Complete the molecular equation possible. Examples of composing equations for ion exchange reactions

IRREVERSIBLE REACTIONS

Reactions with the formation of poorly soluble substances (precipitates).

Let's create molecular and ion-molecular equations for the reaction between solutions of iron(III) chloride and sodium hydroxide.

1. Let's write it down molecular equation and select the coefficients:

FeCl 3 + 3NaOH = Fe(OH) 3 + 3NaCl

2. Let's find the substance that causes the reaction to occur. This is Fe(OH) 3. We put the sediment sign ↓:

3. We indicate the strength of the base and the solubility of salts:

FeCl 3 + 3NaOH = Fe(OH) 3 ↓ + 3NaCl

soluble strong sediment soluble

Salt base salt

4. Let’s write down the complete ionic-molecular equation (we represent soluble salts and a strong base in the form of ions):

Fe 3+ + 3Cl – + 3Na + + 3OH – = Fe(OH) 3 ↓ + 3Na + + 3Cl –

5. Let us emphasize the formulas that are not involved in the reaction (these are formulas for identical ions in both sides of the equation):

Fe 3+ + 3Cl – + 3Na++ 3OH – = Fe(OH) 3 ↓ + 3Na+ + 3Cl –

6. Eliminate the underlined formulas and get

Fe 3+ + 3OH – = Fe(OH) 3 ↓

The abbreviated ion-molecular shows that the essence of the reaction is reduced to the interaction of Fe 3+ and OH – ions, resulting in the formation of a precipitate of iron(III) hydroxide Fe(OH) 3.

Reactions with the formation of weakly dissociating substances (weak electrolytes).

Let's create molecular and ion-molecular equations for the reaction between solutions nitric acid and potassium hydroxide.

Molecular equation:

HNO3 + KOH = KNO3 + H2O

strong strong soluble weak

acid base salt electrolyte

H + + NO 3 – + K + + OH – = K + NO 3 – + H 2 O

Abbreviated ion-molecular equation:

H + + OH – = H 2 O

Reactions with the formation of gaseous substances.

Let's create molecular and ion-molecular equations for the reaction between solutions of sodium sulfide and sulfuric acid.

Molecular equation:

Na 2 S + H 2 SO 4 = Na 2 SO 4 + H 2 S

soluble strong soluble gas

salt acid salt

Complete molecular ionic equation:

2Na + + S 2 – + 2H + + SO 4 2 – = 2Na + + SO 4 2 – + H 2 S

Abbreviated ion-molecular equation:

2H + + S 2 – = H 2 S

REVERSIVE REACTIONS

Let us analyze the processes that occur when merging solutions of potassium nitrate and sodium chloride.

Molecular equation:

KNO3 + NaCl = KCl + NaNO3

soluble soluble soluble soluble

salt salt salt salt

Complete molecular ionic equation:

K + + NO 3 – + Na + + Cl – ⇄ K + + Cl – + Na + + NO 3 –

IN in this case the abbreviated ion-molecular equation cannot be written: according to the theory electrolytic dissociation, the reaction does not occur. If such a solution is evaporated, we obtain a mixture of four salts: KNO 3, NaCl, KCl, NaNO 3.

using a) HCl b) H 2 S c) H 2 O d) NaOH

7 points

1. 
   Ratio molecular weights higher chloride and oxide of an element located in group IV periodic table, is 17:6. Identify the element.

1. 
   The dissolution of an aluminum sample in a solution of potassium hydroxide at 20 0 C is completed in 36 minutes, and at 40 0 ​​C – in 4 minutes. How long will it take for the same sample of aluminum to dissolve at 65 0 C? Write the reaction equation.

1. 
   Two glasses of the same mass, one of which contains 100 g of a 18.25% solution of hydrochloric acid, and the other contains 100 g of a 16% solution of copper sulfate, were placed on two pans of scales. The equilibrium of the scales was disturbed by adding 2 g of calcium carbonate to hydrochloric acid. Calculate the mass of a portion of iron that should be added to another glass so that the scales balance again.

1. 
   Write the reaction equations that can be used to carry out the following transformations:

5 points

1. 
   Determine the molecular formula of an alkene if it is known that the same amount of it, interacting with various hydrogen halides, forms, respectively, either 5.23 g of a chlorine derivative or 8.2 g of a bromo derivative.

