Solution:
2Cl2 + 2H2O = 4HCl + O2
mp-pa \u003d m (H2O) + m (Cl2) - m (O2);
Δm = m(Cl2) - m(О2) ;
Let us take n(Cl2) as X, then n(O2) = 0.5x;
Let's compose algebraic equation on the basis of the above equality and find X:
Δm \u003d x M (Cl2) - 0.5 x M (O2) \u003d x (71 - 16) \u003d 55x;
x = 0.04 mol;
V(Cl2) \u003d n (Cl2) Vm \u003d 0.004 22.4 \u003d 0.896 l.
Answer: 0.896 l.
10. Calculate the range of acceptable values for the volume of chlorine (n.a.), which is necessary for the complete chlorination of 10.0 g of a mixture of iron and copper.
Solution:
Since the condition does not say what the ratio of metals in the mixture is, it remains to be assumed that the range of acceptable values for the volume of chlorine in this case will be the range between its volumes required for chlorination of 10 g of each metal separately. And the solution of the problem is reduced to the sequential finding of these volumes.
2Fe + 3Cl2 = 2FeCl3
Cu + Cl′2 = CuCl2
n(Cl2) = 1.5n(Fe) = 1.5 10/56 = 0.26 mol;
V(Cl2) \u003d n(Cl2) Vm \u003d 0.26 22.4 \u003d 5.99 ≈ 6 l;
n(Cl′2) = n(Cu) = 10/63.5 = 0.16 mol;
V(Cl′2) \u003d 22.4 0.16 \u003d 3.5 l.
Answer: 3.5 ≤ V(Cl2) ≤ 6l.
11. Calculate the mass of iodine that is formed when a mixture of sodium iodide dihydrate, potassium iodide and magnesium iodide is treated with an excess of an acidified solution of potassium permanganate, in which the mass fractions of all salts are equal, and the total amount of all substances is 50.0 mmol.
Solution:
Let's write down the equations of the reactions taking place in the solution, and compose the general half-reactions, on the basis of which we will arrange the coefficients:
10NaI 2H2O + 2KMnO4 + 8H2SO4 = 5I2 + 2MnSO4 + 5Na2SO4 + K2SO4 + 28H2O
10KI + 2KMnO4 + 8H2SO4 = 5I2 + 2MnSO4 + 6K2SO4 + 8H2O
5MgI2 + 2KMnO4 + 8H2SO4 = 5I2 + 2MnSO4 + 5MgSO4 + K2SO4 + 8H2O
MnO4¯+ 8H+ + 5ē = Mn2+ + 4H2O 2
2I¯− 2 ē = I2 5
2 MnO4¯+ 16H+ + 10 I¯= 2 Mn2+ + 5I2 + 8H2O
From the equality of the mass fractions of the components of the mixture, it follows that their masses are also equal. Taking them for X we compose an algebraic equation based on the equality:
n1 + n2 + n3 = 50.0 mmol
m1/M(NaI 2H2O) + m2/M(KI) + m3/M(MgI2) = 50.0 mmol
m1 = m2 = m3 = x
x / 186 + x / 166 + x / 278 \u003d 50 10-3 mol
m (I2)1 \u003d 5M (I2) m (NaI 2H2O) / 10M (NaI 2H2O) \u003d (5 254 3.33) / 10 186 \u003d 2.27 g;
m(I2)2 = 5M(I2) m(KI)/10M(KI) = (5 254 3.33)/10 166 = 2.55 g;
m (I2) 3 \u003d 5M (I2) m (MgI2) / 10M (MgI2) \u003d (5 254 3.33) / 10 278 \u003d 3.04 g.
Total: 7.86 g
Answer: 7.86
12. When passing through 200 g of a 5.00% solution of hydrogen peroxide chlorine, the mass of the solution increased by 3.9 g. Calculate the mass fractions of the substances in the resulting solution.
