USE assignments in chemistry from 2. Task C2 on the USE in chemistry
We discussed the general algorithm for solving problem No. 35 (C5). It's time to analyze specific examples and offer you a selection of tasks for independent solution.
Example 2. Complete hydrogenation of 5.4 g of some alkyne consumes 4.48 liters of hydrogen (n.a.) Determine the molecular formula of this alkyne.
Decision. We will act in accordance with the general plan. Let the unknown alkyne molecule contain n carbon atoms. General formula of the homologous series C n H 2n-2 . Hydrogenation of alkynes proceeds in accordance with the equation:
C n H 2n-2 + 2Н 2 = C n H 2n+2.
The amount of hydrogen reacted can be found by the formula n = V/Vm. In this case, n = 4.48 / 22.4 = 0.2 mol.
The equation shows that 1 mol of alkyne adds 2 mol of hydrogen (recall that in the condition of the problem we are talking about complete hydrogenation), therefore, n (C n H 2n-2) = 0.1 mol.
By the mass and amount of alkyne, we find its molar mass: M (C n H 2n-2) \u003d m (mass) / n (amount) \u003d 5.4 / 0.1 \u003d 54 (g / mol).
The relative molecular weight of an alkyne is made up of n atomic masses of carbon and 2n-2 atomic masses of hydrogen. We get the equation:
12n + 2n - 2 = 54.
We solve a linear equation, we get: n \u003d 4. Alkyne formula: C 4 H 6.
Answer: C 4 H 6 .
I would like to draw attention to one significant point: the molecular formula C 4 H 6 corresponds to several isomers, including two alkynes (butyn-1 and butyn-2). Based on these problems, we will not be able to unambiguously establish the structural formula of the substance under study. However, in this case, this is not required!
Example 3. During the combustion of 112 l (n.a.) of an unknown cycloalkane in excess oxygen, 336 l of CO 2 are formed. Set the structural formula of cycloalkane.
Decision. The general formula for the homologous series of cycloalkanes is: C n H 2n. With the complete combustion of cycloalkanes, as with the combustion of any hydrocarbons, carbon dioxide and water are formed:
C n H 2n + 1.5n O 2 \u003d n CO 2 + n H 2 O.
Please note: the coefficients in the reaction equation in this case depend on n!
During the reaction, 336 / 22.4 \u003d 15 mol of carbon dioxide was formed. 112/22.4 = 5 mol of hydrocarbon entered into the reaction.
Further reasoning is obvious: if 15 moles of CO 2 are formed per 5 moles of cycloalkane, then 15 molecules of carbon dioxide are formed per 5 molecules of hydrocarbon, i.e., one molecule of cycloalkane gives 3 molecules of CO 2. Since each molecule of carbon monoxide (IV) contains one carbon atom, we can conclude that one cycloalkane molecule contains 3 carbon atoms.
Conclusion: n \u003d 3, the formula of cycloalkane is C 3 H 6.
As you can see, the solution to this problem does not "fit" into the general algorithm. We did not look for the molar mass of the compound here, did not make any equation. According to formal criteria, this example is not similar to the standard C5 problem. But above, I have already emphasized that it is important not to memorize the algorithm, but to understand the MEANING of the actions performed. If you understand the meaning, you yourself will be able to make changes to the general scheme at the exam, choose the most rational way to solve it.
In this example, there is another "strangeness": it is necessary to find not only the molecular, but also the structural formula of the compound. In the previous task, we failed to do this, but in this example - please! The fact is that the formula C 3 H 6 corresponds to only one isomer - cyclopropane.
Answer: cyclopropane.
Example 4. 116 g of some limiting aldehyde were heated for a long time with an ammonia solution of silver oxide. During the reaction, 432 g of metallic silver was formed. Set the molecular formula of aldehyde.
Decision. The general formula for the homologous series of limiting aldehydes is: C n H 2n+1 COH. Aldehydes are easily oxidized to carboxylic acids, in particular, under the action of an ammonia solution of silver oxide:
C n H 2n + 1 COH + Ag 2 O \u003d C n H 2n + 1 COOH + 2Ag.
Note. In reality, the reaction is described by a more complex equation. When Ag 2 O is added to an aqueous solution of ammonia, a complex compound OH is formed - diammine silver hydroxide. It is this compound that acts as an oxidizing agent. During the reaction, an ammonium salt of a carboxylic acid is formed:
C n H 2n + 1 COH + 2OH \u003d C n H 2n + 1 COONH 4 + 2Ag + 3NH 3 + H 2 O.
Another important point! The oxidation of formaldehyde (HCOH) is not described by the above equation. When HCOH reacts with an ammonia solution of silver oxide, 4 mol of Ag is released per 1 mol of aldehyde:
НCOH + 2Ag 2 O \u003d CO 2 + H 2 O + 4Ag.
Be careful when solving problems related to the oxidation of carbonyl compounds!
Let's go back to our example. By the mass of released silver, you can find the amount of this metal: n(Ag) = m/M = 432/108 = 4 (mol). In accordance with the equation, 2 mol of silver is formed per 1 mol of aldehyde, therefore, n (aldehyde) \u003d 0.5n (Ag) \u003d 0.5 * 4 \u003d 2 mol.
Molar mass of aldehyde = 116/2 = 58 g/mol. Try to do the next steps yourself: you need to make an equation, solve it and draw conclusions.
Answer: C 2 H 5 COH.
Example 5. When 3.1 g of some primary amine is reacted with a sufficient amount of HBr, 11.2 g of salt is formed. Set the amine formula.
Decision. Primary amines (C n H 2n + 1 NH 2) when interacting with acids form alkylammonium salts:
C n H 2n+1 NH 2 + HBr = [C n H 2n+1 NH 3] + Br - .
Unfortunately, by the mass of the amine and the resulting salt, we will not be able to find their quantities (since the molar masses are unknown). Let's go the other way. Recall the law of conservation of mass: m(amine) + m(HBr) = m(salt), therefore, m(HBr) = m(salt) - m(amine) = 11.2 - 3.1 = 8.1.
Pay attention to this trick, which is very often used in solving C 5. Even if the mass of the reagent is not given explicitly in the problem statement, you can try to find it from the masses of other compounds.
So, we are back in the mainstream of the standard algorithm. By the mass of hydrogen bromide we find the amount, n(HBr) = n(amine), M(amine) = 31 g/mol.
Answer: CH 3 NH 2 .
Example 6. A certain amount of alkene X, when interacting with an excess of chlorine, forms 11.3 g of dichloride, and when reacting with an excess of bromine, 20.2 g of dibromide. Determine the molecular formula of X.
Decision. Alkenes add chlorine and bromine to form dihalogen derivatives:
C n H 2n + Cl 2 \u003d C n H 2n Cl 2,
C n H 2n + Br 2 \u003d C n H 2n Br 2.
It is pointless in this problem to try to find the amount of dichloride or dibromide (their molar masses are unknown) or the amounts of chlorine or bromine (their masses are unknown).
We use one non-standard technique. The molar mass of C n H 2n Cl 2 is 12n + 2n + 71 = 14n + 71. M (C n H 2n Br 2) = 14n + 160.
The masses of the dihalides are also known. You can find the amount of substances obtained: n (C n H 2n Cl 2) \u003d m / M \u003d 11.3 / (14n + 71). n (C n H 2n Br 2) \u003d 20.2 / (14n + 160).
By convention, the amount of dichloride is equal to the amount of dibromide. This fact gives us the opportunity to make an equation: 11.3 / (14n + 71) = 20.2 / (14n + 160).
This equation has a unique solution: n = 3.
Answer: C 3 H 6
In the final part, I offer you a selection of problems of the C5 type of varying complexity. Try to solve them yourself - it will be a great workout before passing the exam in chemistry!
The condition of task C2 for the exam in chemistry is a text describing the sequence of experimental actions. This text needs to be converted into reaction equations.
The difficulty of such a task is that schoolchildren have little idea of experimental, not "paper" chemistry. Not everyone understands the terms used and the ongoing processes. Let's try to figure it out.
