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The laboratory has calcium carbonate. Suggest a plan for separating these substances using water and hydrochloric acid. What laboratory equipment would be required to separate this mixture? Write the reaction equations that will be used in the separation

Tasks d la school stage Olympiads for schoolchildren in chemistry grade 11.

Exercise 1.

There are ten elements in the Periodic system, in the designation of which the second letter is "e". For each of these elements, characteristic reactions are given below. capital letter marked with " ? ". Arrange the elements in their places.

1) D + T → ? e+n;

2) ? eO 4 → ? e + 2O 2 ;

3) ? e + O 3, F 2, etc. → does not go;

4) 2Au + 6H2 ? eO 4 (load) → 3 ? eO 2 + Au 2 ( ? eO 4) 3 + 6H 2 0;

5) ? eCl 2 + 4NaOH → Na 2 [ ? e(OH) 4 ] + 2NaCl;

6) ? eO 3 + 3H 2 O → H 6 ? eO 6 ;

7) ? e(OH) 4 + 4HCl → ? eCl 3 + 1/2Cl 2 + 4H 2 0;

8) ? eCl 4 + Li → ? eH 4 + Li;

9) ? eCl 2 + K 3 [ ? e(CN) 6 ] → K ? e[ ? e(CN) 6 ]↓ (blue precipitate);

10) 2H ? eO 4 + 7H 2 S → ? e 2 S 7 ↓ + 8H 2 0.

Exercise 2.

The chemist obtained samples of three silvery-white metals and found a way to quickly distinguish between them. To do this, he exposed the samples to the action of acids and sodium hydroxide solution. The results of his research are presented below.

Symbols: "+" - the reaction is in progress, « –» - the metal does not react. Determine what metals could be obtained by the chemist and write the corresponding reaction equations.

Task 3.

A small piece of a plastic disposable cup was heated without air to 400°C. As a result of heating, a hydrocarbon was obtained X(carbon content 92.26% by mass, its vapor density in oxygen 3.25). It is known that during the oxidation of a hydrocarbon X With a solution of potassium permanganate in an acidic medium, benzoic acid is formed as the only organic product.

1. Calculate molecular formula X.

2. Give the structural formula and the name of the hydrocarbon X. What is the name of the original polymer?

3. Write the reaction equation (with all products and stoichiometric coefficients) for hydrocarbon oxidation X potassium permanganate solution, acidified with sulfuric acid.

Task 4.

The young chemist Petya received from his mother the task to buy 1 liter of food vinegar in the store (mass fraction of acetic acid CH3COOH 9%) for home canning. Arriving at the store, he discovered that only acetic essence was on sale (the mass fraction of acetic acid was 70%). Petya decided that he could make food vinegar out of it on his own. At home, in the reference book, he managed to find the density values ​​\u200b\u200bfor a 9% solution of acetic acid - 1.012 g / ml and for a 70% solution - 1.069 g / ml. Of the equipment, Petya has only measuring cylinders of various sizes.

What safety rule should be observed when preparing dilute solutions from concentrated acids?

Which of the substances available at home should Petya have on hand if acid gets on his skin? Name this substance and reflect its effect on the acid in the reaction equation.

What volume of acetic essence should Petya measure in order to prepare 1 liter of 9% acetic acid solution?

Task 5.

The laboratory has iron, hydrochloric acid, caustic soda, calcium carbonate, copper oxide (II). Using these substances, as well as the products of their interaction, give at least 10 equations for the reactions of obtaining new organic matter.

Task 6.

When dissolving 2.8 g of an alloy of copper and silver in concentrated nitric acid, 5.28 g of a mixture of nitrates was formed. Determine the mass fractions of metals in the alloy.

Answers grade 11

Exercise 1.

Answer:

Points

Even if the solver cannot determine all the elements by reactions, this can be done using the Periodic Table, given that the maximum oxidation state of an element cannot be greater than the group number.

1) D + T → He+n;

2) Xe O 4 → Xe+2O2;

3)Ne+ O 3, F 2, etc. → does not go;

4) 2Au + 6H2 Se O 4 (load) → 3 Se O 2 + Au 2 ( Se O 4) 3 + 6H 2 0;

5) Be Cl 2 + 4NaOH → Na 2 [ Be(OH) 4 ] + 2NaCl;

6) Te O 3 + 3H 2 O → H 6 Te O 6 ;

7) Ce(OH)4 + 4HCl → Ce Cl 3 + 1/2Cl 2 + 4H 2 0;

8) Ge Cl 4 + Li → Ge H4 + Li;

9) Fe Cl 2 + K 3 → K Fe[Fe(CN) 6 ]↓ (blue precipitate);

10) 2H Re O 4 + 7H 2 S → Re 2 S 7 ↓ + 8H 2 0.

Maximum score

Task 2.

By the totality of properties, i.e. according to the behavior in reactions with acids and sodium hydroxide, we conclude: I metal - silver (copper does not match in color), metal II - aluminum, metal III - zinc. Reaction equations: 2Al + 6HCl = 2AlCl 3 + 3H 2 Zn + 2HCl = ZnCl 2 + 2H 2 Ag + 2HNO 3 (conc.) = AgNO 3 + NO 2 + H 2 OAl + HNO 3 (conc.) No reaction Zn + 4HNO 3 (conc.) + Zn(NO 3) 2 + 2NO 2 + 2H 2 OZn + 2NaOH + 2H 2 O = Na 2 + H 2 2Al + 6NaOH + 6H 2 O = 2Na 3 + 3H 2 (7 points)

Task 3.

Answer:

(Other formulations of the answer are allowed that do not distort its meaning)

Points

1. Mr(X) = 3.25 32 = 104 a.m.u.

Find the molecular formula of hydrocarbon X:

C: H \u003d 0.9226 / 12.01: 0.0774 / 1.008 \u003d 1: 1, taking into account molecular weight we get C 8 H 8.

2. Since the oxidation of hydrocarbon X with a solution of potassium permanganate in an acidic medium produces benzoic acid (C 6 H 5 COOH) as the only organic product, its molecule contains a benzene ring with one substituent.

Subtracting the C 6 H 5 fragment from the C 8 H 8 empirical formula, we obtain the C 2 H 3 substituent. Only possible variant the substituent is vinyl, and the hydrocarbon X is styrene (vinylbenzene).

Therefore, the polymer from which the disposable cup was made is polystyrene.

3. The reaction equation for the oxidation of styrene with a solution of KMnO 4 acidified with H 2 SO 4:

2KMnO 4 + 3H 2 SO 4 → + CO 2 + 2 MnSO 4 + K 2 SO 4 + 4H 2 O

All elements of the answer are written incorrectly

Maximum score

Task 4.

Pour water into acid (1 point).

Baking soda or sodium bicarbonate (1 point).

NaHCO3 + CH3COOH = CH3COOHa + H2O neutralization reaction (2 points).

The mass of acetic acid in a 9% solution was calculated as 91.08 g (1 point).

The mass of the vinegar essence solution was calculated - 130.1 g (1 point).

The volume of vinegar essence was calculated - 121.7 ml or ≈ 122 ml (1 point).

Total: 7 points.

Task 5.

Possible answers:

Fe + 2HCl \u003d FeCl 2 + H 2

HCl + NaOH = NaCl + H2O

2HCl + CaCO 3 \u003d CaCl 2 + H 2 O + CO 2

CuO + 2HCl \u003d CuCl 2 + H 2 O

2NaOH + CO 2 \u003d Na 2 CO 3 + H 2 O; NaOH + CO 2 \u003d NaHCO 3 + H 2 O

CuO + H 2 \u003d Cu + H 2 O

FeCl 2 + 2NaOH \u003d Fe (OH) 2 + 2NaCl

3Fe + 4H 2 O \u003d Fe 3 O 4 + 4H 2

CaCO 3 \u003d CaO + CO 2

CuO + CO 2 \u003d CuCO 3

Fe 3 O 4 + 8HCl \u003d FeCl 2 + 2FeCl 3 + 4H 2 O, etc.

Task 6.

Reaction equations:

Cu + 4 HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

Ag + 2HNO 3 \u003d AgNO 3 + NO 2 + H 2 O (1 point)

We introduce the notation: n(Cu)=xmol n(Ag)= ymol; Then

a) m(Cu)=64x, m(Ag)=108y

m (mixtures)= 64x+108y=2.8

b) according to equation (1) n (Cu (NO 3) 2 \u003d x, m (Cu (NO 3) 2 \u003d 188x;

c) according to equation (2) n (AgNO 3) \u003d y, m (AgNO 3) \u003d 170y

d) m(mixture) = 188x+170y=5.28 (2 points)

3) compose and solve the system of equations:

64x+108y=2.8x=0.01mol Cu

188x+170y=5.28 y=0.02mol Ag (2 points)

4) we calculate the mass fractions of the components of the mixture:

a) m (Cu) \u003d 0.01 * 64 \u003d 0.64 g. w(Cu)= 0.64/2.8= 0.2286 or 22.86%

m (Ag) \u003d 0.02 * 108 \u003d 2.16 g. w(Ag)= 2.16/2.8= 0.7714 or 77.14% (2 points)

maximum score - 7 points

Tasks for the school stage of the Chemistry Olympiad Grade 10

Exercise 1.

What gases can be obtained, having at its disposal the following substances:

sodium chloride, sulfuric acid, ammonium nitrate, water, ammonium nitrite, hydrochloric acid, potassium permanganate, sodium hydroxide, aluminum carbide, calcium carbide and sodium sulfite?

Write all the equations of possible reactions, indicate the conditions for their occurrence.

Task 2.

For the three chemical elements A, B and C, it is known that they take part in the following transformations:

C 2 + 3B 2 2CB 3

4CB 3 + 5A 2 4CA + 6V 2 A

4CB 3 + 3A 2 = 2C 2 + 6V 2 A

C 2 + A 2 \u003d 2 CA

4CA 2 + 2B 2 A + A 2 \u003d 4VCA 3

What elements are included in the task? Write down the reaction equations.

Exercise3.

When dissolving 11.5 g of a mixture of aluminum, magnesium and copper in hydrochloric acid, 5.6 l (N.O.) of gas was released. The undissolved residue was brought into solution with concentrated nitric acid. At the same time, 4.48 liters (n.o.s.) of gas were released. Determine the mass fractions (in%) of the components of the initial mixture.

Task 4.

The reaction scheme is given:

KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O

1. Determine the oxidation states of the elements.

2. Write an electronic balance equation

3. Determine the oxidizing agent and reducing agent

4. Arrange the coefficients in this equation.

5. List the areas of application of the substance, the formula of which is KMnO 4

Task 5.

IN periodic system there are nine elements whose names in Russian are nouns "not male". For each of these elements, characteristic reactions are given below, in which they are encrypted with the letter " E". Define elements for each reaction:

1) 2H 2 E+ E O 2 → 3 E+ 2H 2 O;

2) E Cl 3 + KI → E Cl 2 + KCl +1/2 I 2;

3) E NO 3 + KCl → E Cl↓ + KNO3;

4) E+ 3HNO 3 conc. + 4HCl conc. → H[ E Cl 4 ] + 3NO 2 + 3H 2 O;

5) E Cl 2 + 4NH 3 →[ E(NH 3) 4 ]Cl 2 (blue);

E(catalyst), 800°С

6) 4NH 3 + 5O 2 4NO + 6H 2 O;

7) 2 E+O2 2EO

8) E Cl 2 + Cl 2 → E Cl4;

9) E 2 O 3 + 4 HNO 3 conc. +( X-2) H 2 O → E 2 O 5 X H 2 O + 4NO 2

Task 6.

As a result of the reaction of calcium phosphate with magnesium at

heating produces two substances, one of which

reacts with water, producing a colorless toxic

garlic-smelling gas. Last oxidize

air oxygen.

Write the equations of all these chemical processes, name

their products.

Calculate the volume of air (N.O.) required to oxidize the gas,

if 2.4 g of magnesium was used in the first of these reactions.

Answers grade 10

Exercise 1.

1) 2NaCl (solid) + H 2 SO 4 (conc.) Na 2 SO 4 + 2HCl

or NaCl (solid) + H 2 SO 4 (conc.) NaHSO 4 + HCl

NH 4 NO 3 = N 2 O+ 2H2O

According to Le Chatelier's principle: if an external influence is exerted on a system in equilibrium, then as a result of the processes occurring in it, the equilibrium position shifts to the side that weakens this effect, with a decrease in pressure, the equilibrium in a reversible process shifts towards an increase in pressure.

a) N 2 O 4 (g) ⇄ 2NO 2 (g),

the process proceeds with an increase in volume (1

b) 2NO (g) + O 2 (g) ⇄ 2NO 2 (g),

the process proceeds with a decrease in volume (3\u003e 2), i.e., an increase in pressure, so the equilibrium is shifted to the starting substances.

c) 3Fe 2 O 3 (c) + CO (g) ⇄ 2Fe 3 O 4 (c) + CO 2 (g),

the process proceeds without a change in volume, a decrease in pressure does not affect the state of equilibrium.

Example 8.

Explain why it is impossible to extinguish a ignited metallic calcium with water. Write reaction equations.

Solution:

The calcium metal reacts with water, so adding water to the burning calcium will only speed up the process. We write the equation for the reaction of interaction of calcium with water:

Ca + 2H 2 O \u003d Ca (OH) 2 + H 2.

Example 9

Write the reaction equations in molecular and ionic form, with the help of which the following transformations can be carried out: Cl -  Cl 2  Cl -  AgCl.

Solution:

1. Interaction of manganese (IV) oxide with concentrated hydrochloric acid:

molecular form:

4HCl + MnO 2 \u003d MnCl 2 + Cl 2 + 2H 2 O,

ionic form

4H + + 2Cl - + MnO 2 = Mn 2+ + Cl 2 + 2H 2 O.

Sodium Chloride Violence Electrolysis:

molecular form:

molecular form:

2NaCl 2Na + Cl 2,

ionic form

2Na + + 2Cl - 2Na + Cl 2.

2. Interaction of chlorine with sodium bromide solution:

molecular form:

Cl 2 + 2NaBr \u003d 2NaCl + Br 2,

ionic form:

Cl 2 + 2Br - \u003d 2Cl - + Br 2.

Interaction of metallic sodium and gaseous chlorine:

molecular form:

2Na + Cl 2 \u003d 2NaCl.

3. Interaction of sodium chloride solution with silver nitrate solution:

NaCl + AgNO 3 \u003d AgCl + NaNO 3,

ionic form:

Cl - + Ag + = AgCl.

Example 10

Solution:

1. Interaction of copper with dilute nitric acid:

Cu + 6HNO 3 \u003d Cu (NO 3) 2 + 4NO + H 2 O.

2. Oxidation of copper with oxygen:

2Cu + O 2 \u003d 2CuO.

3. Interaction of copper with concentrated hydrochloric acid in the presence of oxygen:

2Cu + O 2 + 4HCl \u003d 2CuCl 2 + 2H 2 O.

4. Interaction of copper (II) chloride solution with sodium hydroxide solution:

CuCl 2 + 2NaOH \u003d Cu (OH) 2 + 2NaCl.

5. Thermal decomposition of copper (II) hydroxide:

Cu (OH) 2 \u003d CuO + H 2 O.

6. Interaction of copper (II) nitrate with sodium hydroxide solution:

Cu (NO 3) 2 + 2NaOH \u003d Cu (OH) 2 + 2NaNO 3.

7. Thermal decomposition of copper (II) nitrate:

2Cu(NO 3) 2 \u003d 2CuO + 4NO 2 + O 2.

8. Interaction of copper (II) oxide with hydrochloric acid:

CuO + 2HCl \u003d CuCl 2 + H 2 O.

Tasks for independent solution

Option 1

    Write reaction equations showing the properties of aluminum hydroxide.

    Show how, using one reagent, to determine which of the flasks contains dry salts: sodium chloride, sodium carbonate, sodium sulfide. Write the equations for the corresponding reactions.

    Prepare basic copper carbonate from metallic copper using as few reagents as possible. Write down the equations of the corresponding reactions.

    To neutralize 10.0 g of a solution containing a mixture of hydrochloric and hydrobromic acids, 2.5 g of a 3.2% NaOH solution was required, and when the solution of the same mass was treated with a solution of silver nitrate, 0.3315 g of a precipitate precipitated. Determine the mass fractions (%) of acids in the initial solution.

    Write the equations for the interaction of iron (III) hydroxide with concentrated hydrochloric and dilute sulfuric and nitric acids.

    Calculate how much industrial zinc containing 96% zinc and 27.5% HCl solution must be consumed to obtain 1 ton of 45% zinc chloride solution.

    Specify which element is oxidized and which is reduced in the following reactions: a) NH 3 + O 2  N 2 + H 2 O; b) KI + Cu(NO 3) 2  CuI + I 2 + KNO 3. Arrange the coefficients, indicate the transition of electrons.

    On one collective farm, hemp was applied per hectare: phosphate fertilizers - 60 kg (in terms of P 2 O 5), potash fertilizers - 150 kg (in terms of K 2 O) and blue vitriol- 10 kg. Assuming, for simplicity, that the latter does not contain impurities, indicate how many moles of each of the remaining oxides fall on 1 mole of copper (II) oxide.

    Indicate how it is necessary to change the temperature and pressure (increase or decrease) in order to shift the equilibrium in the decomposition reaction of calcium carbonate: CaCO 3 (c) ⇄ CaO (c) + CO 2 (g) - 178 kJ to the side of decomposition products.

    Explain why solutions of sodium and potassium hydroxides destroy glassware, especially when boiled for a long time. Write reaction equations.

    Write the reaction equations in molecular and ionic forms, with the help of which the following transformations can be carried out: CO 3 2-  CaCO 3  Ca 2+  CaSO 4 .

    Write the reaction equations that can be used to carry out the following transformations:

Option 2

    Write the equations for the interaction of dilute and concentrated sulfuric acid: a) with copper; b) with zinc; c) lead.

    Show what reactions can be used to distinguish between solutions of sulfuric, nitric and hydrochloric acids. Write the equations for the corresponding reactions.

    You have been given substances: calcium nitrate, sulfuric acid, caustic soda, potassium carbonate. How, using only these reagents, to obtain sodium nitrate in two ways? Write down the equations of the corresponding reactions.

    Sulfur and aluminum contained in the mixture interacted with each other. The reaction product was processed hot water. Part of the evolved gas was passed through chlorine water, the precipitate that formed was separated, and an excess of silver nitrate was added to the solution. 8.61 g of a white cheesy precipitate formed. Another part of the gas was passed through 145 ml of a 10% copper sulfate solution (density 1.1 g / ml), as a result of which the concentration of copper sulfate in the solution became 6.09%. Calculate the mass of sulfur that reacted. Write the equations for all reactions.

    Indicate what "lime" is known. Write them chemical composition and receiving equations.

    Calculate how many grams of chromium can be obtained by reacting chromium (III) oxide with silicon, the mass of which is 10 g. The product yield is 90%.

    Determine the oxidation states of each element and arrange the coefficients in the following schemes: a) Fe + FeI 3  FeI 2 ; b) H 2 S + I 2 + H 2 O  H 2 SO 4 + HI.

    In one collective farm, fertilizers were applied for spring wheat at the rate of: ammonium nitrate - 150 kg, superphosphate (containing 30% of assimilable P 2 O 5) - 300 kg and potassium chloride - 100 kg per hectare. Calculate how much this is in terms of nitrogen, calcium dihydrogen phosphate and potassium oxide.

    Indicate how the increase in pressure will affect the equilibrium in the systems: a) SO 2 (g) + Cl 2 (g) ⇄ SO 2 Cl 2 (g); b) H 2 (d) + Br 2 (d) ⇄ 2HBr (d).

    Show how to chemically remove corrosion products (aluminum oxide and hydroxide) from an aluminum product without damaging the metal. Write reaction equations.

    Show which of the mixtures: a) metal oxide and hydroxide; b) metal and metal oxide - when interacting with water, they give a solution of only one substance. Give examples, write reaction equations.

    Write the reaction equations that can be used to carry out the following transformations:

Ca  Ca(OH) 2  CaCO 3  CaO  Ca(OH) 2  CaCl 2  Ca.

Option 3

    Write ionic equations exchange reactions if:

    one of the resulting substances dissociates little into ions, the second is insoluble;

    one of the formed substances is soluble, the second is not;

    the reaction is reversible;

    one of the resulting substances is soluble, the second is released in the form of a volatile substance.

    Three test tubes contain dry substances: calcium oxide, aluminum oxide, phosphorus oxide. Show which reagents can be used to distinguish between these substances. Write reaction equations.

    Concentrated hydrochloric acid, water, manganese dioxide, copper and zinc were dispensed in the form of thin wires. How, using these substances, to obtain zinc chloride and copper (II) chloride in the form of crystalline hydrates? Describe the progress of the work, make equations chemical reactions, specify the conditions for their flow.

    When burning 0.896 l (n.a.) of a mixture of CO and CO 2 in excess oxygen, 0.112 l of oxygen was consumed, the resulting gas mixture was passed through a solution containing 2.96 g of slaked lime. Determine the composition of the initial gas mixture (in % by volume), as well as the composition and mass of the formed precipitate.

    Show how alkaline earth metal hydroxides can be obtained. What element hydroxide is the strongest alkali? Specify the technical names for calcium and barium hydroxides.

    In the industrial production of copper sulfate, copper scrap is oxidized when heated with atmospheric oxygen and the resulting copper (II) oxide is dissolved in sulfuric acid. Calculate the consumption of copper and 80% H 2 SO 4 per 1 ton of CuSO 4  5H 2 O if the product yield is 75%.

    Write the reaction equations: a) magnesium iodide with bromine; b) magnesium with a solution of hydrobromic acid. Indicate which element in each case is an oxidizing agent and which reducing agent, show the transition of electrons.

    In one collective farm, in addition to manure, the following masses of mineral fertilizers were applied for potatoes per hectare: granulated superphosphate containing 12.5% ​​of assimilable P 2 O 5 - 0.15 tons, ammonium nitrate - 0.1 tons and potassium chloride containing 90% KCl - 0.1 tons. Recalculate what masses of calcium hydrogen phosphate, nitrogen and potassium oxide this corresponds to.

    The reaction proceeds according to the equation: 2SO 2 (g) + O 2 (g) ⇄ 2SO 3 (l) + 284.2 kJ. By changing what parameters can the equilibrium be shifted towards the formation of sulfur oxide (VI)?

    Write the equations for all reactions that can occur when metal lithium and sodium are stored in air.

    Will sodium ions be preserved: a) during the interaction of sodium hydroxide with hydrochloric acid; b) in the interaction of sodium hydroxide with copper (II) chloride. Write the equations for the corresponding reactions.

    Write the reaction equations that can be used to carry out the following transformations:

Option 4

    Explain how they dissociate into ions aqueous solutions amphoteric bases. Give examples of such bases and show their dissociation into ions.

    Specify which one reagent can be used to recognize solutions of three substances: potassium chloride, aluminum chloride and magnesium chloride.

    The laboratory has iron, hydrochloric acid, caustic soda, calcium carbonate, copper oxide (II). Is it possible to obtain 12 new inorganic substances if these reagents and their interaction products are used as starting materials? Write the equations for the corresponding reactions.

    There is a mixture of nitrogen and hydrogen. Nitrogen was obtained by thermal decomposition of 12.8 g of ammonium nitrite, hydrogen - by "dissolving" 19.5 g of zinc in an excess of dilute sulfuric acid. Under appropriate conditions, the gases reacted, and then they were passed through 100 ml of a 32% sulfuric acid solution (density 1.22 g / ml). Determine which gas was in excess and what is the mass fraction (%) of salt in the solution. Assume that all reactions proceed in 100% yield.

    Indicate the composition of "caustic soda", "crystal soda", "soda ash", "drinking soda". Write down the reaction equations for their preparation.

    Calculate how much copper (in g) will be obtained from 500 g of chalcopyrite CuFeS 2 when it interacts with silicon (IV) oxide in an oxygen atmosphere. Product yield 75%.

    Specify which element is oxidized and which is reduced in the following reactions: a) MnS + HNO 3 (conc.)  MnSO 4 + NO 2 + H 2 O; b) Al + V 2 O 5  V + Al 2 O 3. Arrange the coefficients, indicate the transition of electrons.

    Explain why potassium nitrate is called a non-ballast fertilizer. Calculate the content of nutrients in it.

    Under certain conditions, the reaction of hydrogen chloride with oxygen is reversible: 4HCl (g) + O 2 (g) ⇄ 2Cl 2 (g) + 2H 2 O (g) + 116.4 kJ. Show what effect the following will have on the equilibrium state of the system: a) an increase in pressure; b) temperature increase; c) introduction of a catalyst.

    Explain the basis for the use of heated calcium to purify argon from oxygen and nitrogen impurities. Write reaction equations.

    Give examples of reactions in which processes occur, expressed by the following schemes: a) Al 0  Al 3+ ; b) Al 3+ + OH -  Al (OH) 3.

    Write the reaction equations that can be used to carry out the following transformations:

TASKS FOR 10 GRADE STUDENTS

EXERCISE 1

The task includes material on the chemistry of hydrocarbons of the limiting and unsaturated series: structure, isomerism and nomenclature, properties, production. The solution of calculation problems involves the use of the basic concepts of chemistry: relative atomic and molecular mass, mole, molar mass, molar volume, relative gas density.

Examples of problem solving

Example 1

    Give the IUPAC name for each of the following compounds:

A)
b)

Solution:

a) 2methyl3,3dimethylpentane;

b) cis-propylethylethylene (cis-heptene-3)

Example 2

Describe the stages of initiation, growth and termination of the chain reaction:

CH 3 CH 2 CH 3 + Br 2
CH 3 CHBrCH 3 +HBr.

Solution:

a) initiation:

b) chain growth:

CH 3 CH 2 CH 3 + Br  CH 3 HCH3 + HBr.

c) open circuit:

CH 3 HCH 3 + Br  CH 3 CHBrCH 3.

Example 3

Complete the equations for the following reactions and indicate the reactions proceeding according to the radical mechanism:

a) C 2 H 6 + Cl 2;

b) C 2 H 4 + HBr ;

c) H 2 C \u003d CH - CH 3 + HBr ;

d) C 3 H 8 + HNO 3 (razb.)
.

Name the reaction products.

Solution:

a) C 2 H 6 + Cl 2 C 2 H 5 Cl + HCl;

radical mechanism, reaction products: C 2 H 5 Cl - chloroethane and HCl - hydrogen chloride.

b) C 2 H 4 + HBr  C 2 H 5 Br;

reaction product: C 2 H 5 Br - bromoethane.

c) H 2 C \u003d CH - CH 3 + HBr  H 3 C - CHBr - CH 3;

reaction product: H 3 C - CHBr - CH 3 - 2-bromopropane.

d) C 3 H 8 + HNO 3 (razb.) H 3 C - CH (NO 2) - CH 3 + H 2 O;

radical mechanism, reaction products: H 3 C - CH (NO 2) - CH 3 - 2-nitropropane and H 2 O - water.

Example 4

Write a scheme of chemical reactions that make it possible to obtain chloroprene from methane:

.

Solution:

CH 4
CH3Cl
C 2 H 6 C 2 H 5 Cl C 4 H 10


.

Example 5

When passing 2 liters of a mixture of propane and propylene through liquid bromine, the mass of the flask with bromine increased by 1.1 g. Determine the volumetric composition of the mixture and the mass of the resulting products.

Given:

V(mixture) = 2 l

m(vials) = 1.1 g

M(Br 2) \u003d 160 g / mol

M(C 3 H 6 Br 2) = 202 g/mol

Find:

V(propane mixed)

V(propylene in mixture)

m(products)

Solution:

Let's find the amount of substance in the mixture, using the corollary from Avogadro's law:

1 mol - 22.4 l

n mol - 2 l

T(mixtures) = 0.0892 mol.

With bromine at n. y. only propylene reacts. Let's write the reaction equation:

C 3 H 6 + Br 2 \u003d C 3 H 6 Br 2.

Let it react X g of bromine, then the mass of the resulting dibromopropane is (1.1 + X). Then the amount of substance bromine is
, and the amount of substance dibromopropane is
. According to the reaction equation, 1 mol of bromine gives 1 mol of dibromopropane, therefore: chemist"

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  • Tasks for the school stage of the Olympiad for schoolchildren in chemistry Grade 11.

    Exercise 1.

    There are ten elements in the Periodic system, in the designation of which the second letter is "e". For each of these elements, characteristic reactions are given below. The capital letter is indicated by the sign "? ". Arrange the elements in their places.

    1) D + T → ? e+n;

    2) ? eO 4 → ? e + 2O 2 ;

    3) ? e + O 3, F 2, etc. → does not go;

    4) 2Au + 6H2? eO 4 (load) → 3 ? eO 2 + Au 2 (? eO 4 ) 3 + 6H 2 0;

    5) ? eCl 2 + 4NaOH → Na 2 [ ? e(OH) 4 ] + 2NaCl;

    6) ? eO 3 + 3H 2 O → H 6 ? eO 6 ;

    7) ? e(OH) 4 + 4HCl → ? eCl 3 + 1/2Cl 2 + 4H 2 0;

    8) ? eCl4 + Li → ? eH 4 + Li;

    9) ? eCl 2 + K 3 [ ? e(CN) 6 ] → K ? e[ ? e(CN) 6 ]↓ (blue precipitate);

    10) 2H? eO 4 + 7H 2 S → ? e 2 S 7 ↓ + 8H 2 0.

    Task 2.

    The chemist obtained samples of three silvery-white metals and found a way to quickly distinguish between them. To do this, he exposed the samples to the action of acids and sodium hydroxide solution. The results of his research are presented below.

    Symbols: "+" - the reaction is on, "-" - the metal does not react.

    Determine what metals could be obtained by the chemist and write the corresponding reaction equations.

    Task 3.

    A small piece of a plastic disposable cup was heated without air to 400°C. As a result of heating, a hydrocarbon was obtained X (carbon content 92.26% by mass, its vapor density in oxygen 3.25). It is known that during the oxidation of a hydrocarbon X With a solution of potassium permanganate in an acidic medium, benzoic acid is formed as the only organic product.

    1. Calculate the molecular formula X.

    2. Give the structural formula and the name of the hydrocarbon X . What is the name of the original polymer?

    3. Write the reaction equation (with all products and stoichiometric coefficients) for hydrocarbon oxidation X potassium permanganate solution, acidified with sulfuric acid.

    Task 4.

    The young chemist Petya received from his mother the task to buy 1 liter of food vinegar in the store (mass fraction of acetic acid CH3COOH 9%) for home canning. Arriving at the store, he discovered that only acetic essence was on sale (the mass fraction of acetic acid was 70%). Petya decided that he could make food vinegar out of it on his own. At home, in the reference book, he managed to find the density values ​​\u200b\u200bfor a 9% solution of acetic acid - 1.012 g / ml and for a 70% solution - 1.069 g / ml. Of the equipment, Petya has only measuring cylinders of various sizes.

    1. What safety rule should be observed when preparing dilute solutions from concentrated acids?
    2. Which of the substances available at home should Petya have on hand if acid gets on his skin? Name this substance and reflect its effect on the acid in the reaction equation.
    3. What volume of acetic essence should Petya measure in order to prepare 1 liter of 9% acetic acid solution?

    Task 5.

    The laboratory has iron, hydrochloric acid, caustic soda, calcium carbonate, copper oxide (II). Using these substances, as well as the products of their interaction, give at least 10 equations for the reactions of obtaining new inorganic substances.

    Task 6.

    When dissolving 2.8 g of an alloy of copper and silver in concentrated nitric acid, 5.28 g of a mixture of nitrates was formed. Determine the mass fractions of metals in the alloy.

    Answers grade 11

    Exercise 1.

    Answer:

    Points

    Even if the solver cannot determine all the elements by reactions, this can be done using the Periodic Table, given that the maximum oxidation state of an element cannot be greater than the group number.

    1) D + T → He + n;

    2) Xe O 4 → Xe + 2O 2;

    3) Ne + O 3, F 2, etc. → does not go;

    4) 2Au + 6H 2 Se O 4 (loaded) → 3 Se O 2 + Au 2 (Se O 4 ) 3 + 6H 2 0;

    5) BeCl 2 + 4NaOH → Na 2 [Be (OH) 4] + 2NaCl;

    6) Te O 3 + 3H 2 O → H 6 Te O 6;

    7) Ce (OH) 4 + 4HCl → Ce Cl 3 + 1/2Cl 2 + 4H 2 0;

    8) GeCl 4 + Li → Ge H 4 + Li;

    9) Fe Cl 2 + K 3 → K Fe [Fe (CN) 6 ]↓ (blue precipitate);

    10) 2H Re O 4 + 7H 2 S → Re 2 S 7 ↓ + 8H 2 0.

    Maximum score

    Task 2.

    By the totality of properties, i.e. according to the behavior in reactions with acids and sodium hydroxide, we conclude: I metal - silver (copper does not match in color), metal II - aluminum, metal III - zinc.
    Reaction equations:
    2Al + 6HCl = 2AlCl 3+3H2
    Zn + 2HCl = ZnCl 2 + 2H 2
    Ag + 2HNO 3 (conc.) = AgNO 3 + NO 2 + H 2 O
    Al + HNO 3 (conc.) No reaction
    Zn + 4HNO
    3 (conc.) + Zn(NO 3 )2 + 2NO 2 + 2H 2 O
    Zn + 2NaOH + 2H 2 O \u003d Na 2 + H 2
    2Al + 6NaOH + 6H 2 O = 2Na 3 + 3H 2 (7 points)

    Task 3.

    Answer:

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    1. Mr(X) = 3.25 32 = 104 a.m.u.

    Find the molecular formula of hydrocarbon X:

    C: H \u003d 0.9226 / 12.01: 0.0774 / 1.008 \u003d 1: 1, taking into account the molecular weight, we get C 8 H 8 .

    2. Since the oxidation of hydrocarbon X with a solution of potassium permanganate in an acidic medium produces benzoic acid (C 6 H 5 COOH), then its molecule contains a benzene ring with one substituent.

    Subtracting from the gross formula C 8 H 8 fragment C 6 H 5 , we get the substituent С 2 H 3 . The only possible substituent variant is vinyl, and the hydrocarbon X is styrene (vinylbenzene).

    Therefore, the polymer from which the disposable cup was made is polystyrene.

    3. Equation for the reaction of styrene oxidation with KMnO solution 4 , acidified H 2 SO 4 :

    2KMnO 4 + 3H 2 SO 4 → + CO 2 + 2 MnSO 4 + K 2 SO 4 + 4H 2 O

    All elements of the answer are written incorrectly

    Maximum score

    Task 4.

    1. Pour water into acid (1 point).
    2. Baking soda or sodium bicarbonate (1 point).

    NaHCO3 + CH3COOH = CH3COOHa + H2O neutralization reaction (2 points).

    1. The mass of acetic acid in a 9% solution was calculated - 91.08 g (1 point).

    The mass of the vinegar essence solution was calculated - 130.1 g (1 point).

    The volume of vinegar essence was calculated - 121.7 ml or ≈ 122 ml (1 point).

    Total: 7 points.

    Task 5.

    Possible answers:

    Fe + 2HCl \u003d FeCl 2 + H 2

    HCl + NaOH = NaCl + H2O

    2HCl + CaCO 3 \u003d CaCl 2 + H 2 O + CO 2

    CuO + 2HCl \u003d CuCl 2 + H 2 O

    2NaOH + CO 2 \u003d Na 2 CO 3 + H 2 O; NaOH + CO 2 \u003d NaHCO 3 + H 2 O

    CuO + H 2 \u003d Cu + H 2 O

    FeCl 2 + 2NaOH \u003d Fe (OH) 2 + 2NaCl

    3Fe + 4H 2 O \u003d Fe 3 O 4 + 4H 2

    CaCO 3 \u003d CaO + CO 2

    CuO + CO 2 \u003d CuCO 3

    Fe 3 O 4 + 8HCl \u003d FeCl 2 + 2FeCl 3 + 4H 2 O, etc.

    Task 6.

    1. Reaction equations:

    Cu + 4 HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

    Ag + 2HNO 3 \u003d AgNO 3 + NO 2 + H 2 O (1 point)

    1. We introduce the notation: n(Cu)=xmol n(Ag)= ymol; Then

    a) m(Cu)=64x, m(Ag)=108y

    m (mixtures)= 64x+108y=2.8

    b) according to equation (1) n(Cu(NO 3 ) 2 =x, m(Cu(NO 3 ) 2 = 188x;

    c) according to equation (2) n (AgNO 3 )=y, ​​m(AgNO 3 )= 170y

    d) m(mixture) = 188x+170y=5.28 (2 points)

    3) compose and solve the system of equations:

    64x+108y=2.8x=0.01mol Cu

    188x+170y=5.28 y=0.02mol Ag (2 points)

    4) we calculate the mass fractions of the components of the mixture:

    a) m (Cu) \u003d 0.01 * 64 \u003d 0.64 g. w(Cu)= 0.64/2.8= 0.2286 or 22.86%

    M (Ag) \u003d 0.02 * 108 \u003d 2.16 g. w(Ag)= 2.16/2.8= 0.7714 or 77.14% (2 points)

    Maximum score - 7 points

    Tasks for the school stage of the Chemistry Olympiad Grade 10

    Exercise 1.

    What gases can be obtained, having at its disposal the following substances:

    sodium chloride, sulfuric acid, ammonium nitrate, water, ammonium nitrite, hydrochloric acid, potassium permanganate, sodium hydroxide, aluminum carbide, calcium carbide, and sodium sulfite?

    Write all the equations of possible reactions, indicate the conditions for their occurrence.

    Task 2.

    For the three chemical elements A, B and C, it is known that they take part in the following transformations:

    1. C 2 + 3B 2 2CB 3
    2. 4CB 3 + 5A 2 4CA + 6V 2 A
    3. 4CB 3 + 3A 2 = 2C 2 + 6V 2 A
    4. C 2 + A 2 \u003d 2 CA
    5. 4CA 2 + 2B 2 A + A 2 \u003d 4VCA 3

    What elements are included in the task? Write down the reaction equations.

    Task 3.

    When dissolving 11.5 g of a mixture of aluminum, magnesium and copper in hydrochloric acid, 5.6 l (N.O.) of gas was released. The undissolved residue was brought into solution with concentrated nitric acid. At the same time, 4.48 liters (n.o.s.) of gas were released. Determine the mass fractions (in%) of the components of the initial mixture.

    Task 4.

    The reaction scheme is given:

    KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O

    1. Determine the oxidation states of the elements.

    2. Write an electronic balance equation

    3. Determine the oxidizing agent and reducing agent

    4. Arrange the coefficients in this equation.

    5. List the areas of application of the substance whose formula is KMnO 4

    Task 5.

    There are nine elements in the periodic system, whose names in Russian are nouns "not masculine". For each of these elements, characteristic reactions are given below, in which they are encrypted with the letter " E ". Define elements for each reaction:

    1) 2H 2 E + E O 2 → 3 E + 2H 2 O;

    2) E Cl 3 + KI → E Cl 2 + KCl + 1/2 I 2;

    3) E NO 3 + KCl → E Cl↓ + KNO 3;

    4) E + 3HNO 3 conc. + 4HCl conc. → H [ E Cl 4 ] + 3NO 2 + 3H 2 O;

    5) E Cl 2 + 4NH 3 → [ E (NH 3 ) 4 ]Cl 2 (blue);

    E (catalyst), 800°С

    300°C

    7) 2 E + O 2 2EO

    8) E Cl 2 + Cl 2 → E Cl 4;

    9) E 2 O 3 + 4 HNO 3 conc. + (x -2) H 2 O → E 2 O 5 x H 2 O + 4NO 2

    Task 6.

    As a result of the reaction of calcium phosphate with magnesium at

    heating produces two substances, one of which

    reacts with water, producing a colorless toxic

    garlic-smelling gas. Last oxidize

    air oxygen.

    Write the equations of all these chemical processes, name

    their products.

    Calculate the volume of air (N.O.) required to oxidize the gas,

    if 2.4 g of magnesium was used in the first of these reactions.

    Answers grade 10

    Exercise 1.

    1) 2NaCl (solid) + H 2 SO 4 (conc.) Na 2 SO 4 + 2HCl

    Or NaCl (solid) + H 2 SO 4 (conc.) NaHSO 4 + HCl

    1. NH 4 NO 3 \u003d N 2 O + 2H 2 O
    2. NH 4 NO 2 \u003d N 2 + 2H 2 O
    3. 2KMnO 4 \u003d K 2 MnO 4 + MnO 2 + O 2
    4. 2KMnO 4 + 16HCl = 2KCl + 2MnCl 2 + 5Cl 2 + 8H 2 O

    or 2NaCl + 2H 2 O 2NaOH + H 2 + Cl 2

    1. Al 4 C 3 + 12H 2 O \u003d 4Al (OH) 3 + 3CH 4
    2. CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
    3. 2H 2 O 2H 2 + O 2
    4. NH 4 NO 3 + NaOH \u003d NaNO 3 + NH 3 + H 2 O
    5. Na 2 SO 3 + H 2 SO 4 \u003d Na 2 SO 4 + H 2 O + SO 2

    Total: 10 points

    Task 2.

    1. N 2 + 3H 2 2NH 3
    2. 4NH 3 + 5O 2 4NO + 6H 2 O
    3. 4NH 3 + 3O 2 \u003d 2N 2 + 6H 2 O
    4. N 2 + O 2 \u003d 2NO
    5. 4NO 2 + 2H 2 O + O 2 \u003d 4HNO 3

    A is oxygen, B is hydrogen, C is nitrogen.

    1 point for each reaction equation

    Total: 5 points

    Task3.

    1) Copper does not interact with hydrochloric acid.

    4.48 l

    Cu + 4 HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

    n(NO 2 ) = 4.48/22.4 = 0.2 mol;

    N(Cu) = 0.1 mol; m(Cu) = 64 x 0.1 = 6.4 g 1 point

    X mol 1.5 x mol

    2) 2Al + 6HCl = 2AlCl 3 + 3H 2

    Y mol y mol

    Mg + 2HCl = MgCl 2 + H 2 2 points

    3) Determination of the amount of hydrogen substance: n (H 2 ) = 5.6 / 22.4 = 0.25 mol; 1 point

    4) A system has been compiled with two unknowns:

    x + 1.5y = 0.25

    24x + 27y = 5.1 2 points

    5) The system with two unknowns is solved (the values ​​"x" and "y" are determined) 2 points

    X = 0.1; y = 0.1

    6) The masses and mass fractions of the mixture components are determined.

    M(Al) = 2.7g; m(Mg) = 2.4g 1 point

    ώ (Al) = 2.7 / 11.5 = 0.2348 or 23.48%

    ώ(Mg) = 2.4/11.5 = 0.2087 or 20.87%. 1 point

    ώ (Cu) = 6.4/11.5 = 0.5565 or 55.65%

    Total: 10 points

    Task 4.

    1) The oxidation states of the elements are determined

    1-1 +1 +7 -2 +1 +6 -2 0 +1 +6 -2 +2 +6 -2 +1 -2

    KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O 1 point

    2) The equation of the electronic balance was compiled and the oxidizing agent and

    Reducing agent:

    2 points

    3) The coefficients are placed in the reaction equation:

    2KMnO 4 + 10KI + 8H 2 SO 4 = 5I 2 + 6K 2 SO 4 + 2MnSO 4 + 8H 2 O 2 points

    4) Applications of potassium permanganate are listed:

    (laboratory method for obtaining oxygen and chlorine, chemical analysis(permanganatometry), oxidation of organic substances, etc.) only 2 points

    Total: 7 points

    Task 5.

    Answer:

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    These elements are sulfur, iron, silver, gold, copper, platinum, mercury, tin and antimony. All of them, except for sulfur, are metals, and all of them are easily recognizable by their characteristic transformations, or by their oxidation states, which are characteristic only for their position in the Periodic Table.

    1) 2H 2 S + SO 2 → 3S + 2H 2 O;

    2) FeCl 3 + KI → FeCl 2 + KCl +1/2 I 2;

    3) AgNO 3 + KCl → AgCl↓ + KNO 3;

    4) Au + 3HNO 3 conc. + 4HCl conc. → H + 3NO 2 + 3H 2 O;

    5) CuCl 2 + 4NH 3 →Cl 2 (blue);

    Pt (catalyst), 800°C

    6) 4NH 3 + 5O 2 4NO + 6H 2 O;

    300°C 400°C

    7) 2Hg + O 2 2Hg O 2Hg + O 2;

    8) SnCl 2 + Cl 2 → SnCl 4;

    9) Sb 2 O 3 + 4 HNO 3 conc. + (x-2) H 2 O → Sb 2 O 5 x H 2 O + 4NO 2

    All elements of the answer are written incorrectly

    Maximum score

    Task 6.

    Ca 3 (PO 4 ) 2 + 8 Mg \u003d Ca 3 P 2 + 8 MgO (calcium phosphide) (1 point)

    Ca 3 P 2 + 6 H 2 O \u003d 3 Ca (OH) 2 + 2 PH 3 (d) (phosphine) (1 point)

    2 PH 3 + 4 O 2 \u003d 2 H 3 PO 4 (ortho-phosphoric acid) (1 point)

    n (Mg) = 0.1 mol. n(О 2 ) = 0.05 mol. V (O 2) \u003d 1.12 l. (2 points)

    V (air containing 21% oxygen) = 5.3 liters.

    Maximum score - 5 points

    Tasks for the school stage of the Chemistry Olympiad Grade 9

    Exercise 1.

    Task 2.

    Task 3.

    When solving problems, we often encounter the expression of the molar mass of a substance in terms of the molar mass of another substance, most often air and hydrogen.

    Imagine that you have discovered a new planet. An analysis of the composition of the atmosphere of this planet is given below: 48% N 2, 19% O 2, 15% Cl 2 , 8% Ne, 7% Ar, 2.9% Kr and 0.1% CO 2 .

    Calculate the molar mass of your planet's air and the density of each component of the atmosphere from that planet's air.

    Task 4.

    Task 6.

    To obtain a solution of sodium chloride, the mass of sodium carbonate calculated in advance was dissolved in 5.0% hydrochloric acid.

    Determine the mass fraction of sodium chloride in the resulting solution.

    Answers to assignments Grade 9

    Exercise 1.

    In bottles without labels are the following substances: dry silver nitrate, sodium sulfide, calcium chloride; silver and aluminum powders, as well as hydrochloric and concentrated nitric acids. At your disposal are water, a burner and any number of test tubes.

    Make up the reaction equations and indicate the signs by which each of the indicated substances can be determined.

    Number of points -10

    Answer:

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    1) AgNO 3 + HCl \u003d AgCl ↓ + HNO 3

    White

    2) CaCl2 + 2AgNO3 = 2AgCl↓ + Ca(NO3 ) 2

    White

    3) Na2 S + 2HCl = 2NaCl + H2 S

    The smell of rotten eggs

    4) 2Al + 6HCl = 2AlCl3 + 3H2

    Colorless, odorless

    5) Ag + 2HNO3 = AgNO3 + NO2 + H2 O

    Brown, with a pungent odor

    6) Na2 S+4HNO3 = 2NaNO3 + S↓ + 2NO2 + 2H2 O

    Yellow

    All elements of the answer are written incorrectly

    Maximum score

    Task2.

    When sulfur oxide (IV) was passed through a solution of potassium permanganate, a solution was formed in which the mass fraction of sulfuric acid is 5%. Calculate the mass fractions of the remaining reaction products in the resulting solution.

    Number of points -10

    Answer:

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    1) Compose the reaction equation:

    5SO2 + 2KMnO4 + 2H2 O=K2 SO4 + 2MnSO4 + 2H2 SO4

    2) Find the mass of 2mol H2 SO4 – m(H2 SO4 ) = 2mol ∙ 98g/mol = 196g.

    3) Find the mass of 2mol MnSO4 – m(MnSO4 ) \u003d 2mol ∙ 151g / mol \u003d 302g.

    4) Find the mass of 1mol K2 SO4 – m(K2 SO4 ) \u003d 1 mol ∙ 174 g / mol \u003d 174 g.

    5) The mass fraction of each substance is equal to: ω (in-va) \u003d m (in-va): m (p-ra). Since all these substances are in the same solution (i.e., the mass of the solution is the same for them), the ratio of their mass fractions is equal to the mass ratio: ω (K2 SO4 ) : ω(H2 SO4 ) = m(K2 SO4 ) : m(H2 SO4 ) = 174: 196;

    whence ω (K2 SO4 ) = 0.05 ∙ (174: 196) = 0.04 or 4.4%.

    6) ω(MnSO4 ) : ω(H2 SO4 ) = m(MnSO4 ) : m(H2 SO4 ) = 302: 196, whence

    ω(MnSO4 ) = 0.05 ∙ (302:196) = 0.077 or 7.7%.

    All elements of the answer are written incorrectly

    Maximum score

    Task 3.

    M (planet air) \u003d 0.48 * 28 + 0.19 * 32 + 0.15 * 71 + 0.08 * 20 + 0.07 * 40 + 0.029 * 83 + 0.001 * 44 \u003d

    37.021g/mol. (2 points)

    D(N2 )=28/37.021=0.7563

    D(O2 )=0.8644; D(Cl2 )=1.9178; D(Ne)=0.5402; D(Ar)=1.0805; D(Kr)=2.2420; D(CO2 )=1.1858. (8 points)

    Maximum score - 10

    Task 4.

    Consider a sample of ethyl mercaptan weighing 100 g.
    In such a sample, m(C) = 38.7 g, m(H) = 9.6 g, m(S) = 51.7 g.(1 point)
    n (C) \u003d 38.7 g: 12 g / mol \u003d 3.2 mol(1 point)
    n(H) = 9.6 g: 1 g/mol = 9.6 mol
    (1 point)
    n(S) = 51.7 g: 32 g/mol = 1.6 mol
    (1 point)
    n(C) : n(H) : n(S)= 3.2: 9.6: 1.6 = 2:6:1 => C2H6S(2 points)
    Maximum score - 6

    Task 5.

    3Fe + 2O2 = Fe3 O4
    Fe
    3 O4 + 8HCl = FeCl2 + 2FeCl3 + 4H2 O
    FeCl
    2 + 2NaOH = Fe(OH)2 ↓ + 2NaCl
    FeCl
    3 + 3NaOH = Fe(OH)3 ↓ + 3NaCl
    2Fe(OH)
    3 + Fe(OH)2 → Fe3 O4 + 4H2 O (t°, inert atmosphere)
    Substances: A - Fe
    3 O4 ; B and C - FeCl2 or FeCl3 ; D and E - Fe(OH)2 or Fe(OH)3 .

    Maximum score -5

    Task 6.

    106 73 117 44

    Na2 CO3 + 2 HCl = 2 NaCl + CO2 + H2 O

    X 5 Z Y

    Take 100 g of 5% hydrochloric acid.

    X = (106 * 5) / 73 = 7.26 g Na reacts with 5 g HCl2 CO3 , while the solution is released

    Y \u003d (44 * 5) / 73 \u003d 3.01 g CO2 You get sodium chloride Z = (117 * 5) / 73 = 8.01 g

    Mass fraction of NaCl 100% (8.01 / (100 + 7.26 - 3.01)) =7,7 %

    Maximum score -6 points


    8th grade

    Which chemical elements named after countries? Give at least four names.

    Which element was first discovered in the sun?

    Indicate the number of protons and neutrons contained in the nuclei of atoms, the elements you named.

    Number of points -10

    A rain drop has a mass of about 10 -4 g. Calculate the number of water molecules and the total number of atoms of all elements contained in this drop.

    Number of points - 10

    What is percentage isotopes 35 Cl and 37 Cl in natural chlorine having a relative molecular weight of 70.90?

    Number of points - 10

    You have been given a mixture of the following substances: iron, soot, copper, chalk, salt.

      Suggest a plan for separating these substances using water and hydrochloric acid.

      What laboratory equipment would be required to separate this mixture?

      Write the reaction equations that will be used in the separation.

      Calculate the mass of chalk in the mixture from the volume of gas released in 5.6 liters.

    Number of points - 20

    Specify the chemical formulas of gases: nitrogen, hydrogen chloride, hydrogen, ammonia, chlorine, carbon monoxide, hydrogen sulfide, carbon dioxide. Which of these gases are simple substances, oxides, have a color, a characteristic smell, are poisonous? Write your answer in the form of a table using the signs "+" and "-".

    Index

    gases

    sulphurous

    carbon dioxide

    Chemical

    substance

    Characteristic odor

    Number of points - 10

    Olympiad tasks of the school stage of the All-Russian Olympiad

    schoolchildren in chemistry 2011-2012

    Grade 9

    9- 1. Write the reaction equations for the following transformations:

    Zn → ZnS → H 2 S → S → SO 2 → SO 3 → H 2 SO 4 → BaSO 4

    Specify the reaction conditions; consider one of the reactions as redox.

    Number of points - 10

    9-2. A sample of technical iron (II) sulfide weighing 5 g, containing 5% metallic iron, interacts with hydrochloric acid. Calculate the volume of evolved gaseous products (at N.C.) and the volumetric composition of the gas mixture.

    9-3 . With partial dissolution of 4 g of a mixture of copper and magnesium filings in an excess of hydrochloric acid, 1.12 liters of hydrogen (n.o.) were released. Set the composition of the initial mixture in percent by weight

    Number of points - 10

    9- 4. liquid waste laboratory work containing acids must be neutralized with alkali or soda.

    1. Determine the masses of sodium hydroxide and sodium carbonate required to neutralize waste containing 0.60 mol of hydrochloric acid.

    2. What volume of gas (n.o.s.) will be released when the specified amount of waste is neutralized?

    3. How much silver nitrate would be required to precipitate the chloride ions contained in 0.6 mol of hydrochloric acid?

    Number of points -20

    9-5. It is known that four test tubes contain solutions of nitric acid, potassium carbonate, silver nitrate and barium chloride. How, without using other reagents, to determine the contents of each test tube? Make a plan for the experiment and write the reaction equations.

    Number of points - 10

    Olympiad tasks of the school stage of the All-Russian Olympiad

    schoolchildren in chemistry 2011-2012

    Grade 10

    When dissolving 11.5 g of a mixture of aluminum, magnesium and copper in hydrochloric acid, 5.6 l (N.O.) of gas was released. The undissolved residue was brought into solution with concentrated nitric acid. At the same time, 4.48 liters (n.o.s.) of gas were released. Determine the mass fractions (in%) of the components of the initial mixture.

    Quantitypoints – 10.

    The reaction scheme is given:

    KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O

    1. Determine the oxidation states of the elements.

    2. Write an electronic balance equation

    3. Determine the oxidizing agent and reducing agent

    4. Arrange the coefficients in this equation.

    5. List the areas of application of the substance, the formula of which is KMnO 4

    Quantitypoints-7.

    Write the reaction equations corresponding to the transformation scheme:

    t°, Pt KMnO 4 , H 2 O ex. HBr 2KOH(alcohol), t°

    C 2 H 5 Cl → C 3 H 8 ───→ X 1 ─────→ X 2 ─────→ X 3 ─────────→ X 4

    Quantitypoints – 10.

    A mixture of alkene and hydrogen with a total volume of 26.88 l (n.o.) was passed over a platinum catalyst at 200°C. In this case, 75% of the alkene reacted, and the volume of the mixture decreased to 20.16 L (N.O.). When the initial mixture was passed through a flask with bromine water, the entire hydrocarbon reacted, and the mass of the flask increased by 16.8 g. Determine the composition of the initial mixture (in % by volume) and the structure of the initial alkene.

    Number of points - 10

    10-5. Four unlabeled tubes contain aqueous solutions of sodium hydroxide, hydrochloric acid, potash, and aluminum sulfate. Suggest a way to determine the contents of each tube without using additional reagents.

    Number of points - 10

    Olympiad tasks of the school stage of the All-Russian Olympiad

    schoolchildren in chemistry 2011-2012

    Grade 11

    11-1. The laboratory has iron, hydrochloric acid, caustic soda, calcium carbonate, copper oxide (II). Using these substances, as well as the products of their interaction, give at least 10 equations for the reactions of obtaining new inorganic substances

    Number of points - 10

    11-2. How many isomers does C 5 H 12 have. write them down structural formulas and give each substance names according to substitutive nomenclature. Which of these isomers has the highest boiling point? Calculate the relative vapor density of this compound in air.

    Number of points - 7

    In 264 g of a solution with a mass fraction of Hg 2 (NO 3) 2 equal to 20% placed iron filings. After some time, the mass fraction of mercury (I) nitrate in the solution became equal to 6% .

    1. What mass of mercury was obtained as a result of the reaction?

    2. What is the mass fraction of iron nitrate in the resulting solution?

    Number of points - 20

    11-4. Propionic acid is contaminated with formic acid and propyl alcohol. When an excess of potassium bicarbonate was added to 150 g of this acid, 44.8 liters of gas (n.o.) were released. When an excess of ammonia solution of silver oxide was added to the same amount of acid, a precipitate was formed weighing 2.16 g.

    1. Give the equations of all reactions.

    2. Determine the mass fractions of impurities in the acid.

    Number of points - 20

    11-5. The test tubes contain concentrated solutions of the following acids: oxalic, sulfuric, hydroiodic, phosphorous, nitric, formic.

    1. How, without involving any other reagents, to determine the contents of each tube?

    2. Write the reaction equations.

    3. Specify signs of reactions

    Number of points - 20

    Answers to the tasks of the school stage of the All-Russian Olympiad

    schoolchildren in chemistry

    2011

    8th grade

    Points

    Ruthenium (Ru) - named after Russia;

    44 protons, 57 neutrons.

    Polonium (Po) - in honor of Poland;

    84 protons, 37 neutrons.

    Francium (Fr) - in honor of France;

    87 protons, 35 neutrons.

    Germanium (Ge) - in honor of Germany;

    32 protons, 40 neutrons.

    Helium (He) - exposed to the sun;

    2 protons, 2 neutrons.

    Maximum score

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    Calculate the number of moles of water -

    n(H 2 O) \u003d m (H 2 O) : M (H 2 O) \u003d 10 -4 g: 18 g / mol \u003d 5.56 10 -6 mol.

    One mole of water contains N A = 6.02 10 23 water molecules.

    Then 5.56 10 -6 mol contains the number of water molecules equal to

    N (H 2 O) \u003d N A n (H 2 O) \u003d 6.02 10 23 5.56 10 -6 \u003d 33.5 10 17 (molecules)

    Water is made up of three atoms: two hydrogen atoms and one oxygen atom.

    The total number of all atoms contained in a raindrop is

    N ∑ \u003d 3N (H 2 O) \u003d 3 33.5 10 17 \u003d 100.4 10 17 \u003d 10 19 (atoms).

    All elements of the answer are written incorrectly

    Maximum score

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    We take for X the percentage of the isotope 35 Cl.

    Then the percentage of the isotope 37 Cl will be (100 - X)

    The mass of atoms of the isotope 35 Cl is 35X.

    The mass of atoms of the isotope 37 Cl is 37 (100 - X)

    Let's make the equation: 35X + 37 (100 - X) \u003d 35.45

    ω(35 Cl) = 77.5%, ω(37 Cl) = 22.5%.

    All elements of the answer are written incorrectly

    Maximum score

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

      We separate the iron with a magnet.

    The remaining mixture is placed in water - table salt will dissolve,

    soot will be on the surface. The solution is filtered. The soot will remain on the filter.

    We evaporate the filtrate, it will be NaCl.

    We will treat copper and chalk with hydrochloric acid.

    Chalk (CaCO 3) dissolve,

    but copper remains.

      Separation required the following equipment: a magnet, a filter device (a stand with a ring, a funnel, a filter, a glass rod, a filtrate collector (beaker)),

    porcelain cup for evaporation, electric stove,

    beaker for dissolution.

    3) CaCO 3 + 2HCl = CaCl 2 + H 2 O + CO 2

      Let's find the amount of CO 2 gas substance:

    n(CO 2) \u003d 5.6 l: 22.4 mol / l \u003d 0.25 mol;

    n (CaCO 3) \u003d n (CO 2) \u003d 0.25 mol;

    m (CaCO 3) \u003d 0.25 mol 100 g / mol \u003d 25 g.

    Maximum score

    Index

    sulphurous

    carbon dioxide

    Chemical

    substance

    Characteristic odor

    Grading Guidelines

    Points

    Chemical formula (8 1 point)

    Simple substance (3 0.1 points)

    Oxide (2 0.1 points)

    Characteristic odor (4 0.125 points)

    Poisonous (5 0.1 points)

    All items are incorrect

    Maximum score

    Grade 9

    9-2. 1. Under the action of acid on technical iron sulfate (II) with an admixture of iron, the following reactions occur:

    FeS + 2HCl = FeCl 2 + H 2 S (1 point)

    Fe + 2HCl = FeCl 2 + H 2 (1 point)

    2. The amount of FeS and Fe in the original sample (95% and 5%, respectively) is equal to:

    n (FeS) \u003d 5 0.95 / 88 \u003d 5.4 10 -2 mol (1 point)

    n (Fe) \u003d 5 ∙ 0.05 / 56 \u003d 4.48 ∙ 10 -3 mol (1 point)

    3. The amount of gaseous reaction products H 2 S and H 2 obtained from the original sample, in accordance with the given chemical equations, will be:

    n(H 2 S) \u003d 5.4 ∙ 10 -2 mol

    n (H 2) \u003d 4.48 ∙ 10 -3 mol (2 points)

    4. Find the volume of released gaseous products:

    V (H 2 S) \u003d 5.4 ∙ 10 -2 mol ∙ 22.4 \u003d 1.21 (l)

    V (H 2) \u003d 4.48 ∙ 10 -3 mol ∙ 22.4 \u003d 0.1 (l) (2 points)

    5. Calculate the volumetric composition of the gas mixture:

    Vtotal \u003d 1.21 + 0.1 \u003d 1.31 (l)

    φ (H 2 S) \u003d 1.21 / 1.31 \u003d 0.9237 or 92.37%

    φ (H 2) \u003d 0.1 / 1.31 \u003d 0.0763 or 7.63% (2 points)

    Total: 10 points

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    Let's compose the reaction equations:

    1) HCl + NaOH \u003d NaCl + H 2 O (1)

    2HCl + Na 2 CO 3 \u003d 2NaCl + H 2 O + CO 2 (2)

      To neutralize 0.60 mol HCl according to the equation

    reaction (1) requires 0.60 mol of NaOH, since n(HCl) = n(NaOH);

    n (Na 2 CO 3) \u003d 1 / 2n (HCl) \u003d 0.60 mol: 2 \u003d 0.30 mol - equation (2).

    m(NaOH) = 0.60 mol 40 g/mol = 24 g;

    m (Na 2 CO 3) \u003d 0.30 mol 106 g / mol \u003d 31.8 g.

    3) Let us calculate the volume of carbon dioxide released during neutralization according to reaction (2):

    n (CO 2) \u003d 1 / 2n (HCl) \u003d 0.60 mol: 2 \u003d 0.30 mol;

    V (CO 2) \u003d n (CO 2) V M \u003d 0.30 mol 22.4 l / mol \u003d 6.72 l.

    4) Cl - + AgNO 3 = AgCl↓ + NO 3 -

    n (AgNO 3) \u003d n (Cl -) \u003d 0.60 mol;

    m (AgNO 3) \u003d 0.60 mol 170 g / mol \u003d 102 g

    Maximum score

    Grade 10

    1) Copper does not interact with hydrochloric acid.

    Cu + 4 HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

    n (NO 2) \u003d 4.48 / 22.4 \u003d 0.2 mol;

    n(Cu) = 0.1 mol; m(Cu) = 64 x 0.1 = 6.4 g 1 point

    x mol 1.5 x mol

    2) 2Al + 6HCl = 2AlCl 3 + 3H 2

    y mol y mol

    Mg + 2HCl = MgCl 2 + H 2 2 points

    3) Determination of the amount of hydrogen substance: n (H 2) \u003d 5.6 / 22.4 \u003d 0.25 mol; 1 point

    4) A system has been compiled with two unknowns:

    24x + 27y = 5.1 2 points

    5) The system with two unknowns is solved (the values ​​"x" and "y" are determined) 2 points

    x = 0.1; y = 0.1

    6) The masses and mass fractions of the mixture components are determined.

    m(Al) = 2.7 g; m(Mg) = 2.4g 1 point

    ώ (Al) = 2.7 / 11.5 = 0.2348 or 23.48%

    ώ(Mg) = 2.4/11.5 = 0.2087 or 20.87%. 1 point

    ώ (Cu) = 6.4/11.5 = 0.5565 or 55.65%

    Total: 10 points

    1) The oxidation states of the elements are determined

    1-1 +1 +7 -2 +1 +6 -2 0 +1 +6 -2 +2 +6 -2 +1 -2

    KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O 1 point

    2) The equation of the electronic balance was compiled and the oxidizing agent and

    reducing agent:

    3) The coefficients are placed in the reaction equation:

    2KMnO 4 + 10KI + 8H 2 SO 4 = 5I 2 + 6K 2 SO 4 + 2MnSO 4 + 8H 2 O

    4) Applications of potassium permanganate are listed:

    (laboratory method for obtaining oxygen and chlorine, chemical analysis (permanganatometry), oxidation of organic substances, etc.) only 2 points

    Total: 7 points

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    1. C 2 H 5 Cl + CH 3 Cl + Mg → MgCl 2 + CH 3 -CH 2 -CH 3 (+ C 4 H 10 + C 2 H 6)

    2. CH 3 -CH 2 -CH 3 ───→ H 3 C-CH \u003d CH 2 + H 2

    3. H 3 C-CH \u003d CH 2 + 2KMnO 4 + 4H 2 O → 3H 3 C-CH (OH) -CH 2 OH + 2MnO 2 + 2KOH

    4. H 3 C-CH(OH)-CH 2 OH + 2HBr → 2H 2 O + H 3 C-CH(Br)-CH 2 Br

    5. H 3 C-CH (Br) -CH 2 Br + 2 KOH (alcohol) → H 3 C-C≡CH + 2KBr + 2H 2 O

    All elements of the answer are written incorrectly

    Maximum score

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    C n H 2n + H 2 = C n H 2n+2

    2 mol of gases (hydrocarbon and hydrogen) enter into the reaction, and one mol (alkane) is formed. Thus, the decrease in the volume of the mixture is equal to the volume of hydrogen that reacted, or the volume of the reacted alkene.

    This volume is 26.88 - 20.16 \u003d 6.72 (l), that is, 0.3 mol.

    Since 75% of the alkene reacted, it was 0.4 mol in total.

    When passing through bromine water the mass of the flask increased by the mass of the hydrocarbon, i.e. 0.4 mol of alkene has a mass of 16.8 g.

    Molar mass 16.8 / 0.4 \u003d 42 (g / mol).

    Alkene with such a molar mass - C 3 H 6, propene: H 3 C-CH \u003d CH 2

    The composition of the mixture: 0.4 mol of propene occupy a volume of 8.96 liters.

    This is 33% (1/3) of the volume of the mixture.

    The rest - 67% (2/3) - hydrogen.

    All elements of the answer are written incorrectly

    Maximum score

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    Let's make a table of possible pairwise interactions of substances, as a result of which we will (or will not) observe certain signs of reactions.

    Substances

    4. Al 2 (SO 4) 3

    Overall result observations

    Al(OH)3
    CO2

    1 sediment and

    4. Al 2 (S0 4) 3

    A1(OH) 3
    CO2

    2 draft and

    NaOH + HCl \u003d NaCl + H 2 O

    K 2 CO 3 + 2HC1 \u003d 2KS1 + H 2 O + CO 2

    3K 2 CO 3 + Al 2 (SO 4) 3 + 3H 2 O \u003d 2Al (OH) 3 + 3CO 2 + 3K 2 SO 4;

    Al 2 (SO 4) 3 + 6NaOH \u003d 2Al (OH) 3 + 3Na 2 SO 4

    Al(OH) 3 + NaOH + 2H 2 O = Na

    (the presence of sediment depends on the order of draining and excess and alkali)

    Based on the presented table, all substances can be determined by the number of precipitation and gas evolution.

    All elements of the answer are written incorrectly

    Maximum score

    Grade 11

    Possible answers:

    Fe + 2HCl \u003d FeCl 2 + H 2

    HCl + NaOH = NaCl + H2O

    2HCl + CaCO 3 \u003d CaCl 2 + H 2 O + CO 2

    CuO + 2HCl \u003d CuCl 2 + H 2 O

    2NaOH + CO 2 \u003d Na 2 CO 3 + H 2 O; NaOH + CO 2 \u003d NaHCO 3 + H 2 O

    CuO + H 2 \u003d Cu + H 2 O

    FeCl 2 + 2NaOH \u003d Fe (OH) 2 + 2NaCl

    3Fe + 4H 2 O \u003d Fe 3 O 4 + 4H 2

    CaCO 3 \u003d CaO + CO 2

    CuO + CO 2 \u003d CuCO 3

    Fe 3 O 4 + 8HCl \u003d FeCl 2 + 2FeCl 3 + 4H 2 O, etc.

    CH 3 -CH 2 -CH 2 -CH 2 -CH 3 - Pentane (0.5+0.5) = 1 point

    CH 3 -CH-CH 2 -CH 3 - 2-methylbutane (0.5 + 0.5) = 1 point

    CH 3 -C-CH 3 - 2,2-dimethylpropane (0.5 + 0.5) = 1 point

    Pentane has a higher boiling point. the length of the molecule is greater and therefore the intermolecular forces are also greater (1+1 = 2 points)

    D air = M C 5 H 12 / M air M C 5 H 12 = 72 g/mol M air = 29 g/mol (1 point)

    D air = 72/29 = 2.48 (1 point) Total 2 points per task

    Total: 7 points

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    Hg 2 (NO 3) 2 + Fe \u003d Fe (NO 3) 2 + 2Hg ↓ (1)

    M (Fe (NO 3) 2 \u003d 180 g / mol; M (Hg) \u003d 201 g / mol; M (Fe) \u003d 56 g / mol;

    M (Hg 2 (NO 3) 2 \u003d 526 g / mol.

    Let's find the mass of Hg 2 (NO 3) 2 in the initial solution: m (Hg 2 (NO 3) 2) = 0.2 264 = 52.8 (g)

    The mass of the solution changes during the reaction. The mass of the solution increases by the mass of iron that has reacted, and decreases by the mass of mercury that precipitates.

    Let X g of iron reacted. Let us find the mass of the obtained mercury according to the reaction equation (1):

    56 g Fe - 2 201 g Hg m 1 (Hg) \u003d 7.18 X

    X– m 1

    The mass of the resulting solution: m (p-pa) \u003d 264 + X– 7.18X = 264 – 6.18 X(G)

    Find the mass of Hg 2 (NO 3) 2 in the resulting solution:

    m (Hg 2 (NO 3) 2) \u003d 0.06 (264 - 6.18 X) = 15,84 – 0,37X

    Find the mass of Hg 2 (NO 3) 2 that reacted:

    m (Hg 2 (NO 3) 2) = 52.8 - (15.84 - 0.37 X) = 36,96 + 0,37X

    Let's find the value X according to the reaction equation (1), solving the proportion:

    56 g Fe react with 526 g Hg 2 (NO 3) 2

    X - (36.96 + 0.37 X) X= 4.1; m(Fe) = 4.1 g.

    The mass of mercury obtained is 29.4 g(7.18 4.1)

    The mass of the resulting solution is 238.7 g (264 - 6.18 4.1)

    Find the mass of the resulting iron(II) nitrate:

    56 g Fe - 180 g Fe (NO 3) 2

    4.1 g - X X= 13.18; m (Fe (NO 3) 2) = 13.18 g.

    Find the mass fraction of iron(II) nitrate in the resulting solution:

    ω (Fe (NO 3) 2) \u003d m (Fe (NO 3) 2): m (p-pa) \u003d 13.18: 238.7 \u003d 0,055 (5,5 %)

    All elements of the answer are written incorrectly

    Maximum score

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    C 2 H 5 COOH + KHCO 3 \u003d C 2 H 5 COOK + H 2 O + CO 2 (1)

    HCOOH + KHCO 3 \u003d NSOOK + H 2 O + CO 2 (2)

    HCOOH + Ag 2 O \u003d 2Ag ↓ + H 2 O + CO 2 (3)

    M(C 2 H 5 COOH) = 74 g/mol; M(HCOOH) = 46 g/mol; M(Ag) = 108 g/mol

    By the mass of the selected silver (equation 3), we find the mass of formic acid:

    45 g HCOOH - 2 108 g Ag X = 0.46 g; m(HCOOH) = 0.46 g.

    Let's find how much CO 2 is released during the interaction of potassium bicarbonate with formic acid (equation 2):

    46 g HCOOH - 22.4 l CO 2 X = 0.224 l; V (CO 2) \u003d 0.224 l.

    Therefore, when interacting with propionic acid (equation 1), 44.576 liters of CO 2 were released (44.8 - 0.224)

    Find the mass of propionic acid:

    74 g C 2 H 5 COOH - 22.4 l CO 2 X = 147.26 g; m (C 2 H 5 COOH) = 147.26 g.

    X - 44.576 l

    Find the mass of propyl alcohol:

    m (C 3 H 7 OH) \u003d 150 - m (C 2 H 5 COOH) - m (HCOOH) \u003d 150 -147.26 - 0.46 \u003d 2.28 (g)

    The mass fraction of formic acid in a solution of propionic acid is:

    ω(HCOOH) = m(HCOOH):m(mixtures) = 0.46: 150 = 0.0031 (0.31%)

    The mass fraction of propyl alcohol in a solution of propionic acid is:

    ω (C 3 H 7 OH) \u003d m (C 3 H 7 OH): m (mixtures) \u003d 2.28: 150 \u003d 0.0152 (1.52%)

    All elements of the answer are written incorrectly

    Maximum score

    11-5.

    (Other formulations of the answer are allowed that do not distort its meaning)

    Points

    Concentrated sulfuric acid as a strong oxidizing agent will interact with hydroiodic acid, which exhibits reducing properties:

    8HI + H 2 SO 4 \u003d 4I 2 ↓ + H 2 S + 4H 2 O

    A brown precipitate of iodine precipitates and hydrogen sulfide is released with an unpleasant smell of rotten eggs.

    A specific property of formic acid is its ability to decompose into carbon monoxide (II) and water under the action of concentrated sulfuric acid, which exhibits water-removing properties:

    HCOOH CO + H 2 O

    a colorless gas with an unpleasant odor is released

    Concentrated sulfuric acid, when heated, takes away water from oxalic acid:

    t 0 , H 2 SO 4 (k)

    H 2 C 2 O 4 CO 2 + CO + H 2 O

    colorless gases are released

    Phosphorous acid as a reducing agent can react with nitric acid, which is a strong oxidizing agent:

    H 3 PO 3 + 2HNO 3 \u003d H 3 PO 4 + 2NO 2 + H 2 O

    brown gas is released

    Nitric and sulfuric acids, which are strong oxidizing agents, do not react with each other.

    All elements of the answer are written incorrectly

    Maximum score

    Document

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