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

Assignments for the school stage of the Olympiad for schoolchildren in chemistry, grade 11.

Exercise 1.

There are ten elements in the Periodic Table, the second letter of which is “e”. For each of these elements, typical reactions are given below. The capital letter is indicated by the sign "? " Place the elements in their place.

1) D + T → ? e+n;

2) ? eO 4 → ? e + 2O 2 ;

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

4) 2Au + 6H 2 ? eO 4 (heat) → 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.

Task 2.

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

Legend: “+” - the reaction occurs, “–” - the metal does not react.

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

Task 3.

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

1. Calculate the molecular formula X.

2. Give the structural formula and 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 the oxidation of a hydrocarbon X a solution of potassium permanganate acidified with sulfuric acid.

Task 4.

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

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 the acid gets on his skin? Name this substance and reflect its effect on the acid in the reaction equation.
What volume of vinegar essence should Petya measure to prepare 1 liter of 9% acetic acid solution?

Task 5.

The laboratory contains iron, hydrochloric acid, sodium hydroxide, calcium carbonate, and copper (II) oxide. Using these substances, as well as the products of their interaction, give at least 10 reaction equations for the production of new inorganic substances.

Task 6.

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

Answers 11th grade

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 work;
4) 2Au + 6H 2 Se O 4 (heat) → 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 + 1/2Сl 2 + 4H 2 0;
8) Ge Cl 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.

Based on the totality of properties, i.e. Based on the behavior in reactions with acids and sodium hydroxide, we conclude: metal I is silver (copper does not match the color), metal II is aluminum, metal III is 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 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 = Na 2 + H 2
2Al + 6NaOH + 6H 2 O = 2Na 3 + 3H 2 (7 points)

Task 3.

Answer:

(other wording of the answer is allowed that does not distort its meaning)	Points
1. Мr(X) = 3.25·32 = 104 a.m.u. Let's find the molecular formula of hydrocarbon X: C: H = 0.9226/12.01: 0.0774/1.008 = 1:1, taking into account the molecular weight we get C 8 N 8 .
2. Since during the oxidation of hydrocarbon X with a solution of potassium permanganate in an acidic medium, benzoic acid (C) is formed as the only organic product 6 N 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 substituent C 2 H 3 . 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. Equation for the oxidation of styrene with a KMnO solution 4 , acidified with H 2SO4: 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 = CH3COONa + H2O neutralization reaction (2 points).

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

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

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

Total: 7 points.

Task 5.

Possible answers:

Fe + 2HCl = FeCl 2 + H 2

HCl + NaOH = NaCl + H 2 O

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

CuO + 2HCl = CuCl 2 + H 2 O

2NaOH + CO 2 = Na 2 CO 3 + H 2 O; NaOH + CO 2 = NaHCO 3 + H 2 O

CuO + H 2 = Cu + H 2 O

FeCl 2 + 2NaOH = Fe(OH) 2 + 2NaCl

3Fe + 4H 2 O = Fe 3 O 4 + 4H 2

CaCO 3 = CaO + CO 2

CuO + CO 2 = CuCO 3

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

Task 6.

Reaction equations:

Cu+ 4 HNO 3 = Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O

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

We introduce the following 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(mixtures) = 188x+170y=5.28 (2 points)

3) compose and solve a system of equations:

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

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

4) calculate the mass fractions of the mixture components:

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

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

Maximum score – 7 points

Assignments for the school stage of the Chemistry Olympiad, grade 10

Exercise 1.

What gases can be obtained by having the following substances at your disposal:

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

C 2 + 3B 2 2CB 3
4СВ 3 + 5А 2 4СА + 6В 2 А
4СВ 3 + 3А 2 = 2С 2 + 6В 2 А
C 2 + A 2 = 2 CA
4CA 2 + 2B 2 A + A 2 = 4BCA 3

About what elements we're talking about on a mission? Write down the reaction equations.

Task 3.

When 11.5 g of a mixture of aluminum, magnesium and copper was dissolved in hydrochloric acid, 5.6 liters (n.s.) of gas were released. The undissolved residue is transferred into solution with concentrated nitric acid. In this case, 4.48 liters (n.s.) of gas were released. Determine the mass fractions (in%) of the components of the original 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 elements.

