1) 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.
2) 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 caustic soda. A solution of barium chloride was added to the resulting solution. The resulting precipitate was treated with an excess of phosphoric acid.
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Ca 3 (PO 4) 2 → P → P 2 O 5 → Na 3 PO 4 → Ba 3 (PO 4) 2 → BaHPO 4 or Ba (H 2 PO 4) 2 Ca 3 (PO 4) 2 + 5C + 3SiO 2 → 3CaSiO 3 + 2P + 5CO |
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3) 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.
4) Dry sodium chloride was treated with concentrated sulfuric acid at low heating, the resulting gas was passed into a solution of barium hydroxide. 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.
5) A sample 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.
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Al(OH) 3 ←AlCl 3 ←Al 2 S 3 → H 2 S → PbS → PbSO 4 Al 2 S 3 + 6HCl → 3H 2 S + 2AlCl 3 |
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6) Aluminum powder was mixed with sulfur powder, the mixture was heated, the resulting substance was treated with water, while 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.
7) A solution of potassium iodide was treated with a solution of chlorine. 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.
8) A 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.
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Cr 2 O 3 → KCrO 2 → K → K 2 CrO 4 → K 2 Cr 2 O 7 → Cr 2 (SO 4) 3 Cr 2 O 3 + 2KOH → 2KCrO 2 + H 2 O |
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9) 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.
10) 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.
11) Copper(II) oxide was heated in a stream of carbon monoxide. The resulting substance was burned in an atmosphere of chlorine. The reaction product was dissolved in water. The resulting solution was divided into two parts. A solution of potassium iodide was added to one part, a solution of silver nitrate was added to the second. In both cases, the formation of a precipitate was observed. Write the equations for the four described reactions.
12) Copper nitrate was calcined, the resulting solid was dissolved in dilute sulfuric acid. The resulting salt solution was subjected to electrolysis. The substance released at the cathode was dissolved in concentrated nitric acid. The dissolution proceeded with evolution of brown gas. Write the equations for the four described reactions.
13) Iron was burned in an atmosphere of chlorine. The resulting material was treated with an excess of sodium hydroxide solution. A brown precipitate formed, which was filtered off and calcined. The residue after calcination was dissolved in hydroiodic acid. Write the equations for the four described reactions.
14) Powder of metallic aluminum was mixed with solid iodine and a few drops of water were added. Sodium hydroxide solution was added to the resulting salt until a precipitate formed. The resulting precipitate was dissolved in hydrochloric acid. Upon subsequent addition of sodium carbonate solution, precipitation was again observed. Write the equations for the four described reactions.
15) As a result of incomplete combustion of coal, a gas was obtained, in the flow of which iron oxide (III) was heated. The resulting substance was dissolved in hot concentrated sulfuric acid. The resulting salt solution was subjected to electrolysis. Write the equations for the four described reactions.
16) Some amount of zinc sulfide was divided into two parts. One of them was treated with nitric acid, and the other was fired in air. During the interaction of the evolved gases, a simple substance was formed. This substance was heated with concentrated nitric acid, and a brown gas was released. Write the equations for the four described reactions.
17) Potassium chlorate was heated in the presence of a catalyst, and a colorless gas was released. By burning iron in an atmosphere of this gas, iron scale was obtained. It was dissolved in an excess of hydrochloric acid. To the solution thus obtained was added a solution containing sodium dichromate and hydrochloric acid.
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1) 2KClO 3 → 2KCl + 3O 2 2) ЗFe + 2O 2 → Fe 3 O 4 3) Fe 3 O 4 + 8HCI → FeCl 2 + 2FeCl 3 + 4H 2 O 4) 6 FeCl 2 + Na 2 Cr 2 O 7 + 14 HCI → 6 FeCl 3 + 2 CrCl 3 + 2NaCl + 7H 2 O 18) Iron burned in chlorine. The resulting salt was added to a solution of sodium carbonate, and a brown precipitate fell out. This precipitate was filtered off and calcined. The resulting substance was dissolved in hydroiodic acid. Write the equations for the four described reactions. 1) 2Fe + 3Cl 2 → 2FeCl 3 2) 2FeCl 3 + 3Na 2 CO 3 → 2Fe (OH) 3 + 6NaCl + 3CO 2 3) 2Fe(OH) 3 Fe 2 O 3 + 3H 2 O 4) Fe 2 O 3 + 6HI → 2FeI 2 + I 2 + 3H 2 O |
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19) A solution of potassium iodide was treated with an excess of chlorine water, while observing first the formation of a precipitate, and then its complete dissolution. The iodine-containing acid thus formed was isolated from the solution, dried, and gently heated. The resulting oxide reacted with carbon monoxide. Write down the equations of the described reactions.
20) Chromium(III) sulfide powder was dissolved in sulfuric acid. In this case, gas was released and a colored solution was formed. An excess of ammonia solution was added to the resulting solution, and the gas was passed through lead nitrate. The resulting black precipitate turned white after treatment with hydrogen peroxide. Write down the equations of the described reactions.
21) Aluminum powder was heated with sulfur powder, the resulting substance was treated with water. The resulting precipitate was treated with an excess of concentrated potassium hydroxide solution until it was completely dissolved. A solution of aluminum chloride was added to the resulting solution, and the formation of a white precipitate was again observed. Write down the equations of the described reactions.
22) Potassium nitrate was heated with powdered lead until the reaction ceased. The mixture of products was treated with water, and then the resulting solution was filtered. The filtrate was acidified with sulfuric acid and treated with potassium iodide. The released simple substance was heated with concentrated nitric acid. In the atmosphere of the resulting brown gas, red phosphorus was burned. Write down the equations of the described reactions.
23) Copper was dissolved in dilute nitric acid. An excess of ammonia solution was added to the resulting solution, observing first the formation of a precipitate, and then its complete dissolution with the formation of a dark blue solution. The resulting solution was treated with sulfuric acid until the characteristic blue color of copper salts appeared. Write down the equations of the described reactions.
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1) 3Cu + 8HNO 3 → 3Cu (NO 3) 2 + 2NO + 4H 2 O 2) Cu (NO 3) 2 + 2NH 3 H 2 O → Cu (OH) 2 + 2NH 4 NO 3 3) Cu (OH) 2 + 4NH 3 H 2 O → (OH) 2 + 4H 2 O 4) (OH) 2 + 3H 2 SO 4 → CuSO 4 + 2 (NH 4) 2 SO 4 + 2H 2 O |
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24) Magnesium was dissolved in dilute nitric acid, and no evolution of gas was observed. The resulting solution was treated with an excess of potassium hydroxide solution while heating. The resulting gas was burned in oxygen. Write down the equations of the described reactions.
25) A mixture of potassium nitrite and ammonium chloride powders was dissolved in water and the solution heated gently. The released gas reacted with magnesium. The reaction product was added to an excess of hydrochloric acid solution, and no gas evolution was observed. The resulting magnesium salt in solution was treated with sodium carbonate. Write down the equations of the described reactions.
26) Aluminum oxide was fused with sodium hydroxide. The reaction product was added to an ammonium chloride solution. The released gas with a pungent odor is absorbed by sulfuric acid. The middle salt thus formed was calcined. Write down the equations of the described reactions.
27) Chlorine reacted with a hot solution of potassium hydroxide. When the solution was cooled, crystals of Berthollet salt precipitated. The resulting crystals were added to a hydrochloric acid solution. The resulting simple substance reacted with metallic iron. The reaction product was heated with a new sample of iron. Write down the equations of the described reactions.
28) Copper was dissolved in concentrated nitric acid. An excess of ammonia solution was added to the resulting solution, observing first the formation of a precipitate, and then its complete dissolution. The resulting solution was treated with an excess of hydrochloric acid. Write down the equations of the described reactions.
29) Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The brown precipitate formed was filtered off and dried. The resulting substance was fused with iron. Write the equations for the four described reactions.
30) As a result of incomplete combustion of coal, a gas was obtained, in the flow of which iron oxide (III) was heated. The resulting substance was dissolved in hot concentrated sulfuric acid. The resulting salt solution was treated with an excess of potassium sulfide solution.
31) Some amount of zinc sulfide was divided into two parts. One of them was treated with hydrochloric acid, and the other was fired in air. During the interaction of the evolved gases, a simple substance was formed. This substance was heated with concentrated nitric acid, and a brown gas was released.
32) Sulfur was fused with iron. The reaction product was treated with hydrochloric acid. The resulting gas was burned in an excess of oxygen. The combustion products were absorbed by an aqueous solution of iron(III) sulfate.
The proposed material presents the methodological development of practical work for the 9th grade: “Solving experimental problems on the topic “Nitrogen and phosphorus”, “Determination of mineral fertilizers”, as well as laboratory experiments on the topic “Exchange reactions between electrolyte solutions”.
Exchange reactions between electrolyte solutions
Methodical development consists of three parts: theory, practical work, control. In the theoretical part, some examples of molecular, complete and reduced ionic equations of chemical reactions occurring with the formation of a precipitate, a low-dissociating substance, and gas evolution are given. In the practical part, tasks and recommendations for students on how to perform laboratory experiments are given. The control consists of test tasks with the choice of the correct answer.
Theory
1. Reactions proceeding with the formation of a precipitate.
a) When copper (II) sulfate reacts with sodium hydroxide, a blue precipitate of copper (II) hydroxide is formed.
CuSO 4 + 2NaOH \u003d Cu (OH) 2 + Na 2 SO 4.
Cu 2+ + + 2Na + + 2OH - \u003d Cu (OH) 2 + 2Na + +,
Cu 2+ + 2OH - \u003d Cu (OH) 2.
b) When barium chloride reacts with sodium sulfate, a white milky precipitate of barium sulfate precipitates.
Molecular equation of a chemical reaction:
BaCl 2 + Na 2 SO 4 \u003d 2NaCl + BaSO 4.
Full and reduced ionic reaction equations:
Ba 2+ + 2Cl - + 2Na + + = 2Na + + 2Cl - + BaSO 4,
Ba 2+ + \u003d BaSO 4.
2.
When sodium carbonate or bicarbonate (baking soda) reacts with hydrochloric or other soluble acid, effervescence or intense release of gas bubbles is observed. This releases carbon dioxide CO 2 , causing the clear solution of lime water (calcium hydroxide) to become cloudy. Lime water becomes cloudy, because. insoluble calcium carbonate is formed.
a) Na 2 CO 3 + 2HCl \u003d 2NaCl + H 2 O + CO 2;
b) NaHCO 3 + HCl = NaCl + CO 2 + H 2 O;
Ca (OH) 2 + CO 2 \u003d CaCO 3 + H 2 O.
a) 2Na + + + 2H + + 2Cl - = 2Na + + 2Cl - + CO 2 + H 2 O,
2H + = CO 2 + H 2 O;
b) Na + + + H + + Cl - = Na + + Cl - + CO 2 + H 2 O,
H + \u003d CO 2 + H 2 O.
3. Reactions proceeding with the formation of a low-dissociating substance.
When sodium or potassium hydroxide reacts with hydrochloric acid or other soluble acids in the presence of the phenolphthalein indicator, the alkali solution becomes colorless, and as a result of the neutralization reaction, a slightly dissociating substance H 2 O is formed.
Molecular equations of chemical reactions:
a) NaOH + HCl = NaCl + H 2 O;
c) 3KOH + H 3 PO 4 = K 3 PO 4 + 3H 2 O.
Full and reduced ionic reaction equations:
a) Na + + OH - + H + + Cl - = Na + + Cl - + H 2 O,
OH - + H + \u003d H 2 O;
b) 2Na + + 2OH - + 2H + + = 2Na + + + 2H 2 O,
2OH - + 2H + = 2H 2 O;
c) 3K + + 3OH - + 3H + + = 3K + + + 3H 2 O,
3OH - + 3H + \u003d 3H 2 O.
Workshop
1. Exchange reactions between electrolyte solutions, proceeding with the formation of a precipitate.
a) Conduct a reaction between solutions of copper(II) sulfate and sodium hydroxide. Write the molecular, complete and abbreviated ionic equations of chemical reactions, note the signs of a chemical reaction.
b) Conduct a reaction between solutions of barium chloride and sodium sulfate. Write the molecular, complete and abbreviated ionic equations of chemical reactions, note the signs of a chemical reaction.
2. Reactions that go with the release of gas.
Carry out reactions between solutions of sodium carbonate or sodium bicarbonate (baking soda) with hydrochloric or other soluble acid. Pass the released gas (using a gas outlet tube) through transparent lime water poured into another test tube until it becomes cloudy. Write molecular, complete and abbreviated ionic equations of chemical reactions, note the signs of these reactions.
3. Reactions proceeding with the formation of a low-dissociating substance.
Conduct neutralization reactions between alkali (NaOH or KOH) and acid (HCl, HNO 3 or H 2 SO 4), after placing phenolphthalein in the alkali solution. Mark observations and write molecular, complete and abbreviated ionic equations of chemical reactions.
signs, accompanying these reactions, can be selected from the following list:
1) release of gas bubbles; 2) precipitation; 3) the appearance of a smell; 4) dissolution of the precipitate; 5) heat release; 6) change in the color of the solution.
Control (test)
1. The ionic equation for the reaction in which the blue precipitate is formed is:
a) Cu 2+ + 2OH - = Cu (OH) 2;
c) Fe 3+ + 3OH - = Fe (OH) 3;
d) Al 3+ + 3OH - \u003d Al (OH) 3.
2. The ionic equation for the reaction in which carbon dioxide is released is:
a) CaCO 3 + CO 2 + H 2 O \u003d Ca 2+ +;
b) 2H + + SO 2- 3 \u003d H 2 O + SO 2;
c) CO 2- 3 + 2H + = CO 2 + H 2 O;
d) 2H + + 2OH - = 2H 2 O.
3. The ionic equation of the reaction in which a low-dissociating substance is formed is:
a) Ag + + Cl - = AgCl;
b) OH - + H + = H 2 O;
c) Zn + 2H + = Zn 2+ + H 2;
d) Fe 3+ + 3OH - \u003d Fe (OH) 3.
4. The ionic equation for the reaction in which a white precipitate is formed is:
a) Cu 2+ + 2OH - = Cu (OH) 2;
b) СuO + 2H + = Cu 2+ + H 2 O;
c) Fe 3+ + 3OH - = Fe (OH) 3;
d) Ba 2+ + SO 2- 4 = BaSO 4.
5. The molecular equation that corresponds to the reduced ionic reaction equation 3OH - + 3H + = 3H 2 O is:
a) NaOH + HCl = NaCl + H 2 O;
b) 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O;
c) 3KOH + H 3 PO 4 = K 3 PO 4 + 3H 2 O;
d) Ba(OH) 2 + 2HCl = BaCl 2 + H 2 O.
6. Molecular equation that corresponds to the reduced ionic reaction equation
H + + \u003d H 2 O + CO 2, -
a) MgCO 3 + 2HCl = MgCl 2 + CO 2 + H 2 O;
b) Na 2 CO 3 + 2HCl = 2NaCl + CO 2 + H 2 O;
c) NaHCO 3 + HCl = NaCl + CO 2 + H 2 O;
d) Ca(OH) 2 + CO 2 = CaCO 3 + H 2 O.
| Answers. 1 -A; 2 -V; 3 -b; 4 -G; 5 -V; 6 -V. |
Solving experimental problems on the topic “Nitrogen and phosphorus”
When studying new material on the topic “Nitrogen and Phosphorus”, students perform a series of experiments related to the production of ammonia, the determination of nitrates, phosphates, ammonium salts, and acquire certain skills and abilities. This methodological development contains six tasks. To perform practical work, three tasks are enough: one - to obtain a substance, two - to recognize substances. When performing practical work, students can be offered tasks in a form that will make it easier for them to prepare a report (see tasks 1, 2). (Answers are for the teacher.)
Exercise 1
Get ammonia and experimentally prove its presence.
a) Getting ammonia.
Heat a mixture of equal portions of solid ammonium chloride and calcium hydroxide powder in a test tube with a gas outlet tube. In this case, ammonia will be released, which must be collected in another dry test tube located with a hole ... ......... ( Why?).
Write the equation for the reaction of obtaining ammonia.
…………………………………………………..
b) Determination of ammonia.
Can be identified by smell………… (name of substance), as well as by changing the color of litmus or phenolphthalein. When ammonia is dissolved in water, ……. (foundation name), so litmus test.……. (specify color), and colorless phenolphthalein becomes …………. (specify color).
Instead of dots, insert words according to the meaning. Write the reaction equation.
…………………………………………………..
* Ammonia smells like ammonia in the first-aid kit - an aqueous solution of ammonia. - Note. ed.
Task 2
Get copper nitrate in two different ways, having the following substances available: concentrated nitric acid, copper shavings, copper (II) sulfate, sodium hydroxide. Write the equations of chemical reactions in molecular form, note the changes. In the 1st method for a redox reaction, write the equations of the electronic balance, determine the oxidizing agent and reducing agent. In the 2nd method, write the abbreviated ionic reaction equations.
1st s p o s o b. Copper + nitric acid. Lightly heat the contents of the test tube. The colorless solution becomes….. (specify color), because formed.... (name of substance); gas is released …….. colors with an unpleasant odor, this is ……. (name of substance).
2nd s p o s o b. When copper (II) sulfate reacts with sodium hydroxide, a precipitate is obtained ... .. colors, this is ...... (name of substance). We add nitric acid to it until the precipitate is completely dissolved ......... (name of sediment). A clear blue solution is formed…… (name of salt).
Task 3
Prove empirically that ammonium sulfate contains NH 4 + and SO 2- 4 ions. Note observations, write molecular and abbreviated ionic reaction equations.
Task 4
How to experimentally determine the presence of solutions of sodium orthophosphate, sodium chloride, sodium nitrate in test tubes No. 1, No. 2, No. 3? Note observations, write molecular and abbreviated ionic reaction equations.
Task 5
Having substances: nitric acid, copper shavings or wire, universal indicator paper or methyl orange, prove by experience the composition of nitric acid. Write the equation for the dissociation of nitric acid; molecular equation for the reaction of copper with concentrated nitric acid and electron balance equations, determine the oxidizing agent and reducing agent.
Task 6
Get a solution of copper nitrate in different ways, having substances: nitric acid, copper oxide, basic copper carbonate or hydroxocopper(II) carbonate. Write molecular, complete and abbreviated ionic equations of chemical reactions. Note signs of chemical reactions.
Control tests
1. Write the reaction equation for the yellow precipitate.
2. The ionic equation of the reaction in which a white curdled precipitate is formed is:
3. To prove the presence of a nitrate ion in nitrates, you need to take:
a) hydrochloric acid and zinc;
b) sulfuric acid and sodium chloride;
c) sulfuric acid and copper.
4. The reagent for the chloride ion is:
a) copper and sulfuric acid;
b) silver nitrate;
c) barium chloride.
5. In the reaction equation, the scheme of which
HNO 3 + Cu -> Cu(NO 3) 2 + NO 2 + H 2 O,
before the oxidizing agent, you must put the coefficient:
a) 2; b) 4; at 6.
6. Basic and acidic salts correspond to pairs:
a) Cu (OH) 2, Mg (HCO 3) 2;
b) Cu(NO 3) 2, HNO 3;
c) 2 CO 3 , Ca(HCO 3) 2 .
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Answers. 1 -A; 2 -b; 3 -V; 4 -b; 5 -b; 6 -V. |
Determination of mineral fertilizers
The methodical development of this practical work consists of three parts: theory, practical work, control. The theoretical part provides general information on the qualitative determination of cations and anions that are part of mineral fertilizers. The workshop provides examples of seven mineral fertilizers with a description of their characteristic features, as well as equations for qualitative reactions. In the text, instead of dots and a question mark, you need to insert answers that are appropriate in meaning. To perform practical work at the discretion of the teacher, it is enough to take four fertilizers. The control of students' knowledge consists of test tasks for determining the formulas of fertilizers, which are given in this practical work.
Theory
1. The reagent for the chloride ion is silver nitrate. The reaction proceeds with the formation of a white curdled precipitate:
Ag + + Cl - = AgCl.
2. The ammonium ion can be detected with alkali. When an ammonium salt solution is heated with an alkali solution, ammonia is released, which has a sharp characteristic odor:
NH + 4 + OH - = NH 3 + H 2 O.
You can also use red litmus paper moistened with water, universal indicator or phenolphthalein strip of paper to determine the ammonium ion. The paper must be held over the vapors released from the test tube. Red litmus turns blue, the universal indicator turns purple, and phenolphthalein turns crimson.
3. To determine nitrate ions, shavings or pieces of copper are added to the salt solution, then concentrated sulfuric acid is added and heated. After a while, a brown gas with an unpleasant odor begins to be released. Emission of brown gas NO 2 indicates the presence of ions.
For example:
NaNO 3 + H 2 SO 4 NaHSO 4 + HNO 3,
4HNO 3 + Cu \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O.
4. The reagent for the phosphate ion is silver nitrate. When it is added to a phosphate solution, a yellow precipitate of silver phosphate precipitates:
3Ag + + PO 3- 4 = Ag 3 PO 4.
5. The reactant for the sulfate ion is barium chloride. A white milky precipitate of barium sulfate precipitates, insoluble in acetic acid:
Ba 2+ + SO 2- 4 \u003d BaSO 4.
Workshop
1. Silvinite (NaCl KCl), pink crystals, good water solubility. The flame turns yellow. When viewing the flame through blue glass, a violet color is noticeable. WITH …….. (name of reagent) gives a white precipitate (name of salt).
KCl+? -> KNO 3 + AgCl.
2. Ammonium nitrate NH 4 NO 3, or …….. (name of fertilizer), white crystals, highly soluble in water. With sulfuric acid and copper, a brown gas is released .... (name of substance). With solution……. (name of reagent) when heated, the smell of ammonia is felt, its vapors turn red litmus into ....... color.
NH 4 NO 3 + H 2 SO 4 NH 4 HSO 4 + HNO 3,
HNO 3 + Cu -> Cu(NO 3) 2 + ? +? .
NH4NO3+? -> NH 3 + H 2 O + NaNO 3.
3. Potassium nitrate (KNO 3), or …… (name of fertilizer), with H 2 SO 4 and ……… (name of substance) produces brown gas. The flame turns purple.
KNO 3 + H 2 SO 4 KHSO 4 + HNO 3,
4HNO3 + ? -> Cu(NO 3) 2 + ? + 2H2O.
4. Ammonium chloride NH 4 Cl with solution ……. (name of reagent) when heated, it forms ammonia, its vapor turns red litmus blue. WITH …… (name of reagent anion) silver gives a white cheesy precipitate ...... (name of sediment).
NH4Cl+? \u003d NH 4 NO 3 + AgCl,
NH4Cl+? \u003d NH 3 + H 2 O + NaCl.
5. Ammonium sulfate (NH 4) 2 SO 4 with an alkali solution forms ammonia when heated, its vapor turns red litmus blue. WITH …….. (name of reagent) gives a white milky precipitate. (name of sediment).
(NH 4) 2 SO 4 + 2NaOH \u003d 2NH 3 + 2H 2 O +? ,
(NH 4) 2 SO 4 +? -> NH 4 Cl + ? .
6. Sodium nitrate NaNO 3, or ...... (name of fertilizer), white crystals, good solubility in water, gives brown gas with H 2 SO 4 and Cu. The flame turns yellow.
NaNO 3 + H 2 SO 4 NaHSO 4 + ? ,
Cu -> Cu(NO 3) 2 +? + 2H2O.
7. Calcium dihydrophosphate Ca (H 2 PO 4) 2, or ...... (name of fertilizer), gray fine-grained powder or granules, poorly soluble in water, with ….. (name of reagent) gives ….. (specify color) sediment ……… (name of substance) AgH 2 PO 4 .
Ca(H 2 PO 4) 2 + ? -> 2AgH 2 PO 4 + Ca(NO 3) 2 .
Control (test)
1. Pink crystals, highly soluble in water, color the flame yellow; when interacting with AgNO 3, a white precipitate precipitates - this is:
a) Ca(H 2 PO 4) 2; b) NaCl KCl;
c) KNO 3 ; d) NH 4 Cl.
2. Crystals are highly soluble in water; in reaction with H 2 SO 4 and copper, a brown gas is released, with an alkali solution, when heated, it gives ammonia, the vapor of which turns red litmus blue, is:
a) NaNO 3 ; b) (NH 4) 2 SO 4;
c) NH 4 NO 3; d) KNO 3 .
3. Light crystals, highly soluble in water; when interacting with H 2 SO 4 and Cu, brown gas is released; the flame turns purple - this is:
a) KNO 3 ; b) NH 4 H 2 PO 4;
c) Ca(H 2 PO 4) 2 CaSO 4; d) NH 4 NO 3.
4. Crystals are highly soluble in water; with silver nitrate it gives a white precipitate, with alkali, when heated, it gives ammonia, the vapors of which turn red litmus blue, is:
a) (NH 4) 2 SO 4; b) NH 4 H 2 PO 4;
c) NaCl KCl; d) NH 4 Cl.
5. Light crystals, highly soluble in water; with BaCl 2 it gives a white milky precipitate, with alkali it gives ammonia, the vapors of which turn red litmus blue, is:
c) NH 4 Cl; d) NH 4 H 2 PO 4.
6. Light crystals, highly soluble in water; when interacting with H 2 SO 4 and Cu, it gives a brown gas, the flame turns yellow - this is:
a) NH 4 NO 3; b) (NH 4) 2 SO 4;
c) KNO 3 ; d) NaNO 3 .
7. Gray fine-grained powder or granules, solubility in water is poor, with a solution of silver nitrate gives a yellow precipitate - this is:
a) (NH 4) 2 SO 4; b) NaCl KCl;
c) Ca(H 2 PO 4) 2; d) KNO 3 .
| Answers. 1 -b; 2 -V; 3 -A; 4 -G; 5 -b; 6 -G; 7 -V. |
- Tasks for self-examination are a prerequisite for mastering the material, each section is accompanied by test tasks on the topics covered, which must be solved.
- Having solved all the tasks from the section, you will see your result and be able to see the answers to all the examples, which will help you understand what mistakes you made and where your knowledge needs to be strengthened!
- The test consists of 10 tests of task 8, part 1 of the USE, the answers are mixed randomly, and are taken from the database of questions we created!
- Try to get above 90% correct answers to be sure of your knowledge!
- Use only the reference material below if you want to check the fixing of the material!
- After passing the test, look at the answers to the questions where you made a mistake and consolidate the material before re-passing!
If you are studying with a tutor, then write your real name at the beginning of testing! Based on your name, the tutor will find the test you passed, review your mistakes and take into account your gaps in order to fill them in the future!
Reference material for passing the test:
Mendeleev table
Solubility table
The types of questions that are found in this test (you can see the answers to the questions and the full conditions of the tasks by passing the test above to the end. We advise you to look at how to solve these questions in ours):
- A solution of substance Y was added to a test tube with a solution of salt X. As a result, a reaction occurred, which is described by the abbreviated ionic equation ____. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of substance Y was added to a test tube with a solution of salt X. As a result of the reaction, a white precipitate was observed. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of substance Y was added to a test tube with a solution of potassium salt X. As a result, a reaction occurred, which is described by the following abbreviated ionic equation: ____. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of substance Y was added to a test tube with a solution of salt X. As a result of the reaction, evolution of a colorless gas was observed. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of acid Y was added to a test tube with a solution of substance X. As a result, a reaction occurred, which is described by the following abbreviated ionic equation: ____. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of salt Y was added to a test tube with a solution of substance X. As a result of the reaction, a blue precipitate was observed. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of substance Y was added to a test tube with a solid, water-insoluble substance X. As a result of the reaction, the dissolution of the solid substance was observed without gas evolution. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of salt Y was added to a test tube with a solution of substance X. As a result, a reaction occurred, which is described by the following abbreviated ionic equation: ____. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of salt Y was added to a test tube with a solution of substance X. As a result of the reaction, a brown precipitate was observed. From the proposed list, select substances X and Y that can enter into the described reaction.
- A solution of substance Y was added to a test tube with a solution of acid X. As a result, a reaction occurred, which is described by the following abbreviated ionic equation. From the proposed list, select substances X and Y that can enter into the described reaction.
Colorless bright blue-blue
Analytical features of substances and analytical reactions
When conducting qualitative and quantitative analysis, use analytical features of substances and analytical reactions.
Analytical Features – such properties of the analyte or the products of its transformation, which make it possible to judge the presence of certain components in it. Characteristic analytical features - color, smell, angle of rotation of the plane of polarization of light, radioactivity, ability to interact with electromagnetic radiation (for example, the presence of characteristic bands in the IR absorption spectra or maxima in the absorption spectra in the visible and UV regions of the spectrum), etc.
Analytical reaction - chemical transformation of the analyte under the action of an analytical reagent with the formation of products with noticeable analytical features. As analytical reactions, the most commonly used reactions are the formation of colored compounds, the release or dissolution of precipitates, gases, the formation of crystals of a characteristic shape, the coloring of a gas burner flame, the formation of compounds that luminesce in solutions, etc. The results of analytical reactions are affected by temperature, concentration of solutions, pH environment, the presence of other substances (interfering, masking, catalyzing processes), etc.
Let's illustrate what has been said with some examples.
Formation of colored compounds. Copper ions Сu 2+ in aqueous solutions, in which they exist in the form of almost colorless (pale bluish) aquacomplexes 2+ , when interacting with ammonia, they form a soluble complex (ammonia) 2+ of bright blue-blue, coloring the solution in the same color:
2+ + 4NH 3 \u003d 2+ + P H 2 O
Using this reaction, it is possible to identify (detect) Cu 2+ copper ions in aqueous solutions.
If colorless (pale yellow) ions of ferric iron Fe 3+ (also in the form of an aquocomplex 3+) are present in an aqueous solution, then with the introduction of thiocyanate ions (thiocyanate ions) NCS - , the solution turns into an intense color due to the formation of complexes 3– n Red:
3+ + P NCS-=3- n + P H 2 O
Where P < или = 6. При этом, в зависимости от отношения концентраций 3+ и NCS – , образуется равновесная смесь комплексов с P= 1; 2; 3; 4; 5; 6. All of them are painted red. This reaction is used to discover (detect) iron ions (III).
Note that individual multiply charged ions, for example, Cu 2+, Fe 2+, Fe 3+, Co 3+, Ni 2+, etc., as well as hydrogen ions H + (i.e. protons - nuclei of the hydrogen atom), cannot exist in aqueous solutions under normal conditions, since they are thermodynamically unstable and interact with water molecules or with other particles to form aquocomplexes (or complexes of a different composition):
M m++n H 2 O \u003d [M (H 2 O) n] m +(aquacomplex)
H+ + H 2 O = H 3 O + (hydronium ion)
In what follows, for brevity, in chemical equations we will not always indicate the water molecules that are part of the aquocomplexes, remembering, however, that in fact, the reactions in solutions involve the corresponding aquocomplexes, and not "naked" metal or hydrogen cations. So, for simplicity, we will write H +, Cu 2+, Fe 2+, etc. instead of the more correct H 3 O + , 2+ , 3+ , respectively, etc.
Isolation or dissolution of precipitates. Ba 2+ ions present in an aqueous solution can be precipitated by adding a solution containing sulfate ions SO 4 2+ in the form of a sparingly soluble white precipitate of barium sulfate:
Ba 2+ + SO 4 2+ \u003d BaSO 4. ↓(white precipitate)
A similar picture is observed during the precipitation of calcium ions Ca 2+ by soluble carbonates:
Ca 2+ + CO 3 2– → CaCO 3 ↓(white precipitate)
The white precipitate of calcium carbonate dissolves under the action of acids, according to the scheme:
CaCO 3 + 2HC1 → CaC1 2 + CO 2 + H 2 O
This releases gaseous carbon dioxide.
Chloroplatinate ions 2– form yellow precipitates upon addition of a solution containing potassium cations K + or ammonium NH + . If a solution of sodium chloroplatinate Na 2 (this salt is quite soluble in water) is treated with a solution of potassium chloride KCl or ammonium chloride NH 4 C1, then yellow precipitates of potassium hexachloroplatinate K 2 or ammonium (NH 4) 2, respectively (these salts are slightly soluble in water):
Na 2 + 2KS1 → K 2 ↓ + 2NaCl
Na 2 + Z NH 4 C1 → (NH 4) 2 ↓ + 2NaCl
Reactions with evolution of gases(outgassing reactions). The reaction of dissolution of calcium carbonate in acids has already been cited above, in which gaseous carbon dioxide is released. Let us point out some more gas-evolving reactions.
If alkali is added to a solution of any ammonium salt, then gaseous ammonia is released, which can be easily determined by the smell or by the blue of wet red litmus paper:
NH 4 + + OH - \u003d NH 3 H 2 0 → NH 3 + H 2 0
This reaction is used in both qualitative and quantitative analysis.
Sulfides under the action of acids emit gaseous hydrogen sulfide:
S 2– + 2H + → H 2 S
which is easily felt by the specific smell of rotten eggs.
Formation of characteristic crystals(microcrystalloscopic reactions). Sodium ions Na + in a drop of solution, when interacting with hexahydroxoantibate (V) ions, form white crystals of sodium hexahydroxoantibate (V) Na of a characteristic shape:
Na + + -- = Na
The shape of the crystals is clearly visible when viewed under a microscope. This reaction is sometimes used in qualitative analysis to discover sodium cations.
Potassium ions K + when reacting in neutral or acetic acid solutions with soluble sodium and lead hexanitrocuprate (P) Na 2 Pb form black (or brown) crystals of potassium and lead hexanitrocuprate (P) K 2 Pb [Cu (N0 2) 6] characteristic cubic forms that can also be seen when viewed under a microscope. The reaction proceeds according to the scheme:
2K + + Na 2 Pb \u003d K 2 Pb [Cu (N0 3) 6] + 2Na +
It is used in qualitative analysis to detect ( discoveries) potassium cations. Microcrystalloscopic analysis was first introduced into analytical practice in 1794-1798. member of the St. Petersburg Academy of Sciences T.E. Lovitz.
Coloring the flame of a gas burner. When compounds of some metals are introduced into the flame of a gas burner, the flame is colored in one color or another, depending on the nature of the metal. Thus, lithium salts color the flame carmine-red, sodium salts yellow, potassium salts violet, calcium salts brick-red, barium salts yellow-green, etc.
This phenomenon can be explained as follows. When a compound of a given metal (for example, its salt) is introduced into the flame of a gas burner, this compound decomposes. The metal atoms formed during the thermal decomposition of the compound are excited at a high temperature of the flame of a gas burner, i.e., absorbing a certain portion of thermal energy, they pass into some kind of excited electronic state that has more energy compared to the unexcited (ground) state. The lifetime of the excited electronic states of atoms is negligible (very small fractions of a second), so that the atoms almost instantly return to the unexcited (ground) state, emitting the absorbed energy in the form of light radiation with one or another wavelength, depending on the energy difference between the excited and ground the energy levels of the atom. For atoms of different metals, this energy difference is not the same and corresponds to light radiation of a certain wavelength. If this radiation lies in the visible region of the spectrum (in the red, yellow, green or some other part of it), then the human eye fixes one or another color of the burner flame. The coloration of the flame is short-term, since the metal atoms are carried away with the gaseous products of combustion.
The coloring of a gas burner flame with metal compounds is used in qualitative analysis to discover metal cations that emit radiation in the visible region of the spectrum. The atomic absorption (fluorescent) methods of element analysis are also based on the same physicochemical nature.
In table. 3.1 shows examples of burner flame colors from some elements.
Formation of a gaseous substance
Na 2 S + 2HCl \u003d H 2 S + 2NaCl
2Na + + S 2- + 2H + + 2Cl - \u003d H 2 S + 2Na + + 2Cl -
Ionic-molecular reaction equation,
2H + + S 2- = H 2 S is a short form of the reaction equation.
Precipitation formation
with the formation of poorly soluble substances:
a) NaCl + AgNO 3 = NaNO 3 + AgCl
Cl - + Ag + = AgCl - reduced ion-molecular equation.
Reactions in which weak electrolytes or poorly soluble substances are part of both the products and the starting materials, as a rule, do not proceed to the end, i.e. are reversible. The equilibrium of the reversible process in these cases is shifted towards the formation of the least dissociated or least soluble particles..
BaCl 2 + Na 2 SO 4 = BaSO 4 ↓ + 2NaCl
Molecular reaction equation,
Ba 2+ + 2Cl - + 2Na + + SO= BaSO 4 ↓ + 2Na + + 2Cl -
Ionic-molecular reaction equation,
Ba 2+ + SO \u003d BaSO 4 ↓ - a short form of the reaction equation.
Precipitation condition. Solubility product
There are no absolutely insoluble substances. Most solids have limited solubility. In saturated electrolyte solutions of sparingly soluble substances, the precipitate and the saturated electrolyte solution are in a state of dynamic equilibrium. For example, in a saturated solution of barium sulfate, which is in contact with the crystals of this substance, a dynamic equilibrium is established:
BaSO 4 (t) \u003d Ba 2+ (p) + SO 4 2- (p).
For this equilibrium process, we can write the expression for the equilibrium constant, taking into account that the concentration of the solid phase is not included in the expression for the equilibrium constant: Kp =
This value is called the solubility product of a sparingly soluble substance (PR). Thus, in a saturated solution of a poorly soluble compound, the product of the concentrations of its ions to the power of stoichiometric coefficients is equal to the value of the solubility product. In the considered example
PR BaSO4 = .
The solubility product characterizes the solubility of a poorly soluble substance at a given temperature: the smaller the solubility product, the worse the solubility of the compound. Knowing the solubility product, one can determine the solubility of a sparingly soluble electrolyte and its content in a certain volume of a saturated solution.
In a saturated solution of a strong, sparingly soluble electrolyte, the product of the concentrations of its ions in powers equal to the stoichiometric coefficients for given ions (at a given temperature) is a constant value called the solubility product.
The value of PR characterizes the comparative solubility of substances of the same type (forming the same number of ions during dissociation) substances. The greater the PR of a given substance, the greater its solubility. For example:
In this case, the least soluble is iron (II) hydroxide.
Precipitation condition :
X y > PR(K x A y).
This condition is achieved by introducing the ion of the same name into the saturated solution-precipitate system. Such a solution is supersaturated relative to this electrolyte, so it will precipitate.
Precipitate dissolution condition:
X y< ПР(K x A y).
This condition is achieved by binding one of the ions sent by the precipitate into the solution. The solution in this case is unsaturated. When crystals of a sparingly soluble electrolyte are introduced into it, they will dissolve. The equilibrium molar concentrations of K y+ and A x- ions are proportional to the solubility S (mol/l) of the substance K x A y:
X S and = y S
PR = (x S) x (y S) y = x x y y S x+y
The relations obtained above make it possible to calculate the values of SP from the known solubility of substances (and, consequently, the equilibrium concentrations of ions) from the known values of SP at T = const.