Very often, concepts that seem completely clear to a chemist are misunderstood by applicants. Here is a short glossary of such terms.
Dictionary of obscure terms.
- Hinge- it's just a certain portion of a substance of a certain mass (it was weighed on the scales). It has nothing to do with the canopy over the porch :-)
- Ignite- heat the substance to a high temperature and heat until the end of chemical reactions. This is not "potassium mixing" or "piercing with a nail."
- "Blow up a mixture of gases"- this means that the substances reacted with an explosion. Usually an electric spark is used for this. The flask or vessel at the same time do not explode!
- Filter- separate the precipitate from the solution.
- Filter- pass the solution through a filter to separate the precipitate.
- Filtrate- it's filtered solution.
- Dissolution of a substance is the transition of a substance into a solution. It can occur without chemical reactions (for example, when sodium chloride NaCl is dissolved in water, a solution of sodium chloride NaCl is obtained, and not alkali and acid separately), or in the process of dissolution, the substance reacts with water and forms a solution of another substance (when barium oxide is dissolved, it will turn out barium hydroxide solution). Substances can be dissolved not only in water, but also in acids, alkalis, etc.
- Evaporation- this is the removal of water and volatile substances from a solution without decomposition of the solids contained in the solution.
- Evaporation- this is simply a decrease in the mass of water in a solution by boiling.
- fusion- this is the joint heating of two or more solids to a temperature when they begin to melt and interact. It has nothing to do with swimming on the river :-)
- Sediment and residue.
These terms are often confused. Although these are completely different concepts.
"The reaction proceeds with the release of a precipitate"- this means that one of the substances obtained in the reaction is slightly soluble. Such substances fall to the bottom of the reaction vessel (tubes or flasks).
"Remainder" is a substance that left, was not spent completely or did not react at all. For example, if a mixture of several metals was treated with acid, and one of the metals did not react, it can be called remainder. - Saturated A solution is a solution in which, at a given temperature, the concentration of a substance is the highest possible and no longer dissolves.
unsaturated a solution is a solution in which the concentration of a substance is not the maximum possible; in such a solution, some more amount of this substance can be additionally dissolved until it becomes saturated.
Diluted and "very" diluted solution - these are very conditional concepts, rather qualitative than quantitative. It is assumed that the concentration of the substance is low.
The term is also used for acids and bases. "concentrated" solution. This is also conditional. For example, concentrated hydrochloric acid has a concentration of only about 40%. And concentrated sulfuric is an anhydrous, 100% acid.
In order to solve such problems, it is necessary to clearly know the properties of most metals, non-metals and their compounds: oxides, hydroxides, salts. It is necessary to repeat the properties of nitric and sulfuric acids, potassium permanganate and dichromate, redox properties of various compounds, electrolysis of solutions and melts of various substances, decomposition reactions of compounds of different classes, amphotericity, hydrolysis of salts and other compounds, mutual hydrolysis of two salts.
In addition, it is necessary to have an idea about the color and state of aggregation of most of the studied substances - metals, non-metals, oxides, salts.
That is why we analyze this type of tasks at the very end of the study of general and inorganic chemistry.
Let's look at some examples of such tasks.
Example 1: The reaction product of lithium with nitrogen was treated with water. The resulting gas was passed through a solution of sulfuric acid until the chemical reactions ceased. The resulting solution was treated with barium chloride. The solution was filtered and the filtrate was mixed with sodium nitrite solution and heated.
Decision:
Example 2:Hinge aluminum was dissolved in dilute nitric acid, and a gaseous simple substance was released. Sodium carbonate was added to the resulting solution until the gas evolution ceased completely. dropped out the precipitate was filtered and calcined, filtrate evaporated, the resulting solid the rest was fused with ammonium chloride. The evolved gas was mixed with ammonia and the resulting mixture was heated.
Decision:
Example 3: Aluminum oxide was fused with sodium carbonate, the resulting solid was dissolved in water. Sulfur dioxide was passed through the resulting solution until the complete cessation of interaction. The precipitate formed was filtered off, and bromine water was added to the filtered solution. The resulting solution was neutralized with sodium hydroxide.
Decision:
Example 4: Zinc sulfide was treated with a hydrochloric acid solution, the resulting gas was passed through an excess of sodium hydroxide solution, then a solution of iron (II) chloride was added. The precipitate obtained was calcined. The resulting gas was mixed with oxygen and passed over the catalyst.
Decision:
Example 5: Silicon oxide was calcined with a large excess of magnesium. The resulting mixture of substances was treated with water. At the same time, a gas was released, which was burned in oxygen. The solid combustion product was dissolved in a concentrated solution of cesium hydroxide. Hydrochloric acid was added to the resulting solution.
Decision:
Tasks C2 from the USE options in chemistry for independent work.
- Copper nitrate was calcined, the resulting solid precipitate was dissolved in sulfuric acid. Hydrogen sulfide was passed through the solution, the resulting black precipitate was calcined, and the solid residue was dissolved by heating in concentrated nitric acid.
- Calcium phosphate was fused with coal and sand, then the resulting simple substance was burned in an excess of oxygen, the combustion product was dissolved in an excess of sodium hydroxide. A solution of barium chloride was added to the resulting solution. The resulting precipitate was treated with an excess of phosphoric acid.
- Copper was dissolved in concentrated nitric acid, the resulting gas was mixed with oxygen and dissolved in water. Zinc oxide was dissolved in the resulting solution, then a large excess of sodium hydroxide solution was added to the solution.
- Dry sodium chloride was treated with concentrated sulfuric acid at low heating, the resulting gas was passed into a barium hydroxide solution. A solution of potassium sulfate was added to the resulting solution. The resulting precipitate was fused with coal. The resulting substance was treated with hydrochloric acid.
- A weighed portion of aluminum sulfide was treated with hydrochloric acid. In this case, gas was released and a colorless solution was formed. An ammonia solution was added to the resulting solution, and the gas was passed through a solution of lead nitrate. The precipitate thus obtained was treated with a solution of hydrogen peroxide.
- Aluminum powder was mixed with sulfur powder, the mixture was heated, the resulting substance was treated with water, gas was released and a precipitate formed, to which an excess of potassium hydroxide solution was added until complete dissolution. This solution was evaporated and calcined. An excess of hydrochloric acid solution was added to the resulting solid.
- The potassium iodide solution was treated with a chlorine solution. The resulting precipitate was treated with sodium sulfite solution. First, a solution of barium chloride was added to the resulting solution, and after separating the precipitate, a solution of silver nitrate was added.
- Gray-green powder of chromium (III) oxide was fused with an excess of alkali, the resulting substance was dissolved in water, and a dark green solution was obtained. Hydrogen peroxide was added to the resulting alkaline solution. A yellow solution was obtained, which turns orange when sulfuric acid is added. When hydrogen sulfide is passed through the resulting acidified orange solution, it becomes cloudy and turns green again.
- (MIOO 2011, training work) Aluminum was dissolved in a concentrated solution of potassium hydroxide. Carbon dioxide was passed through the resulting solution until the precipitation ceased. The precipitate was filtered off and calcined. The resulting solid residue was fused with sodium carbonate.
- (MIOO 2011, training work) Silicon was dissolved in a concentrated solution of potassium hydroxide. An excess of hydrochloric acid was added to the resulting solution. The cloudy solution was heated. The separated precipitate was filtered off and calcined with calcium carbonate. Write the equations of the described reactions.
Answers to tasks for independent solution:
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Dmitry Ivanovich Mendeleev discovered the periodic law, according to which the properties of the elements and the elements they form change periodically. This discovery was graphically displayed in the periodic table. The table shows very well and clearly how the properties of the elements change over the period, after which they are repeated in the next period.
To solve task No. 2 of the Unified State Exam in chemistry, we just need to understand and remember which properties of the elements change in which directions and how.
All this is shown in the figure below.
From left to right, electronegativity, non-metallic properties, higher oxidation states, etc. increase. And the metallic properties and radii decrease.
From top to bottom, vice versa: the metallic properties and radii of atoms increase, while the electronegativity decreases. The highest oxidation state, corresponding to the number of electrons in the outer energy level, does not change in this direction.
Let's look at examples.
Example 1 In the series of elements Na→Mg→Al→Si
A) the radii of atoms decrease;
B) the number of protons in the nuclei of atoms decreases;
C) the number of electron layers in atoms increases;
D) the highest degree of oxidation of atoms decreases;If we look at the periodic table, we will see that all the elements of this series are in the same period and are listed in the order in which they appear in the table from left to right. To answer this kind of question, you just need to know a few patterns of changes in properties in the periodic table. So from left to right along the period, metallic properties decrease, non-metallic ones increase, electronegativity increases, ionization energy increases, and the radius of atoms decreases. From top to bottom, metallic and reducing properties increase in a group, electronegativity decreases, ionization energy decreases, and the radius of atoms increases.
If you were attentive, you already understood that in this case the atomic radii decrease. Answer A.
Example 2 In order of increasing oxidizing properties, the elements are arranged in the following order:
A. F→O→N
B. I→Br→Cl
B. Cl→S→P
D. F→Cl→BrAs you know, in Mendeleev's periodic table, oxidizing properties increase from left to right in a period and from bottom to top in a group. Option B just shows the elements of one group in order from bottom to top. So B fits.
Example 3 The valence of elements in the higher oxide increases in the series:
A. Cl→Br→I
B. Cs→K→Li
B. Cl→S→P
D. Al→C→NIn higher oxides, the elements show their highest oxidation state, which will coincide with the valency. And the highest degree of oxidation grows from left to right in the table. We look: in the first and second versions, we are given elements that are in the same groups, where the highest degree of oxidation and, accordingly, the valence in oxides does not change. Cl → S → P - are located from right to left, that is, on the contrary, their valence in the higher oxide will fall. But in the row Al→C→N, the elements are located from left to right, the valence in the higher oxide increases in them. Answer: G
Example 4 In the series of elements S→Se→Te
A) the acidity of hydrogen compounds increases;
B) the highest degree of oxidation of elements increases;
C) the valence of elements in hydrogen compounds increases;
D) the number of electrons in the outer level decreases;Immediately look at the location of these elements in the periodic table. Sulfur, selenium and tellurium are in the same group, one subgroup. Listed in order from top to bottom. Look again at the diagram above. From top to bottom in the periodic table, metallic properties increase, radii increase, electronegativity, ionization energy and non-metallic properties decrease, the number of electrons at the outer level does not change. Option D is ruled out immediately. If the number of external electrons does not change, then the valence possibilities and the highest oxidation state also do not change, B and C are excluded.
Option A remains. We check for order. According to the Kossel scheme, the strength of oxygen-free acids increases with a decrease in the oxidation state of an element and an increase in the radius of its ion. The oxidation state of all three elements is the same in hydrogen compounds, but the radius grows from top to bottom, which means that the strength of acids also grows.
The answer is A.Example 5 In order of weakening of the main properties, the oxides are arranged in the following order:
A. Na 2 O → K 2 O → Rb 2 O
B. Na 2 O → MgO → Al 2 O 3
B. BeO→BaO→CaO
G. SO 3 → P 2 O 5 → SiO 2The main properties of oxides weaken synchronously with the weakening of the metallic properties of the elements forming them. And Me-properties weaken from left to right or from bottom to top. Na, Mg and Al are just arranged from left to right. Answer B.
ASSIGNMENTS C2 USE IN CHEMISTRY
An analysis of the content of the task shows that the first substance is unknown, but the characteristic properties of the substance itself (color) and the reaction products (color and state of aggregation) are known. For all other reactions, the reagent and conditions are indicated. Tips can be considered as indications of the class of the obtained substance, its state of aggregation, characteristic features (color, smell). Note that two reaction equations characterize the special properties of substances (1 - decomposition of ammonium dichromate; 4 - reducing properties of ammonia), two equations characterize the typical properties of the most important classes of inorganic substances (2 - reaction between metal and non-metal, 3 - hydrolysis of nitrides).
When solving these tasks, students can be recommended to draw diagrams:
t o C Li H 2 O CuO
(NH 4) 2 Cr 2 O 7 → gas → X → gas with a pungent odor → Сu
Highlight clues, key points, for example: an orange substance that decomposes with the release of nitrogen (colorless gas) and Cr 2 O 3 (green substance) - ammonium dichromate (NH 4) 2 Cr 2 O 7.
t o C
(NH 4) 2 Cr 2 O 7 → N 2 + Cr 2 O 3 + 4H 2 O
N 2 + 6Li → 2 Li 3 N
t o C
Li 3 N+ 3H 2 O → NH 3 + 3LiOH
t o C
NH 3 + 3CuO → 3Cu + N 2 + 3H2O
Filtration - a method for separating heterogeneous mixtures using filters - porous materials that pass liquid or gas, but retain solids. When separating mixtures containing a liquid phase, a solid remains on the filter, filtrate .
Evaporation -
Ignition -
CuSO 4 ∙5H 2 O → CuSO 4 + 5H 2 O
Thermally unstable substances decompose (insoluble bases, some salts, acids, oxides): Cu (OH) 2 →CuO + H 2 O; CaCO 3 → CaO + CO 2
Substances that are unstable to the action of air components oxidize when calcined, react with air components: 2Cu + O 2 → 2CuO;
4Fe (OH) 2 + O 2 → 2Fe 2 O 3 + 4H 2 O
In order to prevent oxidation during calcination, the process is carried out in an inert atmosphere: Fe (OH) 2 → FeO + H 2 O
Sintering, fusion -
Al 2 O 3 + Na 2 CO 3 → 2NaAlO 2 + CO 2
If one of the reactants or the reaction product can be oxidized by air components, the process is carried out with an inert atmosphere, for example: Сu + CuO → Cu 2 O
Burning
4FeS 2 + 11O 2 → 2Fe 2 O 3 + 8SO 2
GASES:
Painted : Cl 2 - yellow-green;NO 2 - brown; O 3 - blue (all have smells). All are poisonous, dissolve in water,Cl 2 and NO 2 react with her.
Colorless, odorless : H 2 , N 2 , O 2 , CO 2 , CO (poison), NO (poison), inert gases. All are poorly soluble in water.
Colorless with an odor : HF, HCl, HBr, HI, SO 2 (pungent odors), NH 3 (ammonia) - highly soluble in water and poisonous,
PH 3 (garlic), H 2 S (rotten eggs) - slightly soluble in water, poisonous.
COLORED SOLUTIONS:
yellow
Chromates, e.g. K 2 CrO 4
Solutions of iron salts (III), for example, FeCl 3,
bromine water,
cyellow before brown
orange
Dichromates, e.g. K 2 Cr 2 O 7
green
Hydroxo complexes of chromium (III), for example, K 3, salts of nickel (II), for example NiSO 4,
manganates, e.g. K 2 MnO 4
blue
Copper salts ( II), for example СuSO 4
From pink before purple
Permanganates, e.g. KMnO 4
From green before blue
Salts of chromium (III), for example, CrCl 3
PAINTED DRAINAGE,
yellow
AgBr, AgI, Ag 3 PO 4 , BaCrO 4 , PbI 2 , CdS
brown
Fe(OH) 3 , MnO 2
black, black-brown
blue
Cu(OH) 2 , KF e
green
Cr (OH) 3 - gray-green
Fe (OH) 2 - dirty green, turns brown in air
OTHER COLORED SUBSTANCES
yellow
sulfur, gold, chromates
orange
o copper oxide (I) - Cu 2 O
dichromates
red
Fe 2 O 3 , CrO 3
black
With uO, FeO, CrO
purple
green
Cr 2 O 3, malachite (CuOH) 2 CO 3, Mn 2 O 7 (liquid)
In the process of preparing students for solving tasks C2, you can offer them compose texts of assignments in accordance with the schemes of transformations . This task will allow students to master the terminology and remember the characteristic features of substances.
Example 1:
t o C t o C /H 2 HNO 3 (conc) NaOH, 0 o C
(CuOH) 2 CO 3 → CuO → Cu → NO 2 → X
Text:
Example 2:
O 2 H 2 S R - R t o C/AlH 2 O
ZnS → SO 2 → S → Al 2 S 3 → X
Text: Zinc sulfide was fired. The resulting gas with a pungent odor was passed through a solution of hydrogen sulfide until a yellow precipitate formed. The precipitate was filtered off, dried and fused with aluminum. The resulting compound was placed in water until the reaction terminated.
The next step is to ask students to draw up both schemes for the transformation of substances and texts of tasks. Of course, the "authors" of the tasks must submit and own decision . At the same time, students repeat all the properties of inorganic substances. And the teacher can form a bank of tasks C2.
After that you can go to solving C2 tasks . At the same time, students draw up a scheme of transformations according to the text, and then the corresponding reaction equations. To do this, reference points are highlighted in the text of the task: the names of substances, an indication of their classes, physical properties, conditions for conducting reactions, names of processes.
Example 1 manganese nitrate (II
Decision:
Selection of support moments:
manganese nitrate (II ) - Mn (NO 3) 2,
calcined- heated to decomposition,
solid brown matter- Mn O 2,
– HCl,
Hydrosulphuric acid - solution H 2 S,
barium chloride – BaCl 2 forms a precipitate with the sulfate ion.
t o C HCl H 2 S solution BaCl 2
Mn (NO 3) 2 → Mn O 2 → X → Y → ↓ (BaSO 4 ?)
1) Mn(NO 3) 2 → Mn O 2 + 2NO 2
2) Mn O 2 + 4 HCl → MnCl 2 + 2H 2 O + Cl 2 (gasX)
3) Cl 2 + H 2 S → 2HCl + S (not suitable, because there is no product that precipitates with barium chloride) or 4Cl 2 + H 2 S + 4H 2 O → 8HCl + H 2 SO 4
4) H 2 SO 4 + BaCl 2 → BaSO 4 + 2HCl
Example 2.
Decision:
Selection of support moments:
Orange copper oxide- Cu 2 O,
- H 2 SO 4,
blue solution- salt of copper (II), СuSO 4
Potassium hydroxide– CON,
Blue precipitate - Cu (OH) 2,
Calcined - heated to decomposition
Solid black matter CuO,
Ammonia- NH3.
Drawing up a transformation scheme:
H 2 SO 4 KOH t o C NH 3
Cu 2 O → СuSO 4 → Cu (OH) 2 ↓ → CuO → X
Drawing up reaction equations:
1) Cu 2 O + 3 H 2 SO 4 → 2 СuSO 4 + SO 2 + 3H 2 O
2) СuSO 4 + 2 KOH → Cu (OH) 2 + K 2 SO 4
3) Cu(OH) 2 → CuO + H 2 O
4) 3CuO + 2NH 3 → 3Cu + 3H 2 O + N 2
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SOLUTIONS
1 . Sodium was burned in an excess of oxygen, the resulting crystalline substance was placed in a glass tube and carbon dioxide was passed through it. The gas coming out of the tube was collected and burned in its atmosphere of phosphorus. The resulting substance was neutralized with an excess of sodium hydroxide solution.
1) 2Na + O 2 \u003d Na 2 O 2
2) 2Na 2 O 2 + 2CO 2 \u003d 2Na 2 CO 3 + O 2
3) 4P + 5O 2 \u003d 2P 2 O 5
4) P 2 O 5 + 6 NaOH = 2Na 3 PO 4 + 3H 2 O
2. Aluminum carbide treated with hydrochloric acid. The released gas was burned, the combustion products were passed through lime water until a white precipitate formed, further passing the combustion products into the resulting suspension led to the dissolution of the precipitate.
1) Al 4 C 3 + 12HCl = 3CH 4 + 4AlCl 3
2) CH 4 + 2O 2 \u003d CO 2 + 2H 2 O
3) CO 2 + Ca (OH) 2 \u003d CaCO 3 + H 2 O
4) CaCO 3 + H 2 O + CO 2 \u003d Ca (HCO 3) 2
3. Pyrite was roasted, the resulting gas with a pungent odor was passed through hydrosulfide acid. The resulting yellowish precipitate was filtered off, dried, mixed with concentrated nitric acid and heated. The resulting solution gives a precipitate with barium nitrate.
1) 4FeS 2 + 11O 2 → 2Fe 2 O 3 + 8SO 2
2) SO 2 + 2H 2 S \u003d 3S + 2H 2 O
3) S+ 6HNO 3 = H 2 SO 4 + 6NO 2 + 2H 2 O
4) H 2 SO 4 + Ba(NO 3) 2 = BaSO 4 ↓ + 2 HNO 3
4 . Copper was placed in concentrated nitric acid, the resulting salt was isolated from the solution, dried and calcined. The solid reaction product was mixed with copper shavings and calcined in an inert gas atmosphere. The resulting substance was dissolved in ammonia water.
1) Cu + 4HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
2) 2Cu(NO 3) 2 = 2CuO + 4NO 2 + O 2
3) Cu + CuO = Cu 2 O
4) Cu 2 O + 4NH 3 + H 2 O \u003d 2OH
5 . Iron filings were dissolved in dilute sulfuric acid, the resulting solution was treated with an excess of sodium hydroxide solution. The precipitate formed was filtered and left in air until it turned brown. The brown substance was calcined to constant weight.
1) Fe + H 2 SO 4 \u003d FeSO 4 + H 2
2) FeSO 4 + 2NaOH \u003d Fe (OH) 2 + Na 2 SO 4
3) 4Fe(OH) 2 + 2H 2 O + O 2 = 4Fe(OH) 3
4) 2Fe (OH) 3 \u003d Fe 2 O 3 + 3H 2 O
6 . The zinc sulfide was calcined. The resulting solid reacted completely with the potassium hydroxide solution. Carbon dioxide was passed through the resulting solution until a precipitate formed. The precipitate was dissolved in hydrochloric acid.
1) 2ZnS + 3O 2 = 2ZnO + 2SO 2
2) ZnO + 2NaOH + H 2 O = Na 2
3 Na 2 + CO 2 \u003d Na 2 CO 3 + H 2 O + Zn (OH) 2
4) Zn(OH) 2 + 2 HCl = ZnCl 2 + 2H 2 O
7. The gas released during the interaction of zinc with hydrochloric acid was mixed with chlorine and exploded. The resulting gaseous product was dissolved in water and treated with manganese dioxide. The resulting gas was passed through a hot solution of potassium hydroxide.
1) Zn+ 2HCl = ZnCl 2 + H 2
2) Cl 2 + H 2 \u003d 2HCl
3) 4HCl + MnO 2 = MnCl 2 + 2H 2 O + Cl 2
4) 3Cl 2 + 6KOH = 5KCl + KClO 3 + 3H 2 O
8. Calcium phosphide was treated with hydrochloric acid. The released gas was burned in a closed vessel, the combustion product was completely neutralized with a solution of potassium hydroxide. A solution of silver nitrate was added to the resulting solution.
1) Ca 3 P 2 + 6HCl = 3CaCl 2 + 2PH 3
2) PH 3 + 2O 2 = H 3 PO 4
3) H 3 PO 4 + 3KOH = K 3 PO 4 + 3H 2 O
4) K 3 PO 4 + 3AgNO 3 \u003d 3KNO 3 + Ag 3 PO 4
9
1) (NH 4) 2 Cr 2 O 7 = Cr 2 O 3 + N 2 + 4H 2 O
2) Cr 2 O 3 + 3H 2 SO 4 = Cr 2 (SO 4) 3 + 3H 2 O
3) Cr 2 (SO 4) 3 + 6NaOH \u003d 3Na 2 SO 4 + 2Cr (OH) 3
4) 2Cr(OH) 3 + 3NaOH = Na 3
10 . Calcium orthophosphate was calcined with coal and river sand. The resulting white glow-in-the-dark substance was burned in an atmosphere of chlorine. The product of this reaction was dissolved in an excess of potassium hydroxide. A solution of barium hydroxide was added to the resulting mixture.
1) Ca 3 (PO 4) 2 + 5C + 3SiO 2 = 3CaSiO 3 + 5CO + 2P
2) 2P + 5Cl 2 = 2PCl 5
3) PCl 5 + 8KOH = K 3 PO 4 + 5KCl + 4H 2 O
4) 2K 3 PO 4 + 3Ba(OH) 2 = Ba 3 (PO 4) 2 + 6KOH
11. Aluminum powder was mixed with sulfur and heated. The resulting substance was placed in water. The resulting precipitate was divided into two parts. Hydrochloric acid was added to one part, and sodium hydroxide solution was added to the other until the precipitate was completely dissolved.
1) 2Al + 3S = Al 2 S 3
2) Al 2 S 3 + 6H 2 O \u003d 2Al (OH) 3 + 3H 2 S
3) Al(OH) 3 + 3HCl= AlCl 3 + 3H 2 O
4) Al (OH) 3 + NaOH \u003d Na
12 . Silicon was placed in a solution of potassium hydroxide, after the completion of the reaction, an excess of hydrochloric acid was added to the resulting solution. The precipitate formed was filtered off, dried and calcined. The solid calcination product reacts with hydrogen fluoride.
1) Si + 2KOH + H 2 O = K 2 SiO 3 + 2H 2
2) K 2 SiO 3 + 2HCl = 2KCl + H 2 SiO 3
3) H 2 SiO 3 \u003d SiO 2 + H 2 O
4) SiO 2 + 4HF \u003d SiF 4 + 2H 2 O
V.N. Doronkin, A.G. Berezhnaya, T.V. Sazhnev, V.A. February. Chemistry. Thematic tests. New assignments for the USE-2012. Chemical experiment (C2): teaching aid. - Rostov n / D: Legion, 2012. - 92 p.
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- Chemistry
Description:
METHODOLOGY OF PREPARING STUDENTS FOR THE DECISION
ASSIGNMENTS C2 USE IN CHEMISTRY
When heated, an orange substance decomposes; decomposition products include a colorless gas and a green solid. the released gas reacts with lithium even with slight heating. The product of the latter reaction interacts with water, and a gas with a pungent odor is released, which can reduce metals, such as copper, from their oxides.
An analysis of the content of the task shows that the first substance is unknown, but the characteristic properties of the substance itself (color) and the reaction products (color and state of aggregation) are known. For all other reactions, the reagent and conditions are indicated. Tips can be considered as indications of the class of the obtained substance, its state of aggregation, characteristic features (color, smell). Note that two reaction equations characterize the special properties of substances (1 - decomposition of ammonium dichromate; 4 - reducing properties of ammonia), two equations characterize the typical properties of the most important classes of inorganic substances (2 - reaction between metal and non-metal, 3 - hydrolysis of nitrides).
toC Li H 2 O CuO
(NH 4 )2 Cr 2 O 7 → gas → X →gas with a pungent odor→C u
Highlight clues, key points, for example: an orange-colored substance that decomposes with the release of nitrogen (colorless gas) and Cr2O3 (green substance) - ammonium dichromate ( NH 4 )2 Cr 2 O 7 .
(NH4)2Cr2O7 →N2 + Cr2O3 + 4H2O
N2 + 6Li→2Li3N
Li3N + 3H2O →NH3+ 3LiOH
NH3 + 3CuO →3Cu + N2 + 3H2O
What difficulties can such tasks cause for students?
1. Description of actions with substances (filtration, evaporation, roasting, calcination, sintering, fusion). Students need to understand where a physical phenomenon occurs with a substance, and where a chemical reaction occurs. The most commonly used actions with substances are described below.
Filtration - a method for separating heterogeneous mixtures using filters - porous materials that pass liquid or gas, but retain solids. When separating mixtures containing a liquid phase, a solid remains on the filter, the filtrate passes through the filter.
Evaporation - the process of concentrating solutions by evaporating the solvent. Sometimes evaporation is carried out until saturated solutions are obtained, in order to further crystallize from them a solid in the form of a crystalline hydrate, or until the solvent is completely evaporated in order to obtain a pure solute.
Ignition - heating a substance to change its chemical composition.
The calcination can be carried out in air and in an inert gas atmosphere.
When calcined in air, crystalline hydrates lose water of crystallization:
CuSO 4 ∙5 H 2 O → CuSO 4 + 5 H 2 O
Thermally unstable substances decompose (insoluble bases, some salts, acids, oxides): Cu (OH) 2 → CuO + H 2 O; CaCO 3 → CaO + CO 2
Substances that are unstable to the action of air components, oxidize when ignited, react with air components: 2C u + O 2 → 2 CuO;
4 Fe (OH) 2 + O 2 → 2 Fe 2 O 3 + 4 H 2 O
In order to prevent oxidation during calcination, the process is carried out in an inert atmosphere: Fe (OH) 2 → FeO + H 2 O
Sintering, fusion -This is the heating of two or more solid reactants, leading to their interaction. If the reagents are resistant to the action of oxidizing agents, then sintering can be carried out in air:
Al 2 O 3 + Na 2 CO 3 → 2 NaAlO 2 + CO 2
If one of the reactants or the reaction product can be oxidized by air components, the process is carried out with an inert atmosphere, for example: C u + CuO → Cu 2 O
Burning - a heat treatment process leading to the combustion of a substance (in the narrow sense. In a broader sense, roasting is a variety of thermal effects on substances in chemical production and metallurgy). It is mainly used in relation to sulfide ores. For example, firing pyrite:
4FeS2 + 11O2 → 2Fe2O3 + 8SO2
2. Description of the characteristic features of substances (color, smell, state of aggregation).
The indication of the characteristic features of substances should serve as a hint for students or to verify the correctness of the actions performed. However, if students are not familiar with the physical properties of substances, such information cannot provide an auxiliary function when performing a thought experiment. Below are the most characteristic features of gases, solutions, solids.
GASES:
Painted : Cl 2 - yellow-green; NO 2 - brown; O 3 - blue (all have smells). All are poisonous, dissolve in the input, Cl 2 and NO 2 react with it.
Colorless, odorless: H2, N 2, O 2, CO 2, CO (poison), NO (poison), inert gases. All are poorly soluble in water.
Colorless with an odor: HF , HCl , HBr , HI , SO 2 (pungent odors), NH 3 (ammonia) - highly soluble in water and poisonous,
PH 3 (garlic), H 2 S (rotten eggs) - slightly soluble in water, poisonous.
COLORED SOLUTIONS:
yellow
Chromates, for example K2CrO4
Solutions of iron salts ( III), for example, FeCl 3,
bromine water,
c alcohol and alcohol-water solutions of iodine - depending on the concentration of yellow to brown
orange
Dichromates, for example, K2Cr2O7
green
Chromium hydroxo complexes ( III), for example, K 3 [Cr (OH) 6], nickel salts (II), for example NiSO 4,
manganates, for example, K2MnO4
blue
Copper (II) salts, for example C uSO 4
pink to purple
Permanganates, for example, KMnO4
From green to blue
Salts of chromium (III), for example, CrCl 3
PAINTED DRAINAGE,
PRODUCED IN THE INTERACTION OF SOLUTIONS
yellow
AgBr, AgI, Ag3PO4, BaCrO4, PbI2, CdS
brown
Fe(OH)3, MnO2
black, black-brown
Sulfides of copper, silver, iron, lead
blue
Cu(OH)2, KF e
green
Cr(OH )3 - gray-green
Fe(OH )2 - dirty green, turns brown in the air
OTHER COLORED SUBSTANCES
yellow
sulfur, gold, chromates
orange
o copper oxide (I) - Cu 2 O
dichromates
red
bromine (liquid), copper (amorphous), red phosphorus,
Fe2O3, CrO3
black
With uO, FeO, CrO
Gray with a metallic sheen
Graphite, crystalline silicon, crystalline iodine (during sublimation - purple vapors), most metals.
green
Cr 2 O 3, malachite (CuOH) 2 CO 3, Mn 2 O 7 (liquid)
This, of course, is the minimum information that can be useful for solving tasks C2.
In the process of preparing students for solving tasks C2, you can offer them to compose texts of tasks in accordance with the transformation schemes. This task will allow students to master the terminology and remember the characteristic features of substances.
Example 1:
toC toC / H 2 HNO 3 (conc) NaOH, 0 o C
(CuOH)2CO3→ CuO →Cu→NO2→ X
Text: Malachite was calcined, the resulting black solid was heated in a stream of hydrogen. The resulting red substance was completely dissolved in concentrated nitric acid. The liberated brown gas was passed through a cold solution of sodium hydroxide.
Example 2:
O2 H2S p - p toC/AlH2O
ZnS→SO2→S→Al2S3→X
Text: Zinc sulfide was fired. The resulting gas with a pungent odor was passed through a solution of hydrogen sulfide until a yellow precipitate formed. The precipitate was filtered off, dried and fused with aluminum. The resulting compound was placed in water until the reaction terminated.
At the next stage, students can be invited to draw up both the schemes for the transformation of substances and the texts of tasks themselves. Of course, the “authors” of the tasks must also present their own solution. At the same time, students repeat all the properties of inorganic substances. And the teacher can form a bank of tasks C2.
After that, you can proceed to the solution of tasks C2. At the same time, students draw up a scheme of transformations according to the text, and then the corresponding reaction equations. To do this, reference points are highlighted in the text of the task: the names of substances, an indication of their classes, physical properties, conditions for conducting reactions, names of processes.
Let's give examples of some tasks.
Example 1 manganese nitrate ( II ) was calcined, concentrated hydrochloric acid was added to the resulting solid brown substance. The evolved gas was passed through hydrosulfide acid. The resulting solution forms a precipitate with barium chloride.
Decision:
· Selection of support moments:
manganese nitrate ( II ) - Mn (NO 3 )2,
calcined - heated to decomposition,
solid brown matter– Mn O2,
Concentrated hydrochloric acid– HCl,
Hydrosulphuric acid - solution H2S,
Barium chloride - BaCl 2 , forms a precipitate with the sulfate ion.
· Drawing up a transformation scheme:
toC HCl H2 S solution BaCl 2
Mn (NO 3 )2 → Mn O2 → X → U → ↓ (BaSO 4 ?)
· Drawing up reaction equations:
1) Mn(NO3)2→Mn О 2 + 2NO2
2) Mn O 2 + 4 HCl → MnCl2 + 2H2O + Cl2 ( gas X)
3) Cl 2 + H2 S → 2 HCl + S (not suitable as there is no product that precipitates with barium chloride) or4 Cl 2 + H2 S + 4H2O → 8 HCl + H2 SO 4
4) H 2 SO4 + BaCl2→BaSO4 + 2HCl
Example 2 Orange copper oxide was placed in concentrated sulfuric acid and heated. An excess of potassium hydroxide solution was added to the resulting blue solution. The resulting blue precipitate was filtered off, dried and calcined. The solid black substance thus obtained was placed in a glass tube, heated, and ammonia was passed over it.
Decision:
· Selection of support moments:
Orange copper oxide- Cu 2 O,
concentrated sulfuric acid- H2 SO 4,
Blue solution - copper salt (II), C uSO 4
Potassium hydroxide -KOH,
Blue precipitate - Cu (OH) 2,
Calcined - heated to decomposition
Solid black matter CuO,
Ammonia - NH3.
· Drawing up a transformation scheme:
H2 SO 4 KOH toC NH3
Cu 2 O → С uSO 4 → Cu (OH) 2 ↓ → CuO → X
· Drawing up reaction equations:
1) Cu2O + 3 H 2 SO4 → 2 C uSO4 + SO2 + 3H2O
2) With uSO4 + 2 KOH → Cu(OH)2+ K2SO4
3) Cu (OH) 2 → CuO + H 2 O
4) 3 CuO + 2 NH 3 → 3 Cu + 3H2O + N 2
EXAMPLES OF TASKS FOR INDEPENDENT SOLUTION
1 . Sodium was burned in an excess of oxygen, the resulting crystalline substance was placed in a glass tube and carbon dioxide was passed through it. The gas coming out of the tube was collected and burned in its atmosphere of phosphorus. The resulting substance was neutralized with an excess of sodium hydroxide solution.
2. Aluminum carbide treated with hydrochloric acid. The released gas was burned, the combustion products were passed through lime water until a white precipitate formed, further passing the combustion products into the resulting suspension led to the dissolution of the precipitate.
3. Pyrite was roasted, the resulting gas with a pungent odor was passed through hydrosulfide acid. The resulting yellowish precipitate was filtered off, dried, mixed with concentrated nitric acid and heated. The resulting solution gives a precipitate with barium nitrate.
4 . Copper was placed in concentrated nitric acid, the resulting salt was isolated from the solution, dried and calcined. The solid reaction product was mixed with copper shavings and calcined in an inert gas atmosphere. The resulting substance was dissolved in ammonia water.
5 . Iron filings were dissolved in dilute sulfuric acid, the resulting solution was treated with an excess of sodium hydroxide solution. The precipitate formed was filtered and left in air until it turned brown. The brown substance was calcined to constant weight.
6 . The zinc sulfide was calcined. The resulting solid reacted completely with the potassium hydroxide solution. Carbon dioxide was passed through the resulting solution until a precipitate formed. The precipitate was dissolved in hydrochloric acid.
7. The gas released during the interaction of zinc with hydrochloric acid was mixed with chlorine and exploded. The resulting gaseous product was dissolved in water and treated with manganese dioxide. The resulting gas was passed through a hot solution of potassium hydroxide.
8. Calcium phosphide was treated with hydrochloric acid. The released gas was burned in a closed vessel, the combustion product was completely neutralized with a solution of potassium hydroxide. A solution of silver nitrate was added to the resulting solution.
9 . Ammonium dichromate decomposed on heating. The solid decomposition product was dissolved in sulfuric acid. Sodium hydroxide solution was added to the resulting solution until a precipitate formed. Upon further addition of sodium hydroxide solution to the precipitate, it dissolved.
10 . Calcium orthophosphate was calcined with coal and river sand. The resulting white glow-in-the-dark substance was burned in an atmosphere of chlorine. The product of this reaction was dissolved in an excess of potassium hydroxide. A solution of barium hydroxide was added to the resulting mixture.
12 . Silicon was placed in a solution of potassium hydroxide, after the completion of the reaction, an excess of hydrochloric acid was added to the resulting solution. The precipitate formed was filtered off, dried and calcined. The solid calcination product reacts with hydrogen fluoride.
SOLUTIONS
1 . Sodium was burned in an excess of oxygen, the resulting crystalline substance was placed in a glass tube and carbon dioxide was passed through it. The gas coming out of the tube was collected and burned in its atmosphere of phosphorus. The resulting substance was neutralized with an excess of sodium hydroxide solution.
1) 2 Na + O 2 = Na 2 O 2
2) 2 Na 2 O 2 + 2 CO 2 = 2 Na 2 CO 3 + O 2
3) 4P + 5O2 = 2P2O5
4) P2O5 + 6 NaOH = 2Na3PO4 + 3H2O
2. Aluminum carbide treated with hydrochloric acid. The released gas was burned, the combustion products were passed through lime water until a white precipitate formed, further passing the combustion products into the resulting suspension led to the dissolution of the precipitate.
1) Al4C3 + 12HCl = 3CH4 + 4AlCl3
2) CH4 + 2O2 = CO2 + 2H2O
3) CO2 + Ca(OH)2 = CaCO3 + H2O
4) CaCO3 + H2O + CO2 = Ca(HCO3)2
3. Pyrite was roasted, the resulting gas with a pungent odor was passed through hydrosulfide acid. The resulting yellowish precipitate was filtered off, dried, mixed with concentrated nitric acid and heated. The resulting solution gives a precipitate with barium nitrate.
1) 4FeS2 + 11O2 → 2Fe2O3 + 8SO2
2) SO2 + 2H2 S= 3S + 2H2O
3) S+ 6HNO3 = H2SO4+ 6NO2 + 2H2O
4) H2SO4+ Ba(NO3)2 = BaSO4↓ + 2 HNO3
4 . Copper was placed in concentrated nitric acid, the resulting salt was isolated from the solution, dried and calcined. The solid reaction product was mixed with copper shavings and calcined in an inert gas atmosphere. The resulting substance was dissolved in ammonia water.
1) Cu + 4HNO3 = Cu(NO3)2+ 2NO2 + 2H2O
2) 2Cu(NO3)2 = 2CuO + 4NO2 + O2
3) Cu + CuO= Cu2O
4) Cu2O + 4NH3 + H2O = 2OH
5 . Iron filings were dissolved in dilute sulfuric acid, the resulting solution was treated with an excess of sodium hydroxide solution. The precipitate formed was filtered and left in air until it turned brown. The brown substance was calcined to constant weight.
1) Fe + H2SO4 = FeSO4 + H2
2) FeSO4 + 2NaOH= Fe(OH)2 + Na2SO4
3) 4Fe(OH)2 + 2H2O + O2 = 4Fe(OH)3
4) 2 Fe (OH) 3 \u003d Fe 2 O 3 + 3 H 2 O
6 . The zinc sulfide was calcined. The resulting solid reacted completely with the potassium hydroxide solution. Carbon dioxide was passed through the resulting solution until a precipitate formed. The precipitate was dissolved in hydrochloric acid.
1) 2ZnS + 3O2 = 2ZnO + 2SO2
2) ZnO+ 2NaOH + H2O = Na2
3 Na2 + CO2 = Na2CO3 + H2O + Zn(OH)2
4) Zn(OH)2 + 2HCl= ZnCl2 + 2H2O
7. The gas released during the interaction of zinc with hydrochloric acid was mixed with chlorine and exploded. The resulting gaseous product was dissolved in water and treated with manganese dioxide. The resulting gas was passed through a hot solution of potassium hydroxide.
1) Zn+ 2HCl= ZnCl2 + H2
2) Cl2 + H2 = 2HCl
3) 4HCl + MnO2 = MnCl2 + 2H2O + Cl2
4) 3Cl2 + 6KOH= 5KCl + KClO3 + 3H2O
8. Calcium phosphide was treated with hydrochloric acid. The released gas was burned in a closed vessel, the combustion product was completely neutralized with a solution of potassium hydroxide. A solution of silver nitrate was added to the resulting solution.
1) Ca3P2 + 6HCl = 3CaCl2 + 2PH3
2) PH3 + 2O2 = H3PO4
3) H3PO4 + 3KOH= K3PO4 + 3H2O
4) K 3 PO 4 + 3 AgNO 3 = 3 KNO 3 + Ag 3 PO 4
9 . Ammonium dichromate decomposed on heating. The solid decomposition product was dissolved in sulfuric acid. Sodium hydroxide solution was added to the resulting solution until a precipitate formed. On further addition of sodium hydroxide to the precipitate, it dissolved.
1) (NH4)2Cr2O7 = Cr2O3 + N2 + 4H2O
2) Cr2O3 + 3H2SO4 = Cr2(SO4)3 + 3H2O
3) Cr2(SO4)3 + 6NaOH= 3Na2SO4 + 2Cr(OH)3
4) 2Cr(OH)3 + 3NaOH = Na3
10 . Calcium orthophosphate was calcined with coal and river sand. The resulting white glow-in-the-dark substance was burned in an atmosphere of chlorine. The product of this reaction was dissolved in an excess of potassium hydroxide. A solution of barium hydroxide was added to the resulting mixture.
1) Ca3(PO4)2 + 5C + 3SiO2 = 3CaSiO3 + 5CO + 2P
2) 2P + 5Cl2 = 2PCl5
3) PCl5 + 8KOH= K3PO4 + 5KCl + 4H2O
4) 2K3PO4 + 3Ba(OH)2 = Ba3(PO4)2 + 6KOH
11. Aluminum powder was mixed with sulfur and heated. The resulting substance was placed in water. The resulting precipitate was divided into two parts. Hydrochloric acid was added to one part, and sodium hydroxide solution was added to the other until the precipitate was completely dissolved.
1) 2Al + 3S= Al2S3
2) Al2S3 + 6H2O = 2Al(OH)3 + 3H2S
3) Al(OH)3 + 3HCl= AlCl3 + 3H2O
4) Al(OH)3 + NaOH= Na
12 . Silicon was placed in a solution of potassium hydroxide, after the completion of the reaction, an excess of hydrochloric acid was added to the resulting solution. The precipitate formed was filtered off, dried and calcined. The solid calcination product reacts with hydrogen fluoride.
1) Si + 2KOH + H2O= K2SiO3+ 2H2
2) K2SiO3 + 2HCl = 2KCl + H2SiO3
3) H2SiO3 = SiO2 + H2O
4) SiO 2 + 4 HF \u003d SiF 4 + 2 H 2 O
Tasks C2 USE in chemistry: execution algorithm
Tasks C2 of the Unified State Examination in Chemistry ("Set of Substances") have been the most difficult tasks of Part C for a number of years. And this is no coincidence. In this task, the graduate must be able to apply his knowledge of the properties of chemicals, types of chemical reactions, as well as the ability to arrange coefficients in equations using the example of a variety of, sometimes unfamiliar substances. How to get the maximum number of points on this task? One of the possible algorithms for its implementation can be represented by the following four points:
Let us consider in more detail the application of this algorithm on one of the examples.
Exercise(2011 wording):
The first problem that arises when completing a task is to understand what is hidden under the names of substances. If a person writes the formula of hydrochloric acid instead of perchloric acid, and sulfite instead of potassium sulfide, he drastically reduces the number of correctly written reaction equations. Therefore, the knowledge of the nomenclature should be given the closest attention. It should be taken into account that the trivial names of some substances can also be used in the task: lime water, iron oxide, copper sulphate, etc.
The result of this stage is the recording of the formulas of the proposed set of substances.
To characterize the chemical properties of the proposed substances helps assign them to a specific group or class. In this case, for each substance, it is necessary to give characteristics in two directions. The first is the acid-base, exchange characteristic, which determines the ability to enter into reactions without changing the degree of oxidation.
According to the acid-base properties of substances, substances can be distinguished acidic nature (acids, acid oxides, acid salts), basic nature (bases, basic oxides, basic salts), amphoteric connections, medium salt. When performing a task, these properties can be abbreviated: " To", "O", "BUT", "With"
According to the redox properties of the substance can be classified into oxidizers and reducing agents. However, there are often substances that exhibit redox duality (ORD). This duality may be due to the fact that one of the elements is in an intermediate oxidation state. So, nitrogen is characterized by an oxidation scale from -3 to +5. Therefore, for potassium nitrite KNO 2, where nitrogen is in the +3 oxidation state, the properties of both an oxidizing agent and a reducing agent are characteristic. In addition, in one compound, atoms of different elements can exhibit different properties, as a result, the substance as a whole also exhibits ATS. An example is hydrochloric acid, which can be both an oxidizing agent, due to the H + ion, and a reducing agent, due to the chloride ion.
Duality does not mean the same properties. As a rule, either oxidizing or reducing properties predominate. There are also substances for which redox properties are uncharacteristic. This is observed when the atoms of all elements are in their most stable oxidation states. An example is, for example, sodium fluoride NaF. And, finally, the redox properties of a substance can strongly depend on the conditions under which the reaction is carried out. So, concentrated sulfuric acid is a strong oxidizing agent due to S +6, and the same acid in solution is an oxidizing agent of medium strength due to the H + ion
This feature can also be abbreviated OK","Sun","ATS".
Let's define the characteristics of the substances in our task:
- potassium chromate, salt, oxidizing agent (Cr +6 - the highest oxidation state)
- sulfuric acid, solution: acid, oxidizer (H+)
- sodium sulfide: salt, reducing agent (S -2 - lowest oxidation state)
- copper (II) sulfate, salt, oxidizing agent (Cu +2 - the highest oxidation state)Briefly, it could be written like this:
The juice(Cr+6)
K, ok(H+)
From, Sun(S-2)
The juice(Cu+2
At this stage, it is necessary to determine which reactions are possible between specific substances, as well as the possible products of these reactions. The already defined characteristics of substances will help in this. Since we have given two characteristics for each substance, it is necessary to consider the possibility of two groups of reactions: exchange, without changing the oxidation state, and OVR.
Between substances of basic and acidic nature is characteristic neutralization reaction, the usual product of which is salt and water (in the reaction of two oxides - only salt). In the same reaction, amphoteric compounds can participate in the role of an acid or base. In some rather rare cases, the neutralization reaction is impossible, which is usually indicated by a dash in the solubility table. The reason for this is either the weakness of the manifestation of acidic and basic properties in the original compounds, or the occurrence of a redox reaction between them (for example: Fe 2 O 3 + HI).
In addition to coupling reactions between oxides, one must also take into account the possibility compound reactions oxides with water. Many acid oxides and oxides of the most active metals enter into it, and the corresponding soluble acids and alkalis are the products. However, water is rarely given as a separate substance in item C2.
Salts are characterized exchange reaction, into which they can enter both among themselves and with acids and alkalis. As a rule, it proceeds in solution, and the criterion for the possibility of its occurrence is the RIO rule - precipitation, gas evolution, and the formation of a weak electrolyte. In some cases, the exchange reaction between salts can be complicated hydrolysis reaction, resulting in the formation of basic salts. The complete hydrolysis of the salt or the redox interaction between them can prevent the exchange reaction. The special nature of the interaction of salts is indicated by a dash in the solubility table for the intended product.
Separately, the hydrolysis reaction can be counted as the correct answer to task C2, if the set of substances contains water and salt undergoing complete hydrolysis (Al 2 S 3).
Insoluble salts can enter into exchange reactions usually only with acids. It is also possible to react insoluble salts with acids to form acid salts (Ca 3 (PO 4) 2 + H 3 PO 4 => Ca (H 2 PO 4) 2)
Another relatively rare reaction is the exchange reaction between the salt and the acid oxide. In this case, the more volatile oxide is replaced by the less volatile one (CaСO 3 + SiO 2 => CaSiO 3 + CO 2).
AT redox reactions oxidizing and reducing agents can enter. The possibility of this is determined by the strength of their redox properties. In some cases, the possibility of a reaction can be determined using a series of metal stresses (reactions of metals with salt solutions, acids). Sometimes the relative strength of oxidizing agents can be estimated using the regularities of the Periodic Table (displacement of one halogen by another). However, most often it will require knowledge of specific factual material, the properties of the most characteristic oxidizing and reducing agents (compounds of manganese, chromium, nitrogen, sulfur ...), training in writing OVR equations.
It is also difficult to identify possible RIA products. In general, two rules can be proposed to help make a choice:
- reaction products should not interact with the starting substances, with the environment, in which the reaction is carried out: if sulfuric acid is poured into the test tube, KOH cannot be obtained there, if the reaction is carried out in an aqueous solution, sodium will not precipitate there;
- reaction products should not interact with each other: CuSO 4 and KOH, Cl 2 and KI cannot be obtained simultaneously in a test tube.Consideration must also be given to the type of disproportionation reactions(self-oxidation-self-healing). Such reactions are possible for substances where the element is in an intermediate oxidation state, which means that it can simultaneously be oxidized and reduced. The second participant in such a reaction plays the role of a medium. An example is the disproportionation of halogens in an alkaline medium.
Chemistry is so complex and interesting that it is impossible to give general recipes for all occasions in it. Therefore, along with these two groups of reactions, one more can be named: specific reactions individual substances. The success of writing such reaction equations will be determined by actual knowledge of the chemistry of individual chemical elements and substances.
In predicting reactions for specific substances, it is desirable to follow a certain order so as not to miss any reaction. You can use the approach represented by the following diagram:
We consider the possibility of reactions of the first substance with three other substances (green arrows), then we consider the possibility of reactions of the second substance with the remaining two (blue arrows), and, finally, we consider the possibility of the interaction of the third substance with the last, fourth (red arrow). If there are five substances in the set, there will be more arrows, but some of them will be crossed out during the analysis.
So, for our set, the first substance:
- K 2 CrO 4 + H 2 SO 4, OVR is impossible (two oxidizing agents), the usual exchange reaction is also impossible, because the intended products are soluble. Here we are faced with a specific reaction: chromates, when interacting with acids, form dichromates: => K 2 Cr 2 O 7 + K 2 SO 4 + H 2 O
- K 2 CrO 4 + Na 2 S, the exchange reaction is also impossible, because the intended products are soluble. But the presence of an oxidizing agent and a reducing agent here allows us to conclude that OVR is possible. With OVR, S -2 will be oxidized to sulfur, Cr +6 will be reduced to Cr +3, in a neutral environment it could be Cr (OH) 3. However, at the same time, KOH is formed in the solution. Taking into account the amphoteric nature of Cr(OH) 3 and the rule that the reaction products should not react with each other, we come to the choice of the following products: => S + K + KOH
- K 2 CrO 4 + CuSO 4, but here, an exchange reaction between salts is possible, because most chromates are insoluble in water: => K 2 SO 4 + CuCrO 4Second substance:
- H 2 SO 4 + Na 2 S, the hydrogen ion is not a strong enough oxidizing agent to oxidize the sulfide ion, OVR is impossible. But an exchange reaction is possible, leading to the formation of a weak electrolyte and a gaseous substance: => H 2 S + Na 2 SO 4;
- H 2 SO 4 + CuSO 4 There are no obvious reactions here.Third substance:
- Na 2 S + CuSO 4, the copper ion is also not a strong enough oxidizing agent to oxidize the sulfide ion, OVR is impossible. The exchange reaction between salts will lead to the formation of insoluble copper sulfide: => CuS + Na 2 SO 4.The result of the third stage should be several schemes of possible reactions. Possible problems:
- too many reactions. Since the experts will only evaluate four first reaction equations, you need to choose the simplest reactions, in the course of which you are 100% sure, and discard too complex ones, or those in which you are not too sure. So in our case, it was possible to score the maximum number of points without knowing the specific reaction of the transition of chromates to dichromates. And if you know this not too complicated reaction, then you can refuse to equalize the rather complex OVR, leaving only simple exchange reactions.
- few reactions, less than four. If, when analyzing the reactions of pairs of substances, the number of reactions turned out to be insufficient, the possibility of the interaction of three substances can be considered. Usually these are OVRs, in which a third substance, the medium, can also take part, and, depending on the medium, the reaction products may be different. So in our case, if the found reactions were not enough, we could additionally suggest the interaction of potassium chromate with sodium sulfide in the presence of sulfuric acid. The reaction products in this case would be sulfur, chromium(III) sulfate and potassium sulfate.
If the state of substances is not clearly indicated, for example, it is simply said "sulfuric acid" instead of "solution (meaning diluted) sulfuric acid", it is possible to analyze the possibility of reactions of a substance in different states. In our case, we could take into account that concentrated sulfuric acid is a strong oxidizing agent due to S +6, and can enter into OVR with sodium sulfide to form sulfur dioxide SO 2 .
Finally, we can take into account the possibility of the reaction proceeding differently depending on the temperature, or on the ratio of the amounts of substances. Thus, the interaction of chlorine with alkali can give hypochlorite in the cold, and when heated, potassium chlorate, aluminum chloride, when reacting with alkali, can give both aluminum hydroxide and hydroxoaluminate. All this allows us to write not one, but two reaction equations for one set of initial substances. But we must take into account that this contradicts the condition of the task: "between all the proposed substances, without repeating reagent pairs". Therefore, whether all such equations will be credited depends on the specific set of substances and the discretion of the expert.