Solution:
H2O2 + Cl2 = O2 + 2HCl
1. Find the initial amount of H2O2 in the solution:
n1 (H2O2) \u003d m / M (H2O2) \u003d mР-RA ω / M (H2O2) \u003d 200 0.05 / 34 \u003d
2. Let's take the amount of absorbed chlorine in solution as X, then nO2 = x, and the increase in the mass of the solution is due to the difference in the masses of absorbed chlorine and released oxygen:
m (Cl 2) - m (O 2) \u003d Δ m or x M (Cl 2) - x M (O 2) \u003d Δ m;
71x - 32x = 3.9; x = 0.1 mol.
3. Calculate the amount of substances remaining in the solution:
n2 (H2O2) OXIDATED \u003d n (Cl 2) \u003d 0.1 mol;
n(H2O2) REMAINING IN SOLUTION \u003d n1 - n2 \u003d 0.294 - 0.1 \u003d 0.194 mol;
n (HCl) \u003d 2n (Cl 2) \u003d 0.2 mol.
4. Find the mass fractions of substances in the resulting solution:
ω (H2O2) \u003d n (H2O2) M (H2O2) / mP-PA \u003d 0.194 34 / 203.9 100% \u003d 3.23%;
ω (HCl) \u003d n (HCl) M (HCl) / mP-RA \u003d 0.2 36.5 / 203.9 100% \u003d 3.58%.
Answer:ω(H2O2) = 3.23%;
ω(HCl) = 3.58%.
13. Manganese (II) bromide tetrahydrate weighing 4.31 g was dissolved in a sufficient volume of water. Chlorine was passed through the resulting solution until the molar concentrations of both salts were equal. Calculate how much chlorine (n.a.) was skipped.
Solution:
MnBr2 4H2O + Cl2 = MnCl2 + Br2 + 4H2O
1. Find the initial amount of manganese (II) bromide tetrahydrate in solution:
n(MnBr2 4H2O)ISH. \u003d m / M \u003d 4.31 / 287 \u003d 1.5 10−2 mol.
2. Equality of the molar concentrations of both salts will come when half of the initial amount of Mn Br2 · 4H2O is consumed. That. the amount of chlorine required can be found from the reaction equation:
n(Cl2) = n(MnCl2) = 0.5 n(Mn Br2 4H2O)ISC. \u003d 7.5 10−3 mol.
V(Cl2) \u003d n Vm \u003d 7.5 10−3 22.4 \u003d 0.168 l.
Answer: 0.168 l.
14. Chlorine was passed through 150 ml of a barium bromide solution with a molar salt concentration of 0.05 mol/l until the mass fractions of both salts were equal. Calculate how much chlorine (200C, 95 kPa) was skipped.
Solution:
BaBr2 + Cl2 = BaCl2 + Br2
1. From the equality of the mass fractions of the formed salts, the equality of their masses follows.
m(BaCl2) = m(BaBr2) or n(BaCl2) М(BaCl2) = n′(BaBr2) М(BaBr2).
2. Take n(BaCl2) as X mol, and n′(BaBr2), remaining in solution, for SM V - x = 0.15 0.05 - x = 7.5 10 -3 - x and we will compose an algebraic equation:
208x = (7.5 10−3 − x) 297;
2.2275 = 297x + 208x;
3. Find the amount of chlorine and its volume:
n(Cl2) = n(BaCl2) = 0.0044 mol;
V(Cl2) \u003d nRT / P \u003d (0.0044 8.314 293) / 95 \u003d 0.113 l.
Answer: 113 ml.
15. A mixture of bromide and potassium fluoride with a total mass of 100 g was dissolved in water; an excess of chlorine was passed through the resulting solution. The mass of the residue after evaporation and calcination is 80.0 g. Calculate the mass fractions of the substances in the resulting mixture.
Solution:
1. After calcination of the reaction products, the residue consists of fluoride and potassium chloride:
2KBr + Cl2 = 2KCl + Br2
2. Let us take the amounts of KF and KBr as X And at respectively, then
n(KCl) = n(KBr) = y mol.
We compose a system of equations based on the equalities:
m(KF) + m(KBr) = 100
m(KF) + m(KCl) = 80
n(KF) М(KF) + n(КBr) М(КBr) = 100
n(KF) M(KF) + n(KCl) M(KCl) = 80
58x + 119y = 100 58x = 100 - 119y
58 x + 74.5y \u003d 80 100 - 119y + 74.5y \u003d 80
44.5y = 20; y = 0.45; x = 0.8.