2. Create an electronic balance equation

3. Identify 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 table, the names of which in Russian are “non-masculine” nouns. For each of these elements, below are characteristic reactions in which they are encrypted with the letter “ E " Identify the 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 3conc. + 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 3conc. + (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

When heated, two substances are formed, one of which

interacts with water, releasing a colorless, toxic

gas with a garlicky odor. The latter is oxidized

oxygen of the air.

Write the equations of all the indicated chemical processes, name

their products.

Calculate the volume of air (no.s.) required to oxidize the gas,

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

Answers 10th grade

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 + 2H 2 O
NH 4 NO 2 = N 2 + 2H 2 O
2KMnO 4 = K 2 MnO 4 + MnO 2 + O 2
2KMnO 4 + 16HCl = 2KCl + 2MnCl 2 + 5Cl 2 + 8H 2 O

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

Al 4 C 3 + 12H 2 O = 4Al(OH) 3 + 3CH 4
CaC 2 + 2H 2 O = Ca(OH) 2 + C 2 H 2
2H 2 O 2H 2 + O 2
NH 4 NO 3 + NaOH = NaNO 3 + NH 3 + H 2 O
Na 2 SO 3 + H 2 SO 4 = Na 2 SO 4 + H 2 O + SO 2

Total: 10 points

Task 2.

N 2 + 3H 2 2NH 3
4NH 3 + 5O 2 4NO + 6H 2 O
4NH 3 + 3O 2 = 2N 2 + 6H 2 O
N2 + O2 = 2NO
4NO 2 + 2H 2 O + O 2 = 4HNO 3

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

1 point for each reaction equation

Total: 5 points

Task 3.

1) Copper does not react with hydrochloric acid.

4.48 l

Cu + 4 HNO 3 = 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 = 2AlCl3 + 3H2

Y mole y mole

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 with two unknowns is compiled:

x + 1.5y = 0.25

24x + 27y = 5.1 2 points

5) A system with two unknowns has been solved (the values of “x” and “y” have been 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 electronic balance equation was compiled and the oxidizing agent and

Reducing agent:

2 points

3) The coefficients in the reaction equation are set:

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

4) The areas of application of potassium permanganate are listed:

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

Total: 7 points

Task 5.

Answer:

(other wording of the answer is allowed that does not distort its meaning)	Points
These elements are sulfur, iron, silver, gold, copper, platinum, mercury, tin and antimony. All of them, except sulfur, are metals, and all are easily recognizable by their characteristic transformations, or by oxidation states, characteristic only of 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 3conc. + 4HCl conc. → H + 3NO 2 +3H 2 O;
5)CuCl 2 + 4NH 3 →Cl 2 (blue);
Pt(catalyst),800°С 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 3conc. + (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 = Ca 3 P 2 + 8 MgO (calcium phosphide) (1 point)

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

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

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

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

Maximum score - 5 points

Assignments for the school stage of the Chemistry Olympiad, grade 9

Exercise 1.

Task 2.

Task 3.

When solving problems, we often come across the expression of the molar mass of a substance through 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 this planet's atmosphere 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 based on the air of that planet.

Task 4.

Task 6.

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

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

Answers to tasks 9th grade

Exercise 1.

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

Draw up reaction equations and indicate the signs by which each of the specified substances can be identified.

Number of points –10

Answer:

(other wording of the answer is allowed that does not distort its meaning)	Points
1) AgNO 3 + HCl = AgCl↓ + HNO 3 White
2) CaCl2 + 2AgNO3 = 2AgCl↓ + Ca(NO3 ) 2 White
3)Na2 S + 2HCl = 2NaCl + H2 S Rotten egg smell
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 (IV) oxide was passed through a solution of potassium permanganate, a solution was formed in which the mass fraction of sulfuric acid was 5%. Calculate the mass fractions of the remaining reaction products in the resulting solution.

Number of points –10

Answer:

(other wording of the answer is allowed that does not distort its meaning)	Points
1) Let's create 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 2 mol MnSO4 – m(MnSO4 ) = 2mol ∙ 151g/mol = 302g.
4) Find the mass of 1 mol K2 SO4 – m(K2 SO4 ) = 1mol ∙ 174g/mol = 174g.
5) The mass fraction of each substance is equal to: ω(in-va) = m(in-va) : m(solution). Since all these substances are in one 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, from where 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)=0.48*28+0.19*32+0.15*71+0.08*20+0.07*40+0.029*83+0.001*44=

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 100 g sample of ethyl mercaptan.
In such a sample m(C) = 38.7 g, m(H) = 9.6 g, m(S) = 51.7 g.(1 point)
n (C) = 38.7 g: 12 g/mol = 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
Fe3 O4 + 8HCl = FeCl2 + 2FeCl3 + 4H2 O
FeCl2 + 2NaOH = Fe(OH)2 ↓ + 2NaCl
FeCl3 + 3NaOH = Fe(OH)3 ↓ + 3NaCl
2Fe(OH)3 + Fe(OH)2 → Fe3 O4 + 4H2 O (t°, inert atmosphere)
Substances: A - Fe3 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 , in this case, it is released from the solution

Y = (44*5)/73 = 3.01 g CO2 You will 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

Tasks d for the school stageOlympiads for schoolchildren in chemistry, grade 11.

Exercise 1.

1) D + T → ? e+n;

2) ? eO 4 → ? e + 2O 2 ;

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

4) 2Au + 6H 2 ? eO 4 (heat) → 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.

Legend: “+” - the reaction is in progress, « -" - the metal does not react. Determine which metals could be obtained by a chemist and write the corresponding reaction equations.

Task 3.

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

1. Calculate the molecular formula X.

2. Give the structural formula and 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 the oxidation of a hydrocarbon X a solution of potassium permanganate acidified with sulfuric acid.

Task 4.

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 the acid gets on his skin? Name this substance and reflect its effect on the acid in the reaction equation.

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

Task 5.

Task 6.

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

Answers 11th grade

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 → Ne+n;
2) Xe O 4 → Xe+ 2O 2 ;
3)Ne+ O 3, F 2, etc. → does not work;
4) 2Au + 6H 2 Se O 4 (heat) → 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 O6;
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.

Based on the totality of properties, i.e. Based on the behavior in reactions with acids and sodium hydroxide, we conclude: metal I is silver (copper does not match the color), metal II is aluminum, metal III is 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 wording of the answer is allowed that does not distort its meaning)	Points
1. Мr(X) = 3.25·32 = 104 a.m.u. Let's find the molecular formula of hydrocarbon X: C: H = 0.9226/12.01: 0.0774/1.008 = 1:1, taking into account the molecular weight we obtain 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 overall formula C 8 H 8, we obtain the substituent C 2 H 3. The only possible substituent option is vinyl, and hydrocarbon X is styrene (vinylbenzene). Therefore, the polymer from which the disposable cup was made is polystyrene.
3. 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 = CH3COONa + H2O neutralization reaction (2 points).

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

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

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

Total: 7 points.

Task 5.

Possible answers:

Fe + 2HCl = FeCl 2 + H 2

HCl + NaOH = NaCl + H 2 O

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

CuO + 2HCl = CuCl 2 + H 2 O

2NaOH + CO 2 = Na 2 CO 3 + H 2 O; NaOH + CO 2 = NaHCO 3 + H 2 O

CuO + H 2 = Cu + H 2 O

FeCl 2 + 2NaOH = Fe(OH) 2 + 2NaCl

3Fe + 4H 2 O = Fe 3 O 4 + 4H 2

CaCO 3 = CaO + CO 2

CuO + CO 2 = CuCO 3

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

Task 6.

Reaction equations:

Cu+ 4 HNO 3 = Cu(NO 3) 2 + 2NO 2 + 2H 2 O

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

We introduce the following 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(mixtures) = 188x+170y=5.28 (2 points)

3) compose and solve a system of equations:

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

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

4) calculate the mass fractions of the mixture components:

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

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

maximum score – 7 points

Assignments for the school stage of the Chemistry Olympiad, grade 10

Exercise 1.

What gases can be obtained by having the following substances at your disposal:

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

C 2 + 3B 2 2CB 3

4СВ 3 + 5А 2 4СА + 6В 2 А

4СВ 3 + 3А 2 = 2С 2 + 6В 2 А

C 2 + A 2 = 2 CA

4CA 2 + 2B 2 A + A 2 = 4BCA 3

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

Exercise3.

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 elements.

2. Create an electronic balance equation

3. Identify the oxidizing agent and reducing agent

4. Arrange the coefficients in this equation.

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

Task 5.

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°C

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

7) 2 E+O2 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

When heated, two substances are formed, one of which

interacts with water, releasing a colorless, toxic

gas with a garlicky odor. The latter is oxidized

oxygen of the air.

Write the equations of all the indicated chemical processes, name

their products.

Calculate the volume of air (no.s.) required to oxidize the gas,

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

Answers 10th grade

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+ 2H 2 O

8th grade

What chemical elements are 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 the atoms of the elements you named.

Number of points –10

A raindrop 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 the percentage of the isotopes 35 Cl and 37 Cl in natural chlorine, which has 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, table salt.

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

What laboratory equipment will be needed to separate this mixture?

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

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

Number of points – 20

Indicate 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 odor, or are poisonous? Present your answer in the form of a table using the signs “+” and “-”.

Index	Gases
Index		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

9th grade

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 5 g sample of technical iron (II) sulfide containing 5% metallic iron reacts with hydrochloric acid. Calculate the volume of released gaseous products (at normal conditions) and the volumetric composition of the gas mixture.

9-3 . When 4 g of a mixture of copper and magnesium sawdust was partially dissolved in excess hydrochloric acid, 1.12 liters of hydrogen (n.s.) were released. Establish the composition of the initial mixture as a percentage 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.) will be released when neutralizing the specified amount of waste?

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 can you determine the contents of each test tube without using other reagents? Design an experiment and write reaction equations.

Number of points – 10

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

schoolchildren in chemistry 2011-2012

Grade 10

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 elements.

2. Create an electronic balance equation

3. Identify the oxidizing agent and reducing agent

4. Arrange the coefficients in this equation.

5. List the areas of application of a substance whose formula 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 (no.) 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 (no.). When the initial mixture was passed through a flask with bromine water, all the 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 original alkene.

Number of points – 10

10-5. Four unmarked test tubes contain aqueous solutions of sodium hydroxide, hydrochloric acid, potash and aluminum sulfate. Suggest a way to determine the contents of each test 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 contains iron, hydrochloric acid, sodium hydroxide, calcium carbonate, and copper (II) oxide. Using these substances, as well as the products of their interaction, give at least 10 reaction equations for the production of new inorganic substances

Number of points – 10

11-2. How many isomers does C5H12 have? Write down their structural formulas and give each substance a name according to the 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 solution with a mass fraction of Hg 2 (NO 3) 2 equal to 20% iron filings were placed. After some time, the mass fraction of mercury (I) nitrate in the solution became equal to 6% .

1. What mass of mercury is 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 excess potassium bicarbonate was added to 150 g of this acid, 44.8 liters of gas (n.s.) were released. When an excess of ammonia solution of silver oxide was added to the same amount of acid, a precipitate weighing 2.16 g was formed.

1. Give equations for 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 can you determine the contents of each test tube without using any other reagents?

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 wording of the answer is allowed that does not distort its meaning)	Points
Let's calculate the number of moles of water - n(H 2 O) = m(H 2 O) : M(H 2 O) = 10 -4 g: 18 g/mol = 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 a number of water molecules equal to N(H 2 O) = N A n(H 2 O) = 6.02 10 23 5.56 10 -6 = 33.5 10 17 (molecules)
Water consists of three atoms: two hydrogen atoms and one oxygen atom.
The total number of all atoms contained in a raindrop is equal to N ∑ = 3N(H 2 O) = 3 33.5 10 17 = 100.4 10 17 = 10 19 (atoms).
All elements of the answer are written incorrectly
Maximum score

(other wording of the answer is allowed that does not distort its meaning)	Points
Let us take X to be the percentage of the isotope 35 Cl.
Then the percentage of the 37 Cl isotope will be (100 – X)
The mass of atoms of the 35 Cl isotope is 35X.
The mass of atoms of the 37 Cl isotope is 37(100 – X)
Let's make an equation: 35X + 37(100 – X) = 35.45
ω(35 Cl) = 77.5%, ω(37 Cl) = 22.5%.
All elements of the answer are written incorrectly
Maximum score

(other wording of the answer is allowed that does not distort its meaning)	Points
We separate the iron with a magnet. Place the remaining mixture in water - table salt will dissolve, soot will be on the surface. Filter the solution. The soot will remain on the filter. Let's evaporate the filtrate, it will be NaCl. Treat copper and chalk with hydrochloric acid. Chalk (CaCO 3) dissolve, but the copper will remain.
The following equipment was required for separation: a magnet, a filtering device (a stand with a ring, a funnel, a filter, a glass rod, a filtrate collector (cup)), porcelain steaming cup, 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) = 5.6 l: 22.4 mol/l = 0.25 mol; n(CaCO 3) = n(CO 2) = 0.25 mol; m(CaCO 3) = 0.25 mol · 100 g/mol = 25 g.
Maximum score

Index
	sulphurous		carbon dioxide
Chemical
substance


Characteristic odor

Assessment Guidelines		Points
Chemical formula (8 1 point)
Simple substance (3 · 0.1 points)
Oxide (2 · 0.1 point)

Characteristic odor (4 · 0.125 points)
Poisonous (5 · 0.1 points)
All elements are written incorrectly
Maximum score

9th grade

9-2. 1. When an acid acts on technical iron (II) sulfate 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 a sample of the original sample (95% and 5%, respectively) is equal to:

n (FeS) = 5∙0.95/88 = 5.4 ∙ 10 -2 mol (1 point)

n (Fe) = 5∙0.05/56 = 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) = 5.4 ∙ 10 -2 mol

n(H 2) = 4.48 ∙ 10 -3 mol (2 points)

4. Find the volume of released gaseous products:

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

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

5. Calculate the volumetric composition of the gas mixture:

V total = 1.21 + 0.1 = 1.31 (l)

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

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

Total: 10 points

(other wording of the answer is allowed that does not distort its meaning)	Points
Let's create reaction equations: 1) HСl + NaOH = NaCl + H 2 O (1) 2HCl + Na 2 CO 3 = 2NaCl + H 2 O + CO 2 (2)
To neutralize 0.60 mol HCl according to Eq. reaction (1) requires 0.60 mol NaOH, since n(HCl) = n(NaOH); n(Na 2 CO 3) = 1/2n(HCl) = 0.60 mol: 2 = 0.30 mol – equation (2). m(NaOH) = 0.60 mol · 40 g/mol = 24 g; m(Na 2 CO 3) = 0.30 mol · 106 g/mol = 31.8 g.
3) Let's calculate the volume of carbon dioxide released during neutralization according to reaction (2): n(CO 2) = 1/2n(HCl) = 0.60 mol: 2 = 0.30 mol; V(CO 2) = n(CO 2) · V M = 0.30 mol · 22.4 l/mol = 6.72 l.
4) Cl - + AgNO 3 = AgCl↓ + NO 3 - n(AgNO 3) = n(Cl -) = 0.60 mol; m(AgNO 3) = 0.60 mol 170 g/mol = 102 g
Maximum score

Grade 10

1) Copper does not react with hydrochloric acid.

Cu + 4 HNO 3 = Cu(NO 3) 2 + 2NO 2 + 2H 2 O

n(NO2) = 4.48/22.4 = 0.2 mol;

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

x mole 1.5 x mole

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

y mole y mole

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 with two unknowns is compiled:

24x + 27y = 5.1 2 points

5) A system with two unknowns has been solved (the values of “x” and “y” have been 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

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 electronic balance equation was compiled and the oxidizing agent and

reducing agent:

3) The coefficients in the reaction equation are set:

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

4) The areas of application of potassium permanganate are listed:

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

Total: 7 points

(other wording of the answer is allowed that does 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=CH 2 + H 2
3. H 3 C-CH=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 wording of the answer is allowed that does not distort its meaning)	Points
C n H 2n + H 2 = C n H 2n+2
2 moles of gases (hydrocarbon and hydrogen) react, and one mole (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 alkene that reacted.
This volume is 26.88 - 20.16 = 6.72 (l), that is, 0.3 mol.
Since 75% of the alkene reacted, there was a total of 0.4 mol.
When passed 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 = 42 (g/mol).
Alkene with this molar mass - C 3 H 6, propene: H 3 C-CH = CH 2
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 wording of the answer is allowed that does 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	General observation result


		Al(OH) 3 CO2	1 sediment and
4. Al 2 (S0 4) 3	A1(OH) 3 CO2		2 drafts and
NaOH + HCl = NaCl + H2O
K 2 CO 3 + 2HC1 = 2KS1 + H 2 O + CO 2
3K 2 CO 3 + Al 2 (SO 4) 3 + 3H 2 O = 2Al(OH) 3 + 3CO 2 + 3K 2 SO 4;
Al 2 (SO 4) 3 + 6NaOH = 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 table presented, 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 = FeCl 2 + H 2

HCl + NaOH = NaCl + H 2 O

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

CuO + 2HCl = CuCl 2 + H 2 O

2NaOH + CO 2 = Na 2 CO 3 + H 2 O; NaOH + CO 2 = NaHCO 3 + H 2 O

CuO + H 2 = Cu + H 2 O

FeCl 2 + 2NaOH = Fe(OH) 2 + 2NaCl

3Fe + 4H 2 O = Fe 3 O 4 + 4H 2

CaCO 3 = CaO + CO 2

CuO + CO 2 = CuCO 3

Fe 3 O 4 + 8HCl = 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 because 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 for the task

Total: 7 points

(other wording of the answer is allowed that does not distort its meaning)	Points
Hg 2 (NO 3) 2 + Fe = Fe (NO 3) 2 + 2Hg↓ (1)
M(Fe(NO 3) 2 = 180 g/mol; M(Hg) = 201 g/mol; M(Fe) = 56 g/mol; M(Hg 2 (NO 3) 2 = 526 g/mol.
Let's find the mass of Hg 2 (NO 3) 2 in the original 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 reacts and decreases by the mass of mercury that precipitates.
Let X g of iron reacted. Let us find the mass of the resulting mercury using reaction equation (1): 56 g Fe – 2 201 g Hg m 1 (Hg) = 7.18 X X– m 1
Mass of the resulting solution: m(p-pa) = 264 + X– 7.18X = 264 – 6.18 X(G)
Let's find the mass of Hg 2 (NO 3) 2 in the resulting solution: m(Hg 2 (NO 3) 2) = 0.06·(264 – 6.18 X) = 15,84 – 0,37X
Let's 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 of Fe react with 526 g of Hg 2 (NO 3) 2 X – (36.96 + 0.37 X) X= 4.1; m(Fe) = 4.1 g.
The mass of the resulting mercury is 29.4 g(7.18 4.1)
The mass of the resulting solution is 238.7 g (264 – 6.18 4.1)
Let's 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.
Let us find the mass fraction of iron(II) nitrate in the resulting solution: ω(Fe(NO 3) 2) = m(Fe(NO 3) 2) : m(p-pa) = 13.18: 238.7 = 0,055 (5,5 %)
All elements of the answer are written incorrectly
Maximum score

(other wording of the answer is allowed that does not distort its meaning)	Points
C 2 H 5 COOH + KHCO 3 = C 2 H 5 COOK + H 2 O + CO 2 (1)
HCOOH + KHCO 3 = NCOOC + H 2 O + CO 2 (2)
HCOOH + Ag 2 O = 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
Based on the mass of isolated 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 what volume of 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) = 0.224 l.
Consequently, when interacting with propionic acid (equation 1), 44.576 l of CO 2 (44.8 - 0.224) were released
Let's 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
Let's find the mass of propyl alcohol: m(C 3 H 7 OH) = 150 – m(C 2 H 5 COOH) – m(HCOOH) = 150 –147.26 – 0.46 = 2.28 (g)
The mass fraction of formic acid in a solution of propionic acid is equal to: 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 equal to: ω(C 3 H 7 OH) = m(C 3 H 7 OH): m(mixtures) = 2.28: 150 = 0.0152 (1.52%)
All elements of the answer are written incorrectly
Maximum score

11-5.

(other wording of the answer is allowed that does 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 = 4I 2 ↓ + H 2 S + 4H 2 O A brown precipitate of iodine forms 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, removes 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 = H 3 PO 4 + 2NO 2 + H 2 O brown gas is released
Nitric and sulfuric acids, which are strong oxidizing agents don't react to each other
All elements of the answer are written incorrectly
Maximum score

Document

... – This a single, interconnected process. Let's consider reaction dissolution of zinc in salt acid: Zn0... which oxidizes to water and nitrogen. Write the equation reactions, determine by calculation For standard states substances direction reactions at ...

At Plant No. 8, foreign guns (Bolshevik, Hotchkiss, Maxim, Rheinmetall, etc.) were assigned their own factory indices, thus the Lender system

Document

From substances, dissolved in water. In sea mines For this traditionally used sugar. ... By plan Status of work Possible implementation at cash equipped. at replenished... which will one hundred on bearings, not in babbitt, How Now. This ...

Thematic planning of chemistry lessons in grades 8-11 6

Thematic planning

8. Which inorganic substances available, using water, air, sulfur and calcium. Write the equation reactions and indicate the conditions for their occurrence. 9. At ...

According to Le Chatelier’s principle: if an external influence is exerted on a system that is in equilibrium, then as a result of the processes occurring in it, the equilibrium position shifts in the direction that weakens this influence; when the pressure decreases, the equilibrium in a reversible process shifts toward 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 > 2), i.e. an increase in pressure, so the equilibrium shifts in the starting substances.

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

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

Example 8.

Explain why it is impossible to extinguish ignited calcium metal with water. Write the reaction equations.

Solution:

Calcium metal reacts with water, so adding water to burning calcium will only enhance the process. Let us write the equation for the reaction between calcium and water:

Ca + 2H 2 O = 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 = MnCl 2 + Cl 2 + 2H 2 O,

ionic form

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

Electrolysis solution of sodium chloride:

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 = 2NaCl + Br 2,

ionic form:

Cl 2 + 2Br - = 2Cl - + Br 2.

Reaction between sodium metal and chlorine gas:

molecular form:

2Na + Cl 2 = 2NaCl.

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

NaCl + AgNO 3 = AgCl + NaNO 3,

ionic form:

Cl - + Ag + = AgCl.

Example 10.

Solution:

1. Interaction of copper with dilute nitric acid:

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

2. Oxidation of copper with oxygen:

2Cu + O 2 = 2CuO.

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

2Cu + O 2 + 4HCl = 2CuCl 2 + 2H 2 O.

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

CuCl 2 + 2NaOH = Cu(OH) 2 + 2NaCl.

5. Thermal decomposition of copper (II) hydroxide:

Cu(OH) 2 = CuO + H 2 O.

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

Cu(NO 3) 2 + 2NaOH = Cu(OH) 2 + 2NaNO 3.

7. Thermal decomposition of copper (II) nitrate:

2Cu(NO3)2 = 2CuO + 4NO2 + O2.

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

CuO + 2HCl = CuCl 2 + H 2 O.

Problems to solve independently

Option 1

Write reaction equations showing the properties of aluminum hydroxide.

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

Prepare basic copper carbonate from copper metal using as few reagents as possible. Write down the equations for 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 exposed to a solution of silver nitrate, 0.3315 g of sediment precipitated. Determine the mass fractions (%) of acids in the original solution.

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

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

Indicate 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 and indicate the transition of electrons.

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

Indicate how the temperature and pressure should be changed (increase or decrease) in order to shift the equilibrium in the decomposition reaction of calcium carbonate: CaCO 3 (k) ⇄ CaO (k) + CO 2 (g) - 178 kJ towards the decomposition products.

Explain why solutions of sodium and potassium hydroxides destroy glassware, especially when boiled for a long time. Write the 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) with 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 can you obtain sodium nitrate using only these reagents in two ways? Write down the equations for the corresponding reactions.

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

Indicate which “limes” are known. Write them chemical composition and obtaining 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 state 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.

On one collective farm, fertilizers were applied to spring wheat at the rate of: ammonium nitrate - 150 kg, superphosphate (containing 30% digestible P2O5) - 300 kg and potassium chloride - 100 kg per hectare. Calculate how much this amounts to in terms of nitrogen, calcium dihydrogen phosphate and potassium oxide.

Indicate how an 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 (g) + Br 2 (g) ⇄ 2HBr (g).

Show how to chemically remove corrosion products (aluminum oxide and hydroxide) from an aluminum product without causing damage to the metal. Write the 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 exchange reaction equations if:

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

one of the resulting 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 what reagents can be used to distinguish these substances. Write the reaction equations.

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

When burning 0.896 l (n.s.) 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 prepared. The hydroxide of which element is the strongest alkali? Indicate the technical names of 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 equations for the reactions: a) magnesium iodide with bromine; b) magnesium with a solution of hydrobromic acid. Indicate which element is an oxidizing agent and which a reducing agent in each case, and show the transition of electrons.

On one collective farm, in addition to manure, the following masses of mineral fertilizers per hectare were applied to potatoes: granulated superphosphate containing 12.5% of digestible P 2 O 5 - 0.15 t, ammonium nitrate - 0.1 t and potassium chloride containing 90% KCl – 0.1 t. 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 one achieve a shift in equilibrium towards the formation of sulfur oxide (VI)?

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

Will sodium ions be preserved: a) when sodium hydroxide reacts with hydrochloric acid; b) when sodium hydroxide reacts 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 aqueous solutions of amphoteric bases dissociate into ions. Give examples of such bases and show their dissociation into ions.

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

The laboratory contains iron, hydrochloric acid, sodium hydroxide, calcium carbonate, and copper (II) oxide. Is it possible to obtain 12 new inorganic substances if we use these reagents and their reaction products 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 were then passed through 100 ml of a 32% sulfuric acid solution (density 1.22 g/ml). Determine which gas is in excess and what is the mass fraction (%) of salt in the solution. Assume that all reactions occur with 100% yield.

Indicate the composition of “caustic soda”, “crystalline soda”, “soda ash”, “baking 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 reacts with silicon (IV) oxide in an oxygen atmosphere. Product yield 75%.

Indicate 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 and indicate the transition of electrons.

Explain why potassium nitrate is called ballast-free 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) increase in temperature; c) introduction of a catalyst.

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

Give examples of reactions in which processes expressed by the following schemes occur: 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:

ASSIGNMENTS FOR 10TH GRADE STUDENTS

EXERCISE 1

The assignment includes material on the chemistry of hydrocarbons in the saturated and unsaturated series: structure, isomerism and nomenclature, properties, preparation. Solving calculation problems involves the use of 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-propylethylene (cis-heptene-3)

Example 2.

Describe the stages of initiation, growth and termination of a 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 HCH 3 + 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 that occur by the radical mechanism:

a) C 2 H 6 + Cl 2;

b) C 2 H 4 + HBr ;

c) H 2 C = CH – CH 3 + HBr ;

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

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 = 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 (diluted) 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 diagram of the 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 bottle with bromine increased by 1.1 g. Determine the volumetric composition of the mixture and the mass of the resulting products.

Given:

V(mixtures) = 2 l

m(flasks) = 1.1 g

M(Br 2) = 160 g/mol

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

Find:

V(propane in the mixture)

V(propylene in the mixture)

m(products)

Solution:

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

1 mol – 22.4 l

n mole – 2 l

T(mixtures) = 0.0892 mol.

With bromine at n. u. Only propylene reacts. Let's create the reaction equation:

C 3 H 6 + Br 2 = C 3 H 6 Br 2.

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

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