1. 
   Products complete combustion a mixture of propane and methylamine in excess oxygen was passed through an excess barium hydroxide solution, resulting in the formation of 13.97 g of precipitate. The gases that were not absorbed were passed over hot copper. Then brought to normal conditions the volume of gas became 2.5 times less than the volume of the original mixture of propane and methylamine (n.o.). Define mass fractions substances in the initial mixture of gases.

Olympiad 11th grade (1 round)

2Ва 2+ + 2S 2– + 2Н + + 2Сl – ​​= Ва 2+ + 2НS – + Ва 2+ + 2Сl –

2S 2– + 2Н + = 2НS –

S 2– + H + = NS –

Can
b) BaS + H 2 S = Ba(HS) 2 2b

Ba 2+ + S 2– + H 2 S = Ba 2+ + 2НS –

S 2– + Н 2 S = 2НS –

Can
c) 2BaS + 2H 2 O = Ba(HS) 2 + Ba(OH) 2 2b

S 2– + H 2 O = NS – + OH –

This is not possible, since only part of the original substance undergoes hydrolysis.

Ba 2+ + S 2– + 2Na + + 2OH – = Ba 2+ + 2OH – + 2Na + + S 2–

It is impossible, because the reaction does not occur.
2. The ratio of molecular weights of the higher chloride and oxide of an element located in group IV of the periodic system is 17: 6. Identify the element.
Group IV element (E) has higher valence IV and forms chloride ESl 4 and oxide EO 2:
M(ESl 4) = x + 4 35.5 = (x + 142) g/mol,

M(EO 2) = x + 2 16 = (x + 32) g/mol. 1b

3. Dissolution of an aluminum sample in a solution of potassium hydroxide at 20 0 C is completed in 36 minutes, and at 40 0 ​​C – in 4 minutes. How long will it take for the same sample of aluminum to dissolve at 65 0 C? Write the reaction equation.
Reaction equation
2Al + 2KOH + 6H 2 O = 2K + 3H 2. 1b
Because average speed reaction is inversely proportional to the reaction time, then when the temperature increases from 20 0 C to 40 0 ​​C, the reaction rate will increase by

once.

According to van't Hoff's rule (
) we obtain the value γ:

γ = 3. 1b
The increase in the reaction rate when the temperature increases from 40 0 ​​C to 65 0 C will be:

once. 1b
Consequently, the reaction time at 65 0 C will be 15.588 times less than at 40 0 ​​C and equal

1b
4. Two glasses of the same mass, one of which contains 100 g of a 18.25% solution of hydrochloric acid, and the other – 100 g of a 16% solution of copper sulfate, were placed on two cups of scales. The equilibrium of the scales was disturbed by adding 2 g of calcium carbonate to hydrochloric acid. Calculate the mass of a portion of iron that should be added to another glass so that the scales balance again.
A glass of hydrochloric acid contains

100·0.1825 = 18.25 g HCl. 1b

After adding CaCO 3 the reaction occurs
2HCl + CaCO 3 = CaCl 2 + CO 2 + H 2 O, 1b

in this case, CaCO 3 is completely consumed, since HCl is contained in excess in the solution (as can be seen from the solution of the proportion):

2g CaCO 3 ─ y g CO 2

To restore the equilibrium of the scales, the same amount (1.12 g) of Fe should be added to the second glass, since as a result of the reaction

Response elements:

Reaction equations corresponding to the transformation scheme were compiled:

4 reaction equations written correctly 4 points

Correctly written 3 reaction equations 3 points

Correctly written 2 reaction equations 2 points

Correctly written 1 reaction equations 1 point

All elements of the answer are written incorrectly 0 points

6. Determine the molecular formula of an alkene if it is known that the same amount of it, interacting with various hydrogen halides, forms, respectively, either 5.23 g of a chlorine derivative or 8.2 g of a bromo derivative.
Response elements:

1) The reaction equations are written and it is indicated that the amounts of haloalkanes are equal to each other:

C n H 2 n + HCl C n H 2 n +1 Cl

C n H 2 n + HBr C n H 2 n +1 Br

n(C n H 2n+1 Cl) = n(C n H 2n+1 Br)

2) Decision algebraic equation The molecular formula of the alkene was found:

5.23/(14n+36.5) = 8.2/(14n+81)

Correctly written first element of the answer 1 point

All elements of the answer are written incorrectly 0 points

Maximum score 2

1. 
   C 3 P 8 + 5O 2 -------- 3CO 2 + 4H 2 O
2. 
   4CH 3 NH 2 + 9O 2 ------ 4CO 2 + 2N 2 + 10 H 2 O
3. 
   CO 2 + Ba(OH) 2 ------ BaCO 3 + H 2 O
4. 
   O 2 + 2Cu ------ 2CuO

Let ν(C 3 H 8) = x mol, ν(CH 3 NH 2) = y mol, then

ν(CO 2) = 3x+y, ν(CO 2) = ν(BaCO 3) = 13.97/197 = 0.0709 mol

After passing over the copper, only nitrogen remained.

ν(N 2) = 0.5у, ν (source gases) = x+y

x+y= 2.5(0.5y) x= 0.25y

3х+y+0.0709 0.75y+y=0.0709

x=0.25*0.0405= 0.0101 y=0.0709/1.75=0.0405

ν(C 3 H 8) = 0.0101 mol; m(C 3 H 8)= 0.0101*44= 0?446u

ν(CH 3 NH 2) = 0.0405 mol; m(CH 3 NH 2)0.0405*31= 1.256g

m(mixture) = 0.446+1.256 = 1.7g

ω(C 3 H 8) = 0.0446/1.7 = 0.262(26.2%)

ω(CH 3 H 2) = 1.256/1.7 = 0.738 (73.8%)

1. 
   Equations drawn up chemical reactions 2 points
2. 
   An equation has been created to calculate 2 points
3. 
   A conclusion was made about the remaining gas and its amount of substance was determined 2 points
4. 
   Compiled and solved a system of equations 3 points
5. 
   Mass fractions of substances determined 1 point

The theory of electrolytic dissociation recognizes that all reactions in aqueous solutions of electrolytes are reactions between ions. Therefore, the reaction equations for these processes, written in molecular form, do not reflect the true state of substances in solutions. In addition to writing reaction equations, in molecular form there is an ionic (ion-molecular) form of representing reaction equations between electrolytes in aqueous solutions. IN ion-molecular equations reactions, substances that are slightly soluble, slightly dissociated and gaseous are written in the form of molecules, and strong electrolytes- in the form of ions into which they dissociate. For example, when solutions of copper (II) chloride and sodium hydroxide react, a precipitate of copper (II) hydroxide is formed: CuCl2 + 2NaOH = Cu(OH)2| + 2NaCl. In ionic-molecular form, the equation for this reaction is written as follows: Cu2+ + 2C1″ + 2Na+ + 20NG = Cu(OH)2i + 2Na+ + 2SG. The concentrations of sodium and chlorine ions remain unchanged during the reaction, so they can be excluded from the reaction equation. Since reactions between ions in solution are an example chemical equilibrium, the Jle Chatelier equilibrium shift principle is applicable to them. According to this principle, the equilibrium can shift if any substance is removed from the sphere of the reaction as it proceeds. Removal of a substance can be carried out in three cases: 1) the formation of a poorly soluble precipitate; 2) selection gaseous substance; 3) formation of a slightly dissociated compound. When a solution of (NH4)2S interacts with hydrochloric acid, hydrogen sulfide gas is formed and the equilibrium of the reaction shifts to the right: (NH4)2S + 2HC1 - 2NH4C1 + H2ST, 2NH4+ + S2″ + 2H4″ + 2SG = 2NH4+ + 2SG + H2Sf or 2H+ + S2″ = H2Sf. An example of a reaction whose equilibrium is shifted towards the formation of a slightly dissociated compound is the interaction between solutions of nitric acid and sodium hydroxide: HN03 + NaOH - NaN03 + H20, H+ + N03″ + Na+ + OH» = Na+ + NO3- + H20 or H + + OH" - H20. The reaction with the formation of a slightly soluble compound was discussed above. We often encounter processes in which not one of the three types of exchange reactions considered is carried out, but one or another combination of them. Thus, when a solution of potassium sulfite interacts with sulfuric acid, the formation of a slightly dissociated substance - water, and the release of a gaseous product simultaneously occurs: K2S03 + H2S04 = K2S04 + S02T + H20, 2K+ + S032″ + 2H+ + S042′ - 2K+ + S042″ + S02t + H20 or 2H+ + S032~ - S02t + H20. And when a solution of barium hydroxide interacts with sulfuric acid, both a precipitate and weak electrolyte: Ba(OH)2 + H2S04 = BaS04i + 2H20, ‘Ba2+ + 20H” + 2H* + S042’ “BaS04i + 2H20. Some reactions occur with the formation of two sparingly soluble substances: CuS04 + BaS = BaS04| + CuSj, Cu2+ + S042″ + Ba2* + S2″ = BaS04l + CuSi. In a number of metabolic processes, slightly dissociated or sparingly soluble compounds are found both among the initial and final products of the reaction: nh4oh + n+ + C1-?± nh4+ + cr + n2o. Due to the formation of sparingly soluble compounds in in some cases possible displacement strong acid from weak compounds, for example: Cu24″ + 2СГ + H2S « CuSJ + 2Н* + 2СГ, Cu2+ + H2S-CuSi + 2Н+. Thus, the examples discussed above confirm general pattern: all exchange reactions in electrolyte solutions proceed in the direction of decreasing the number of free ions.

Getting the job done

Experiment 1. Formation of poorly soluble bases. Pour 3-5 drops of an iron (III) salt solution into one test tube, the same amount of a copper (II) salt solution into another, and a nickel (II) salt solution into a third. Add a few drops of alkali solution to each test tube until precipitation occurs. Save the sediment until the next experiment.

What class do the resulting metal hydroxide precipitates belong to? Are these hydroxides strong bases?

Experiment 2. Dissolution of poorly soluble bases. Add a few drops of solution to the precipitates obtained in the previous experiment. hydrochloric acid concentration of 15% until they are completely dissolved.

What new, slightly dissociated compound is formed when bases are dissolved in an acid?

Experiment 3. Formation of slightly soluble salts.

A. Pour 3-5 drops of lead (II) nitrate solution into two test tubes and add a few drops of potassium iodide to one test tube and barium chloride to the other.

What is observed in each test tube?

B. Pour 3-5 drops of sodium sulfate solution into one test tube, and the same amount of chromium (III) sulfate solution into the other. Add a few drops of barium chloride solution to each test tube until precipitation occurs.

What substance is formed as a precipitate? Will a similar reaction of barium chloride, for example, with iron (III) sulfate occur?

Experiment 4. Study of the properties of amphoteric hydroxides.

A . Add 3 drops of zinc salt solution and a few drops of diluted sodium hydroxide solution (from a rack with reagents) until a precipitate of zinc hydroxide forms. Dissolve the resulting precipitates: in one test tube - in a solution of hydrochloric acid, in another - in excess concentrated

B. Add 3 drops of aluminum salt solution and a few drops into two test tubes diluted caustic soda solution (from a rack with reagents) until a precipitate of aluminum hydroxide forms. Dissolve the resulting precipitates: in one test tube - in a solution of hydrochloric acid, in another - in excess concentrated caustic soda solution (from a fume hood).

B. Add 3 drops of chromium (III) salt solution and a few drops of diluted sodium hydroxide solution (from a rack with reagents) until a precipitate of chromium (III) hydroxide forms. Dissolve the resulting precipitates: in one test tube - in a solution of hydrochloric acid, in another - in excess concentrated caustic soda solution (from a fume hood).

Experiment 5. Formation of slightly dissociated compounds. Add 3-5 drops of ammonium chloride solution into a test tube and add a few drops of sodium hydroxide solution. Pay attention to the smell, explain its appearance based on the reaction equation.

Experiment 6. Formation of complexes. Pour 3-5 drops of copper (II) sulfate solution into a test tube, then add drop by drop diluted(from a rack with reagents!) ammonia solution until a precipitate of hydroxycopper (II) sulfate forms according to the reaction:

2CuSO 4 + 2NH 4 OH = (CuOH) 2 SO 4 ↓ + (NH 4) 2 SO 4

Add excess to sediment concentrated ammonia solution (from a fume hood!). Pay attention to the dissolution of the precipitate according to the reaction:

(CuOH) 2 SO 4 + (NH 4) 2 SO 4 + 6NH 4 OH = 2SO 4 + 8H 2 O

What color does the resulting soluble copper ammine complex have?

Experiment 7. Formation of gases.

A. Pour 3-5 drops of sodium carbonate solution and a few drops of sulfuric acid into a test tube. What is being observed?

B. Pour 3-5 drops of sodium sulfide solution and 1 drop of sulfuric acid into a test tube. Pay attention to the smell of escaping gas.