3. Find the masses of substances in the remainder and their mass fractions:
m(KF) = 58 0.8 = 46.5 g.
m (KCl) \u003d 74.5 0.45 \u003d 33.5 g.
ω(KF) = 46.5/80 100% = 58.1%;
ω(KCl) = 33.5/80 100% = 41.9%.
Answer:ω(KF) = 58.1%;
ω(KCl) = 41.9%.
16. A mixture of bromide and sodium iodide was treated with an excess of chlorine water, the resulting solution was evaporated and calcined. The mass of the dry residue turned out to be 2.363 times less than the mass of the initial mixture. How many times will the mass of the precipitate obtained after treatment of the same mixture with an excess of silver nitrate be greater than the mass of the initial mixture?
Solution:
2NaBr + HClO +HCl = 2NaCl + Br2 + H2O
2NaI + HClO + HCl = 2NaCl + I2 + H2O
1. Let us take the mass of the initial mixture as 100 g, and the amounts of NaBr and NaI salts that form it, as X And at respectively. Then, based on the ratio (m(NaBr) + m(NaI))/ m(NaCl) = 2.363, we compose a system of equations:
103x + 150y = 100
2.363 58.5(x+y) = 100
x = 0.54 mol; y = 0.18 mol.
2. Let's write the second group of reactions:
NaBr + AgNO3 = AgBr↓ + NaNO3
NaI + AgNO3 = AgI↓ + NaNO3
Then, to determine the mass ratio of the precipitate formed and the initial mixture of substances (taken as 100 g), it remains to find the amounts and masses of AgBr and AgI, which are equal to n(NaBr) and n(NaI), respectively, i.e. 0.18 and 0 .54 mol.
3. Find the mass ratio:
(m(AgBr) + m(AgI))/(m(NaBr) + m(NaI)) =
(M(AgBr) x + M(AgI) y)/100 =
(188 0.18 + 235 0.54)/100 =
(126,9 + 34,67)/100 = 1,62.
Answer: 1.62 times.
17. A mixture of magnesium iodide and zinc iodide was treated with an excess of bromine water, the resulting solution was evaporated and calcined at 200-3000C. The mass of the dry residue turned out to be 1.445 times less than the mass of the initial mixture. How many times will the mass of the precipitate obtained after treatment of the same mixture with an excess of sodium carbonate be less than the mass of the initial mixture?
Solution:
1. Let's write both groups of reactions, denoting the masses of the initial mixture of substances and the resulting products as m1, m2, m3.
(MgI2 + ZnI2)+ 2Br2 = (MgBr2 + ZnBr2)+ 2I2
(MgI2 + ZnI2)+ 2 Na2CO3 = (MgCO3 + ZnCO3)↓ + 4NaI
m1/m2 = 1.445; m1/ m3 = ?
2. Let's take the amount of salts in the initial mixture as X(MgI2) and at(ZnI2), then the amounts of products of all reactions can be expressed as
n(MgI2) = n(MgBr2) = n(MgCO3) = x mol;
n(ZnI2) = n(ZnBr2) = n(ZnCO3) = mol.
Hydrochloric acid.
In chemical reactions, hydrochloric acid exhibits all the properties of strong acids: it interacts with metals standing in a series of voltages to the left of hydrogen, with oxides (basic, amphoteric), bases, amphoteric hydroxides and salts:
2HCl + Fe \u003d FeCl 2 + H 2
2HCl + CaO = CaCl 2 + H 2 O
6HCl + Al 2 O 3 \u003d 2AlCl 3 + 3H 2 O
HCl + NaOH = NaCl + H2O
2HCl + Cu(OH) 2 = CuCl 2 + 2H 2 O
2HCl + Zn(OH) 2 = ZnCl 2 + 2H 2 O
HCl + NaHCO 3 \u003d NaCl + CO 2 + H 2 O
HCl + AgNO 3 \u003d AgCl ↓ + HNO 3 (qualitative reaction for halide ions)
6HCl (conc.) + 2HNO 3 (conc.) = 3Cl 2 + 2NO + 4H 2 O
HClO 2 - chloride
HClO 3 - chlorine
HClO 4 - chlorine
HClO HClO 2 HClO 3 HClO 4
strengthening of acidic properties
2HClO 2HCl + O 2
HClO + 2HI \u003d HCl + I 2 + H 2 O
HClO + H 2 O 2 \u003d HCl + H 2 O + O 2
Salt.
Salts of hydrochloric acid are chlorides.
NaCl + AgNO 3 \u003d AgCl ↓ + NaNO 3 (qualitative reaction for halide ions)
AgCl + 2(NH 3 ∙ H 2 O) \u003d Cl + 2H 2 O
2AgCl 2Ag + Cl 2
Salts of oxygen-containing acids.
Ca(ClO) 2 + H 2 SO 4 = CaSO 4 + 2HCl + O 2
Ca(ClO) 2 + CO 2 + H 2 O \u003d CaCO 3 + 2HClO
Ca(ClO) 2 + Na 2 CO 3 \u003d CaCO 3 + 2NaClO
Ca(ClO) 2 CaCl 2 + O 2
4KClO 3 3KClO 4 + KCl
2KClO 3 2KCl + 3O 2
2KClO 3 + 3S 2KCl + 3SO 2
5KClO 3 + 6P 5KCl + 3P 2 O 5
KClO 4 2O 2 + KCl
3KClO 4 + 8Al = 3KCl + 4Al 2 O 3
Bromine. Bromine compounds.
Br 2 + H 2 \u003d 2HBr
Br 2 + 2Na = 2NaBr
Br 2 + Mg = MgBr 2
Br 2 + Cu = CuBr 2
3Br 2 + 2Fe = 2FeBr 3
Br 2 + 2NaOH (diff) = NaBr + NaBrO + H 2 O
3Br 2 + 6NaOH (conc.) = 5NaBr + NaBrO 3 + 3H 2 O
Br 2 + 2NaI \u003d 2NaBr + I 2
3Br 2 + 3Na 2 CO 3 \u003d 5NaBr + NaBrO 3 + 3CO 2
3Br 2 + S + 4H 2 O \u003d 6HBr + H 2 SO 4
Br 2 + H 2 S \u003d S + 2HBr
Br 2 + SO 2 + 2H 2 O \u003d 2HBr + H 2 SO 4
4Br 2 + Na 2 S 2 O 3 + 10NaOH \u003d 2Na 2 SO 4 + 8NaBr + 5H 2 O
14HBr + K 2 Cr 2 O 7 \u003d 2KBr + 2CrBr 3 + 3Br 2 + 7H 2 O
4HBr + MnO 2 \u003d MnBr 2 + Br 2 + 2H 2 O
2HBr + H 2 O 2 \u003d Br 2 + 2H 2 O
2KBr + 2H 2 SO 4 (conc.) = 4K 2 SO 4 + 4Br 2 + SO 2 + 2H 2 O
2KBrO 3 3O 2 + 2KBr
2KBrO 4 O 2 + 2KBrO 3 (up to 275°C)
KBrO 4 2O 2 + KBr (above 390°C)
Iodine. iodine compounds.
3I 2 + 3P = 2PI 3
I 2 + H 2 \u003d 2HI
I 2 + 2Na = 2NaI
I 2 + Mg \u003d MgI 2
I 2 + Cu \u003d CuI 2
3I 2 + 2Al = 2AlI 3
3I 2 + 6NaOH (gor.) \u003d 5NaI + NaIO 3 + 3H 2 O
I 2 + 2NaOH (razb) \u003d NaI + NaIO + H 2 O
3I 2 + 10HNO 3 (razb) \u003d 6HIO 3 + 10NO + 2H 2 O
I 2 + 10HNO 3 (conc.) = 2HIO 3 + 10NO 2 + 4H 2 O
I 2 + 5NaClO + 2NaOH \u003d 5NaCl + 2NaIO 3 + H 2 O
I 2 + 5Cl 2 + 6H 2 O \u003d 10HCl + 2HIO 3
I 2 + Na 2 SO 3 + 2NaOH \u003d 2NaI + Na 2 SO 4 + H 2 O
2HI + Fe 2 (SO 4) 3 \u003d 2FeSO 4 + I 2 + H 2 SO 4
2HI + NO 2 \u003d I 2 + NO + H 2 O
2HI + S = I 2 + H 2 S
8KI + 5H 2 SO 4 (conc.) = 4K 2 SO 4 + 4I 2 + H 2 S + 4H 2 O or
KI + 3H 2 O + 3Cl 2 \u003d HIO 3 + KCl + 5HCl
10KI + 8H 2 SO 4 + 2KMnO 4 = 5I 2 + 2MnSO 4 + 6K 2 SO 4 + 8H 2 O
6KI + 7H 2 SO 4 + K 2 Cr 2 O 7 \u003d Cr 2 (SO 4) 3 + 3I 2 + 4K 2 SO 4 + 7H 2 O
2KI + H 2 SO 4 + H 2 O 2 \u003d I 2 + K 2 SO 4 + 2H 2 O
2KI + Fe 2 (SO 4) 3 \u003d I 2 + 2FeSO 4 + K 2 SO 4
2KI + 2CuSO 4 + K 2 SO 3 + H 2 O \u003d 2CuI + 2K 2 SO 4 + H 2 SO 4
2HIO 3 I 2 O 5 + H 2 O
2HIO 3 + 10HCl \u003d I 2 + 5Cl 2 + 6H 2 O
2HIO 3 + 5Na 2 SO 3 = 5Na 2 SO 4 + I 2 + H 2 O
2HIO 3 + 5H 2 SO 4 + 10FeSO 4 = Fe 2 (SO 4) 3 + I 2 + 6H 2 O
I 2 O 5 + 5CO I 2 + 5CO 2
2KIO 3 3O 2 + 2KI
2KIO 3 + 12HCl (conc.) = I 2 + 5Cl 2 + 2KCl + 6H 2 O
KIO 3 + 3H 2 SO 4 + 5KI = 3I 2 + 3K 2 SO 4 + 3H 2 O
KIO 3 + 3H 2 O 2 \u003d KI + 3O 2 + 3H 2 O
2KIO 4 O 2 + 2KIO 3
5KIO 4 + 3H 2 O + 2MnSO 4 = 2HMnO 4 + 5KIO 3 + 2H 2 SO 4
Halogens.
1. The substance obtained at the anode during the electrolysis of a melt of sodium iodide with inert electrodes was isolated and introduced into interaction with hydrogen sulfide. The gaseous product of the last reaction was dissolved in water, and ferric chloride was added to the resulting solution. The precipitate formed was filtered off and treated with hot sodium hydroxide solution. Write the equations of the described reactions.
2. The substance obtained at the anode during the electrolysis of a sodium iodide solution with inert electrodes was introduced into a reaction with potassium. The reaction product was heated with concentrated sulfuric acid and the evolved gas was passed through a hot solution of potassium chromate. Write the equations of the described reactions.
3. Chlorine water has a chlorine smell. When alkalized, the smell disappears, and when hydrochloric acid is added, it becomes stronger than it was before. Write the equations of the described reactions.
4. Colorless gases are released when concentrated acid is exposed to both sodium chloride and sodium iodide. When these gases are passed through an aqueous solution of ammonia, salts are formed. Write the equations of the described reactions.
5. During the thermal decomposition of salt A in the presence of manganese dioxide, a binary salt B and a gas that supports combustion and is part of the air were formed; when this salt is heated without a catalyst, salt B and a salt of an oxygen-containing acid are formed. When salt A interacts with hydrochloric acid, a yellow-green gas (a simple substance) is released and salt B is formed. Salt B colors the flame purple, and when it interacts with a solution of silver nitrate, a white precipitate forms. Write the equations of the described reactions.
6) When an acid solution A is added to manganese dioxide, a poisonous yellow-green gas is released. By passing the released gas through a hot solution of caustic potash, a substance is obtained that is used in the manufacture of matches and some other incendiary compositions. During the thermal decomposition of the latter in the presence of manganese dioxide, a salt is formed, from which, when interacting with concentrated sulfuric acid, the initial acid A can be obtained, and a colorless gas that is part of atmospheric air. Write the equations of the described reactions.
7) