from 3 moles of gases. Consequently, the amount of gas substance decreases by 3 mol, which

2 is a quarter of the original amount, i.e. before the reaction there were 12 moles of gases. Two options are possible: 1) in excess hydrogen: 11 mol H 2 and 1 mol O 2, volume ratio H 2: O 2 = 11: 1 2) in excess oxygen: 2 mol H 2 and 10 mol O 2, volume ratio H 2: O 2 = 1: 5 Grading system: reaction equation 1 point, mention of the idea of ​​“excess-deficiency” 1 point, each answer option 4 points, of which 3 points for calculating the composition of the mixture in moles, 1 point for the volume ratio. 3. Oxygen from carbon dioxide The compound of an alkali metal with oxygen, substance X, reacts with carbon dioxide. The only reaction products are metal carbonate and oxygen in a molar ratio of 2: 3 and a mass ratio of 23: 8. Establish the formula of X and write the equations for its reactions with carbon dioxide and water. The metal can be determined by comparing the mass ratio with the molar ratio: m(m2co 3) 2 M (M2CO 3) 2 M (M2CO 3) 23 = = =, m(o) 3 M (O) from where M(M 2 CO 3) = 138 g/mol, this is K 2 CO 3, metal K. Formula X is determined by the reaction equation. Let's denote the formula X K a O b and first draw up a diagram and then an equation for the reaction. Scheme: K a O b + CO 2 2K 2 CO 3 + 3O 2. Equation: 4KO 2 + 2CO 2 = 2K 2 CO 3 + 3O 2. Substance X KO 2. Reaction equation with water: 4KO 2 + 2H 2 O = 4KOH + 3O 2. (The reaction equation with the formation of H 2 O 2 is also accepted: 2KO 2 + 2H 2 O = 2KOH + H 2 O 2 + O 2.) 2

3 3 Grading system: determination of metal 2 points, establishment of formula X 4 points, reaction equations, 2 points for each 4 points. 4. Electrons also have mass 2 Which substance, consisting of two elements, has the largest mass fraction of electrons? Briefly justify your answer using calculations or reasoned arguments. An atom consists of protons, neutrons and electrons. The number of electrons in an atom is equal to the serial number of the element Z. Let us assume that the masses of the proton and neutron are approximately equal, and write the expression for the mass fraction of the electron: N(e) m(e) N(e) m(e) ω(e) = = = N(e) m(e) N(p) m(p) N(n) m(n) N(e) m(e) N(p) N(n) m(p) Zm(e) 1 = = Zm(e) Am(p) A m(p) 1 Z m(e) Here N denotes the number of particles of a given type in an atom, and A is the mass number. In order for the electron mass fraction to be greatest, the A/Z ratio must be smallest. It is equal to 1 for the 1 H atom; for all other atoms it is equal to 2 or more. Thus, the compound must contain H atoms and atoms of another element with the minimum (if possible) A / Z ratio. In this case, there should be as much hydrogen as possible. Of all hydrogen compounds, CH 4 has the largest mass fraction of hydrogen. The 12 C carbon atom is suitable as the second element since it has A / Z = 2. Answer: CH 4. Grading system: considerations for what the compound should contain H atom, 4 points, considerations about CH 4 6 points (compounds with a lower mass fraction of hydrogen, for example NH 3, H 2 O, 3 points instead of 6), if only the correct answer is given without any reasoning 0 points.

4 4 5. Study of black powder The young chemist examined black powder using the means available to him. When the powder was heated for a long time in air and in a stream of oxygen, the color of the substance became darker, and a gas with a pungent odor was released, similar to the smell of a burning match. The mass of the substance did not change during calcination. Further studies showed that the original test substance does not dissolve in dilute sulfuric acid even when heated, but the calcination product dissolves well in the acid, forming a blue solution. Based on these observations, the young chemist came to the correct conclusion that the substance consists of two elements. a) What elements are included in the composition of the substance under study? b) What is the formula of the substance under study? c) Write down the equation for the reaction that occurs when a substance is calcined in the presence of oxygen. d) Explain using calculation why the mass of a solid does not change during calcination. a) The composition of the substance under study includes copper and sulfur (1.5 points for each element for a total of 3 points). b) Formula of the substance Cu 2 S (3 points; for CuS 0 points). c) Cu 2 S + 2O 2 = 2СuO + SO 2 (3 points; 3 points are also given if the correct equation for the combustion of CuS is given). d) The mass of 1 mole of Cu 2 S (160 g) is equal to the mass of 2 moles of CuO, the product of the reaction SO 2 is a gaseous substance (1 point).

5 5 6. Chemical fountain The following experiment was carried out in a chemistry lesson. A white crystalline substance was placed in Wurtz flask 1 (see Fig. 1), which colors the flame yellow. Using dropping funnel 2, concentrated sulfuric acid was poured into flask 1, and a colorless gas 1.26 times heavier than air immediately began to evolve. The resulting gas was collected in round-bottomed flask 3. 1) What gas was obtained in the device shown in Fig. 1? Justify your answer, confirm it with calculations and the reaction equation. 2) Why was this gas collected by displacing air rather than water? Figure 1 3) What needs to be done with flask 3 filled with gas so that a fountain begins to “glow” inside it (see Figure 2)? Explain your answer. 4) What will be observed (see Fig. 2) if you pour into crystallizer 4: a) an aqueous solution of methyl orange; b) silver nitrate solution? 5) Why should you use a round-bottomed flask rather than a flat-bottomed flask for a “fountain”? Figure 2 1) Hydrogen chloride was obtained in the device. M = 1.26 29 g/mol = 36.5 g/mol (2 points) Sulfuric acid was added to sodium chloride. It is known that sodium ions color the flame yellow, which proves the presence of sodium in the composition of this salt. (1 point) The reaction occurs in Wurtz flask 1: NaCl + H 2 SO 4 = NaHSO 4 + HCl. (1 point) (The equation is allowed: 2NaCl + H 2 SO 4 = Na 2 SO 4 + 2HCl) 2) Hydrogen chloride cannot be collected by displacing water, since it dissolves well in it. (1 point) 3) In order for the flask to “give a fountain”, it is necessary, without releasing gas, to close it with a stopper with a gas outlet tube. Then add a small amount of water into the flask, in which hydrogen chloride will dissolve,

6 and a vacuum will be created in the flask. If the end of the gas outlet tube is lowered into water, it will begin to rise into the flask under the influence of atmospheric pressure. (2 points) 4) When hydrogen chloride is dissolved in a flask, hydrochloric acid is formed. a) In hydrochloric acid, the orange color of the indicator will change to red. (1 point) b) When hydrochloric acid reacts with a solution of silver nitrate, a white precipitate is formed, the mixture becomes cloudy: HCl + AgNO 3 = HNO 3 + AgCl. (1 point) 5) When demonstrating a “fountain”, a vacuum is created inside the flask, so its walls experience significant pressure from the outside. In the case of a round-bottomed flask, the pressure on its walls is distributed evenly. Thus, when using a round-bottomed flask, there is less chance that it will burst, unable to withstand the pressure difference. It is important to note that the walls of the flask must be completely smooth and intact; there should be no cracks, scratches, or chips on them. (1 point) 6

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Class

Problem 8 – 1(7 points). The mixture contains construction sand, granulated sugar, paraffin, and iron filings. How to separate this mixture of substances?

Problem 8 – 1(7 points). Solution. 1. Using a magnet, extract iron from the mixture (1 point). 2. Add water to the mixture (1 point). 3. Paraffin is insoluble in water. It can be carefully collected from its surface (2 points). 4. Sugar dissolved in water. Separate the aqueous sugar solution from the sand by filtration (2 points). 5. By evaporating water, we get sugar (1 point.).

Problem 8 – 2(4 points). What chemical elements are named after Russia, France, Poland? What element was first discovered in the Sun?

Problem 8 – 2(4 points). Solution. 1. The element ruthenium (Ru) is named after Russia (1 point). 2.The element polonium (Po) is named after Poland (1 point). 3.The element francium (Fr) is named after France (1 point). 4. The element helium (He) was first discovered in the Sun (1 point).

Problem 8 – 3(6 points). Early spring morning, when the ambient temperature was still 0°C and the pressure was 760 mm. rt. Art., three comrades, walking their dogs, saw a bottle on the lawn. "It's empty" – said one of them. “No, it is full to the brim, and I know the formula of the substance with which it is filled,” – said another. "Both of you are wrong" – said the third and, to be convincing, he calculated the amount of substance and the number of particles contained in the bottle. Bottle volume 0.7 l. Repeat the arguments and calculations of the third of your comrades.

Problem 8 – 3(6 points). Solution. 1. There is air in the bottle. Therefore it cannot be considered empty (1 point). 2. Since air is a mixture of substances, it is impossible to talk about one substance that makes up the air (1 point). 3. Air temperature 0°C, and pressure 760 mm. rt. Art., correspond to normal conditions. Under these conditions, in accordance with Avogadro's law, a volume of 0.7 liters can be occupied by a gas with an amount of substance equal to n = = 0.03125 mol (2 points). 4. This amount of substance (air) contains a number of particles (various molecules - mainly oxygen, nitrogen and argon) equal to

N=n ∙ (N A) = 0.03125 ∙ (6,02 ∙ 10 23) = 0,188 ∙ 10 23 particles (2 points).

Problem 8 – 4(2 points). In 1820, a scandal broke out in London. At one of the aristocratic receptions, a famous jeweler said to the countess, the mistress of the house: “Your ring, my lady, is not a diamond, but a fake!” How could the jeweler immediately prove to the countess that he was right?

Problem 8 – 4(2 points). Solution. It is difficult to distinguish a rhinestone from a diamond by its appearance, but its origin is revealed by its insufficient hardness: it does not scratch glass.

Z problem 8 – 5(6 points). Using only 11 letters that are included in the name of element number 91 in D.I. Mendeleev’s periodic table, make up as many names of other chemical elements as possible. By suggesting 6 titles, you will receive the maximum score.

Z problem 8 – 5(6 points).Solution. Actinium, No. 89. Titan, No. 22. Yttrium, No. 39. Thorium, No. 90. Sodium, No. 11. Krypton, No. 36.

Problem 8 – 6 (4 points). Calculate how many times the mass of an oxygen atom is greater than the mass of a helium atom.

Problem 8 – 6 (4 points). Given: elements O (oxygen) and He (helium). Define (m – mass of an atom of an element).

Solution. 1. Ar(O) = And = (2 points)

2. Using the formulas obtained, we obtain = = = 4 (2 points).

Problem 8 – 7(2 points). Magnesium combines with sulfur in a mass ratio of 3:4. Determine the mass of magnesium that will react with 20 g of sulfur.

Problem 8 – 7(2 points). Solution. m(Mg) = ; m(Mg) = = 15 g.

Problem 8 – 8(6 points). What is the mass of a mixture consisting of 10 moles of hydrogen gas and 5 moles of oxygen?

Problem 8 – 8Solution. 1. Determine the mass of oxygen included in the mixture: m(O 2) = n(O 2) ∙ M(O 2); m(O 2) = 5 mol∙32g/mol = 160 g (2 points). 2. Calculate the mass of hydrogen included in the mixture: m(H 2) = n(H 2) ∙ M(H 2); m(H 2) = 10 mol∙2g/mol = 20 g (2 points). 3. Calculate the mass of the mixture of oxygen and hydrogen: m(mixture) = m(H 2) + m(O 2); m(mixture) = 20g + 160g = 180g (2 points).

Problem 8 – 9(6 points). Determine the total amount of substance (in moles) contained in 800 g of a 15% by weight aqueous solution of calcium nitrate.

Problem 8 – 9(6 points).Solution. 1. The task requires determining the number of moles of two substances: potassium nitrate and water, of which the solution consists (2 points). 2. The condition of the task corresponds to the expression for the fraction (wt. %) of the dissolved substance 15 = , from which = 120g. M(KNO 3) = 101 g/mol. Mass of salt m(KNO 3) = 120 g corresponds to its amount equal to n(KNO 3) = = = 1.19 mol (2 points).

The mass of water in a solution weighing 800 g is equal to m(H 2 O) ==800 – m(KNO 3) = 800 – 120 = 680 g. This mass of water corresponds to its amount equal to n(H 2 O) = = 37.8 mol (2 points).

Municipal stage of the All-Russian Chemistry Olympiad for schoolchildren

Class

Problem 9 - 1(5 points). Two test tubes are filled one-third each with transparent solutions of sodium hydroxide and aluminum chloride. The amounts of dissolved substances in solutions are equivalent to each other, i.e. The reagents, after mixing, will completely react with each other without residue, forming the maximum amount of precipitate. Given only these two test tubes of solutions and nothing else, recognize the test tube with an alkali solution and the test tube with a salt solution. Motivate your answer.

Problem 9 - 1(6 points). Solution. 1. It all depends on the order of draining (what to pour into what (2 points). 2. If a solution of sodium hydroxide is added dropwise to a solution of aluminum chloride, a precipitate will form immediately after the first drops of the added alkali solution: AlCl 3 + 3NaOH = Al(OH) 3 ↓ + 3NaCl (2 points). If a solution of aluminum chloride is added dropwise to a solution of sodium hydroxide, then the excess alkali at the initial stage will prevent the formation of a precipitate; a precipitate will form only with the addition of the last portions of the salt solution (2 points).

Problem 9 – 2(6 points). A solid, white substance that smokes in the air was brought from the workshop where ice cream is made and placed in a glass of distilled water. The piece "disappeared". In the resulting solution, litmus changes color to red. It is known that the relative density of this substance in air in the gaseous state is 1.5. Name the substance.

Problem 9 – 2(6 points). Solution. We are talking about solid carbon monoxide (IV) - “dry ice”. It is used for cooling, for example, ice cream, metal (2 points).

Possible reaction equations: CO 2 + H 2 O ⇄H 2 CO 3 (1 point); CO 2 + Ca(OH) 2 = CaCO 3 ↓ + H 2 O (1 point); CaCO 3 + CO 2 + H 2 O = Ca(HCO 3) 2 (1 point); CO 2 + CaO = CaCO 3 (1 point).

Problem 9 – 3(9 points). Determine the mass fractions (in %) of iron (II) sulfate and aluminum sulfide in the mixture if, when treating 25 g of this mixture with water, a gas was released that completely reacted with 960 g of a 5% solution of copper sulfate.

Problem 9 – 3(9 points). Solution. 1) Reaction equations were compiled: Al 2 S 3 + 6H 2 O = 2Al(OH) 3 + 3H 2 S; CuSO 4 + H 2 S = CuS + H 2 SO 4 (2 points). 2) The amount of hydrogen sulfide substance was calculated: n (CuSO4) = 960 ∙ 0.05 / 160 = 0.3 mol; n(H 2 S) = n(CuSO4) = 0.3 mol (2 points). 3) The amount of substance and mass of aluminum sulfide and iron(II) sulfate were calculated: n (Al 2 S 3) = n (H 2 S) = 0.1 mol; m(Al 2 S 3) = 0.1∙150 =15 g; m(FeSO 4) = 25 -15 = 10 g (3 points). 4) The mass fractions of iron(II) sulfate and sulfide were determined

aluminum in the initial mixture: ω(FeSO4) = 10 / 25 = 0.4, or 40%; ω (Al 2 S 3) = 15 / 25 = 0.6, or 60% (2 points).

Problem 9-4.“When iron filings are placed in thick, strong vitriol, with which four parts of water are mixed, poured into a narrow-necked glass, then the steam escaping from the candle flame ignites... Sometimes it happens that the ignited steam bursts the glass with a great crash” (M.V. Lomonosov, Complete Works, - M.: 1953, vol. 1, p. 474).

1. Determine the mass fraction (%) of the dissolved substance in diluted “vodka of vitriol” if the initial mass fraction in “strong vodka of vitriol” was 98%, and the fractions of water during dilution were taken by mass.

2. Write the equations for the reactions of iron with a solution of “vodka vitriol” and the combustion of “escaping steam”.

3. Write 3 equations for reactions that can occur during the interaction of iron filings with a solution of “vodka vitriol” depending on its concentration.

4. Determine the ratio of the volumes of a dilute solution of “vodka vitriol” (density 1.2 g/cm 3) and “outgoing steam” under normal conditions, if we accept the occurrence of chemical processes as quantitative.

Problem 9-4 (author – Zhirov A.I.).Solution. Let there be 100 g of concentrated solution. (The content of “vitriol vodka” - sulfuric acid - 98 g) Then the mass of added water will be 400 g. The total mass of the solution is 500 g. The mass fraction of sulfuric acid will be 98: 5 = 19.6 (%).

1. When iron reacts with dilute sulfuric acid, iron (II) sulfate and hydrogen are formed:

Fe + H 2 SO 4 = FeSO 4 + H 2

2H 2 + O 2 = 2H 2 O

2. At a higher acid concentration, hydrogen sulfide and sulfur can be released along with hydrogen:

4Fe + 5H 2 SO 4 = 4FeSO 4 + H 2 S + 4H 2 O

3Fe + 4H 2 SO 4 = 3FeSO 4 + S + 4H 2 O

Concentrated sulfuric acid forms sulfur (IV) oxide and iron (III) sulfate:

2Fe + 6H 2 SO 4 Fe 2 (SO 4) 3 + 3SO 2 + 6H 2 O

4. A liter of dilute sulfuric acid solution has a mass of 1200 g and contains 0.196 ∙ 1200 = 235.2 g of sulfuric acid, which is 2.4 mol of acid. Then, with complete interaction of the acid with iron, 2.4 moles of hydrogen are released or 2.4 · 22.4 = 53.76 (l). The volume of hydrogen released is 53.76 times greater than the volume of dilute sulfuric acid (or the volume of acid is 53.76 times less than the volume of hydrogen).

Grading system

1. Concentration calculation – 5 points

2. Reaction with iron – 2 points

3. Hydrogen combustion – 1 point

4. Three reactions 2 points each – 6 points

5. Volume ratio – 6 points

TOTAL: 20 points

Problem 9-5. Below is a table describing the interaction of solutions of binary potassium salts and elements X 1, X 2, X 3 And X 4, located in the same group of the periodic table, with solutions of silver, lead and mercury nitrates.

1. Identify salts of elements X 1, X 2, X 3 And X 4.

2. Write the interaction equations for binary salts of elements X 1, X 2, X 3 And X 4 with silver, lead and mercury nitrates. In the equations, be sure to indicate the substance that precipitates.

3. Write the interaction equations for solid binary potassium salts of elements X 1, X 2, X 3 And X 4 with concentrated sulfuric acid.

4. When a mixture of dry salts LiX 2, NaX 2 and KX 2 weighing 5.85 g interacted with concentrated sulfuric acid, 12.0 g of hydrosulfates were formed. Determine the volume (at 30 °C and 130 kPa) of gas that can be released.

Problem 9-5 (author – Antonov A. A.).Solution. 1. Lead nitrate and silver nitrate are high-quality reagents for halogens. In this case, silver fluoride is soluble. This means that the encrypted elements are halogens. Silver fluoride, as stated above, is soluble, which means KX 3 – KF. When interacting with silver nitrate, a white precipitate is formed by chlorides, which means KX 2 – KCl. The most intensely colored are silver and lead iodides, then KX 1 is KI, and KX 4 is KBr.

KX 1 – KI, KX 2 – KCl, KX 3 – KF, KX 4 – KBr.

3. KX 1: 2KI + 3H 2 SO 4 → 2KHSO 4 + I 2 + SO 2 + 2H 2 O or

6KI + 7H 2 SO 4 → 6KHSO 4 + 3I 2 + S + 4H 2 O or

8KI + 9H 2 SO 4 → 8KHSO 4 + 4I 2 + H 2 S + 4H 2 O

KX 2: KCl + H 2 SO 4 → KHSO 4 + HCl

KX 3: KF + H 2 SO 4 → KHSO 4 + HF

KX 4: KBr + H 2 SO 4 → KHSO 4 + HBr or

2KBr + 3H 2 SO 4 → 2KHSO 4 + Br 2 + SO 2 + 2H 2 O

In all cases, an acid salt will be formed, since concentrated sulfuric acid is used, i.e. there is a significant excess of acid.

4. Let's write down the equations of all reactions:

LiCl + H 2 SO 4 → LiHSO 4 + HCl

NaCl + H 2 SO 4 → NaHSO 4 + HCl

KCl + H 2 SO 4 → KHSO 4 + HCl

Let it come into interaction x moles of sulfuric acid, then the result is the release x mole of hydrogen chloride. Mass of the reaction mixture before interaction 5.85 + 98 x, and after interaction 12 + 36.5 x. According to the law of conservation of mass

5,85 + 98x = 12 + 36,5x,

where x= 0.1 mol. Means V = ν RT/p= 0.1∙8.31∙303:130 = 1.94 l

Grading system:

1. 1 point for correct identification of each substance (element) 4 points. Note for examiners: if the group is guessed (i.e. that the halogens are encrypted), but in the wrong order, then no more than 1 point for this item.

2. 9 equations 1 point each. 9 points.

3. 4 equations 1 point each. 4 points

Note to assessors: In the reaction with bromine and iodine, count any one reaction. If sulfates are indicated instead of hydrosulfates, then 0.5 points for the reaction.

4. 0.5 points each for equations with lithium and sodium chlorides. For calculating the number of moles 1.5 points.
For volume calculation 0.5 points. only 3 points.

Task 1

The mixture consists of sand, table salt, iron filings and marble chips. Suggest a method for separating the components of a mixture.

Task 2

Theophrastus Bombastus von Hohenheim was one of the most famous scientists of his time and is credited with pioneering the scientific foundations of chemistry. These scientists discovered “combustible air”.

It is known that “combustible air” is the lightest of all gases.

What do we call “flammable air”?

Give two ways to obtain “combustible air”.

For what practical purposes is this substance used? (give two examples).

Task 3

A mixture of carbon dioxide and carbon monoxide occupies a volume of 15 liters at a temperature

25 0 C and pressure 730 mm Hg. The mass of this mixture is 17.5 g. Determine the volume fraction of CO 2.

Task 4

Determine the volume of carbon dioxide that must be passed through 300 g of a solution containing 15 g of barium hydroxide so that its concentration decreases to 4%.

Task 5

Determine the number of iron atoms in an iron ball whose radius is 2 cm.

The density of iron is 7800 kg/m3.

Task 1

1. Iron filings are separated with a magnet 2b

2. Separate table salt by dissolving the residue in water 2b

3. Process the remainder of marble chips and sand

hydrochloric acid, in which marble 2b dissolves

Reaction equation: CaCO 3 + HCI = CaCI 2 + CO 2 + H 2 O 2b

4. Sand remains in sediment 2b

Task 2

1. Combustible air is hydrogen. 2b

2. 2HCI + Zn = ZnCI 2 + H 2 2b

3. 2 H 2 O → 2H 2 + O 2 (electrolysis) 2b

Hydrogen can be used for hydrogenation of fats, reduction of metals from oxides, etc. for each example of application 2b. 4b.

Task 3

1. Determine the volume under normal conditions

PV/T = P 0 V 0 /T 0 ; hence V 0 = 13.2 l. 2b.

2. Determine the amount of substance in the gas mixture

ν = 13.2/22.4 = 0.59 mol 2b

3. Let's make an equation

44x + 28(0.59x) = 17.50 2b

X=0.06mol 2b

4. Since the volume and quantity of the substance are proportional

let's calculate the volume fraction of CO 2

φ(CO 2) =0.06/0.59*100% = 10.17%. 2b

Task 4

1.Reaction equation:

Ba(OH) 2 + CO 2 = BaCO 3 ↓ + H 2 O 1b

2. Let the mass of barium hydroxide that needs to be removed from the solution be Xg, then, based on the conditions of the problem, we obtain the equation:

15-Х/300-0.9Х = 0.04 4b

Hence X = 3.24g

3. The amount of barium hydroxide substance that reacted will be

ν =3.24/171=0.019mol 1b

4. Since according to the reaction equation for the amount of carbon dioxide substances

and barium hydroxide coincide, the volume of carbon dioxide will be:

V=22.4*0.019=0.42l. 4b

Let's determine the volume of the ball

V=4/3 πR 3 =3.35*10 -5 m 3 2b

Determine the mass of the ball

m=3.35*10 -5 *7780=0.26kg 2b

Let's determine the amount of iron substance

ν= 4.67 mol 2b

Let's determine the number of atoms in the sample

N=4.67mol*6.02*10 23 atoms/mol=2.8*10 24 atoms 4b

1. Municipal stage of the All-Russian Chemistry Olympiad for schoolchildren (2)

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   01 Email: [email protected] Municipal stage All-Russian olympiads schoolchildren By chemistry 2011-2012 academic year 8th grade...PMC Ph.D., Associate Professor of the Department of General and Inorganic chemistry KubSU V.I. Zelenov Task 1 What volume...

2. Municipal stage of the All-Russian Chemistry Olympiad for schoolchildren (6)

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   01 Email: [email protected] Municipal stage All-Russian olympiads schoolchildren By chemistry 2013-2014 academic year 9th grade...

3. Municipal stage of the All-Russian Chemistry Olympiad for schoolchildren (8)

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   ... :[email protected] Municipal stage All-Russian olympiads schoolchildren By chemistry 2013-2014 academic year... After the summer holidays in the office chemistry jars of reagents were discovered, ... an alkane having a vapor density By hydrogen 15. Identify compounds...

Item name CHEMISTRY

Class 10

Job completion time 3 hours

Quests

Task 1 (8 points)

In an aqueous-alcohol solution of barium iodide, 3.005·10 22 iodide ions and 13.1 g of salt that did not disintegrate into ions were found. Determine the degree of salt dissociation (in%) in this solution. Write the dissociation equation for barium iodide.

Task 2 (6 points)

10 ml of methane per second comes out through a kitchen stove burner. Conditions in the kitchen: 20 ° C, 750 mm. rt. Art. How long will it take for a kettle containing 1 liter of water to completely boil away? Heat of formation (kcal/mol): CO 2 – 94; liquid water – 68.3; steam water – 57.8; methane ~0.

Task 3 (9 points)

Identify organic substances A-J . Additionally, it is known that the substance A can be obtained from inorganic substances, substances B and J have a pungent characteristic odor, the substance H - burning taste.

Task 4 (9 points)

The mass of the mixture of sodium and potassium oxides is 6 g. After dissolving the mixture in 100 g of 15% potassium hydroxide solution, 72.89 ml of 20% hydrochloric acid (density 1.1 g/ml) was required to neutralize the resulting mixture. Determine the mass fractions of oxides in the initial mixture. Describe the chemistry of the processes occurring.

Task 5 (18 points)

Suggest a method for producing benzoic acid ethanol ester from sodium acetate without using other organic compounds. Name the intermediate substances and indicate the reaction conditions.

Task 6 (17 points)

Gas A (2.24 L, n.s.) passed through salt solution B , and a colored precipitate formed C (25.4 g) and gas evolved D . Only substance E remains in solution. The resulting solution has a highly alkaline reaction. The solution of substance E was evaporated, and the dry residue was weighed. Mass of E 11.2 g. This amount of E was again dissolved in water. Neutralization of the resulting solution required 2 liters of 0.1 M hydrochloric acid. Sediment C separated and dried. It is a dark solid that sublimes when heated. C forms brightly colored complexes with starch. Substances A and E found in nature, both are strong oxidizing agents.

1. Identify all substances, confirm your answer with calculations. Write the reaction equations.

2. Where do A and E in nature? Specify colors A and E in gaseous and liquid states.

3. Humanity still receives substance C in significant quantities from living nature (i.e., from plants and organisms). Guess where C is obtained from? What color is complex C with starch?

4. Determine the molar concentration of salt B in the original solution.

Schoolchildren in chemistry

Item name CHEMISTRY

Class 11

Job completion time 3 hours

Maximum points 62 points

Task 1 (7 points)

Using the electron balance method, create an equation for the reaction:

NaNO 3 + Cu + … = … + Na 2 SO 4 + NO 2 + H 2 O

1. Identify the oxidizing agent and the reducing agent.

2. Give the electronic configurations of nitrogen and copper atoms before and after the reaction.

Task 2 (16 points)

After heating potassium permanganate, its mass decreased by 0.96 g. When the resulting residue was treated with 36.5% hydrochloric acid (density 1.18 g/ml), 6.496 liters of chlorine were obtained. Determine the initial mass of salt and the volume of acid consumed. Describe the chemical processes occurring.

Task 3 (12 points)

Novocaine is a drug used for local anesthesia. The industrial synthesis of novocaine comes from para-nitrobenzene:

Decipher the structures of intermediates A, B, C and novocaine (D ). Determine the products of the interaction of novocaine with analytical reagents: bromine water, concentrated nitric acid, benzaldehyde in the presence of concentrated sulfuric acid, sodium hydroxide.

Task 4 (7 points)

By reacting a mixture of nitrogen and nitric oxide (IV) with a solution of sodium hydroxide, 77 grams of salts were obtained. Write the equations for the corresponding chemical reactions. Calculate the number of moles of each gas if the average molar mass of their mixture is 40 g/mol.

Task 5 (6 points)

After hydrolysis of 4.8 g of binary phosphorus compound ( V ) with one of the elements VII group, it took 35 ml of potassium hydroxide solution (2 mol/l) to neutralize the resulting solution. Determine the formula of this compound. Confirm your answer with calculations and reaction equations.

Task 6 (14 points)

Excess metal oxide solution ( II ) interacts with unsaturated conjugate hydrocarbon A (vapor density in air 2.69) in an aqueous solution of sulfuric acid, while it is known that substance A attaches only one molecule of water. When A is exposed to an excess of ammonia solution of metal oxide ( I ) a precipitate is formed. When passing a mixture of A and hydrogen over a hot metal oxide ( III ) an alkane with the lowest octane number is obtained.

Give possible structural formulas of hydrocarbon A, establish the desired structure of substance A and name it. Give a reasoned explanation. Write down the equations for these reactions and the conditions for their occurrence.

Municipal stage of the Russian Olympiad schoolchildren in chemistry in the Volgograd region in the 2015-2016 academic year

Item name CHEMISTRY

Class 10

Job completion time 3 hours

Maximum points 67 points

Solutions:

Task 1

Task 2

Task 3

Task 4

Task 5

Task 6

Municipal stage of the Russian Olympiad schoolchildren in chemistry in the Volgograd region in the 2015-2016 academic year

Item name CHEMISTRY

Class 11

Job completion time 3 hours

Maximum points 67 points

Solutions:

Task 1

Task 2

Task 3

Task 4

Task 5

Task 6

Municipal stage of the Russian Olympiad schoolchildren in chemistry in the Volgograd region in the 2015-2016 academic year

Item name CHEMISTRY

Class 9

Job completion time 3 hours

Quests

Task 1 ( 10 points)

The coloring pigment of orange paint - red lead - has the composition Pb 3 O 4.
1.What is the possible oxidation state of lead in this compound?
2.What class of chemical compounds does this substance belong to?
3. When red lead reacts with coal, metallic lead can be obtained. Write the reaction equation.
4. How many g of lead can be obtained from 6.85 g of red lead if the reaction yield of the theoretically possible is 97%?

Task 2 ( 12 points)

A qualitative analysis of the leonite mineral showed the presence of potassium, magnesium and sulfate ions in it. When a mineral sample weighing 7.32 g is calcined, its mass decreases by 1.44 g. The same weighed portion of the mineral, when dissolved in water and subsequent addition of an excess barium chloride solution, forms 9.32 g of sediment.
Determine the formula of leonite.

Task 3 ( 12 points)

A substance weighing 14.9 g, formed by two types of chemical particles with the same electronic structure 1s 2 2s 2 2p 6 3s 2 3p 6, reacted completely with 20.0 g of 98% sulfuric acid.

2. Determine the mass of the salt formed.
3. Write the equation for the reaction of an aqueous solution of the resulting salt with metallic magnesium.

Task 4 ( 10 points)

A vessel at 300 K (degrees absolute temperature) and a pressure of 2 atm contains 1 mole of hydrogen and 1 mole of chlorine. After illuminating the mixture with ultraviolet light and completing the reaction, the temperature of the vessel was 600 K.
1. Write the reaction equation.
2. Why might this reaction occur when illuminated?
3. Which process is the first stage of the reaction when illuminated?
4. What pressure (atm) was in the vessel when measuring the temperature after the end of the reaction?
5. How many g of hydrogen chloride are contained in the vessel after the end of the reaction?

Task 5 ( 10 points)

Which of the following substances are suitable for obtaining oxygen from them in the form of a simple substance in one stage:

a) H2O; b) H 2 O 2; c) KMnO 4; d) KClO 3; e) KPO 3

Write the equations for the corresponding reactions

Municipal stage of the Russian Olympiad schoolchildren in chemistry in the Volgograd region in the 2015-2016 academic year

Item name CHEMISTRY

Class 9

Job completion time 3 hours

Maximum points 54 points

SOLUTIONS

Task 1 10 points)

1) Pb 2 PbO 4 - oxidation states +2 and +4 or (not entirely correct, but estimated) Pb(PbO 2) 2 Oxidation states of lead +2 and +3
2) This is the lead salt of lead (lead) acid
3) Pb 3 O 4 + 2 C = 3 Pb + 2 CO 2
4) 6.85 g is 0.01 mol of red lead; it will yield 0.01 mol or 6.21 g of lead. Taking into account the yield of 6.21 g of lead x 0.97 = 6.02 g

Task 2 ( 12 points)

When the sample is calcined, the water of crystallization is removed.
Let's determine its mass: 7.32 - 1.44 = 5.88 g.
When barium chloride is added, barium sulfate is obtained:
M = 233, = 9.32 / 233 = 0.04 mol.
Thus, the initial sample contains 0.04 mol of sulfate ions. The mass of sulfate ions in the sample is 3.84 g, the mass of the total metals is 2.04 g.
To calculate the formula of a mineral, let’s create a system of equations:
2 (Mg) + (K) = 2 (SO 4 2-) (electroneutrality condition).
24 (Mg) + 39 (K) = 2.04 (mass of metals in a mineral sample).
Solving the system of equations, we obtain (Mg) = 0.02, (K) = 0.04.
The ratio is K: Mg: SO 4 = 2: 1: 2. Thus, the formula is K 2 Mg(SO 4) 2. Now we need to find the amount of water of crystallization. 1.44 g of water is 0.08 moles - twice the number of moles of sulfate.
therefore, the formula of leonite is K 2 Mg(SO 4) 2 4 H 2 O

Task 3 ( 12 points)

The electronic structure corresponds to KCl, CaS, ScP and combinations of these cations and anions.
20 g of sulfuric acid corresponds to 20/98 = 0.204 mol
Then 14.9 g corresponds to 0.2 or 0.4 mol of salt.
Suitable KCl (74.5)
14.9 g of CaS corresponds to 0.207 mol of salt; from it with sulfuric acid only CaSO 4 is formed, which is almost insoluble in water and does not react in solution with magnesium. 14.9 g ScP corresponds to 0.196 mol salt; At the same time, sulfuric acid is in short supply. Formally fit K 2 S- you will get medium and acidic sulfate.
0.2 mol each of KCl and H 2 SO 4
1) KCl + H 2 SO 4 = HCl + KHSO 4
2) You will get 0.2 mol or 27.2 g of acid salt
3) 2 KHSO 4 + Mg = K 2 SO 4 + MgSO 4 + H 2

Task 4 ( 10 points)

1) H 2 + Cl 2 = 2 HCl
2) This is a chain reaction that is initiated by light, causing the chlorine molecule to break down into atoms: Cl 2 [h] 2 Cl .
3) Then comes the reaction: H 2 + Cl . = HCl + H . The chain continues with hydrogen atoms: H . + Cl 2 = HCl + Cl .
4) During the reaction, the volume of the gas does not change, therefore the change in pressure can only be associated with a change in the temperature of the gas: P 1 /T 1 = P 2 /T 2 ; P 2 = P 1 T 2 /T 1 .
Then P 2 = 2*600/300 = 4 (atm)
5) From 1 mole of hydrogen and 1 mole of chlorine you get 2 moles of HCl: 36.5*2 = 73 g

Task 5 ( 10 points)

1. . 2 H 2 O = 2 H 2 + O 2 electrolysis

2 H 2 O 2 = 2 H 2 O + O 2 catalytic decomposition

2 KMnO 4 = K 2 MnO 4 + MnO 2 + O 2 thermal decomposition

2 KClO 3 = 2 KCl + 3 O 2 thermal decomposition

Only potassium phosphate does NOT decompose to release oxygen.

9th grade

The municipal stage of the All-Russian Chemistry Olympiad for schoolchildren was held with a fairly large number of participants - 197 people. but I was not pleased with the results. About a quarter of all participants (50 people) received 0 points for their work, about 100 more people scored from 1 to 10% points, and only five overcame the 50% barrier.

The right to take part in the regional stage of the All-Russian Olympiad was granted to the ten best participants. Among them are two winners of last year’s Olympiad for 8th grade schoolchildren: Alexander Zdanovich and Fedor Koryakin.

It turned out that out of the two hundred best 9th grade chemists, a quarter do not know and cannot

Write the reaction equation: sulfuric acid with alkali (task 1);
- dissociation of aluminum chloride;
- aluminum chloride with silver nitrate;
- aluminum chloride with alkali solution (task 2);
- magnesium carbonate with sulfuric acid (task 3);
- formation of carbon monoxide (IV) from simple substances - thermochemical;
- formation of calcium oxide from simple substances - thermochemical (task 4);
Calculate
- mass of the dissolved substance by mass fraction (task 1);
- the amount of substance by the number of particles (task 2);
- thermal effect of the reaction in terms of mass of matter and heat (task 4);
- amount of substance by mass of substance (task 5);

Because if they knew and were able to perform at least one of these simple elements of solving a problem, they would no longer have 0 points. And if they could do all this, they would already have at least 25% of these points. And the list of these simplest elements can be continued.

10th grade

There were significantly fewer tenth-graders among the participants in the municipal stage of the All-Russian Chemistry Olympiad for schoolchildren - 114 people. But here, too, a third of the participants (36 people) received zero points. Only four people passed the 50% barrier. Among them are two participants in the regional Olympiad last year: Salikov Zufar and Gavrilova Alisa. The best 10th grade students received the right to compete at the regional Olympiad:

What a third of the 114 best 10th grade chemists could not do:

Write the reaction equation: obtaining HCl from chlorides (task 1);
- alkene with chlorine (in general form) (task 2);
- ammonium sulfate and NaOH;
- ammonia with acid solutions (task 3);
- decomposition of sodium nitrate;
- decomposition of copper nitrate (task 4);
- benzene combustion - thermochemical;
- acetylene combustion - thermochemical;
- formation of benzene from acetylene (task 5);
Calculate mass of solution by volume and density;
- mass of solute by mass fraction;
- amount of substance by mass of substance (task 1.3);
- amount of substance by volume of gas (task 4);
- thermal effect of the reaction in terms of mass of matter and heat (task 5);
Explain formation of ammonium ion (task 3);
- purpose of the acetylene torch (task 5);

Any one of these elements would give the participant 1 point, and a combination of them would earn 15 points. Only 9 people were able to score 15 or more points.

11th grade

Students in the 11th grade had to be the most determined in their love for chemistry. There were 109 of them, and there were only 20 zero results. But three were able to overcome the 50% barrier. Nevertheless, the nine best participants were given the opportunity to prove themselves at the regional Olympiad:

What you had to not know and not be able to do in order to be able to get 0 points in 11th grade:

Write the reaction equation: burning sucrose according to a known formula;
- interaction of KOH with carbon dioxide (task 1);
- aluminum oxide with soda during fusion (task 2);
- obtaining Berthollet salt (task 3);
- electrolysis of copper sulfate (at least cathodic process) (task 4);
- dissociation of sulfuric acid;
- obtaining sulfuric acid (task 5);
Calculate the amount of dissolved substance by volume and density of the solution and mass fraction (task 1,2,4);
- amount of substance by molar concentration (task 5);
- pH based on the concentration of hydrogen ions (task 5);
Explain what is Berthollet's salt,
- what substance is included in the matchbox lubricant;
- what device protects the lungs from harmful gas (task 3);
- what acid is contained in “acid rain”;
- what is pH (task 5);

This would be enough to score 17 points; only 6 out of 109 people were able to cope with this.

So, the dates for the regional stage of the All-Russian Chemistry Olympiad for schoolchildren are known - January 30-February 1, 2010. The composition of its future participants has been determined. We wish them fruitful preparations and successful performances!

Other materials on the holding of regional Olympiads for schoolchildren in 2010 are presented at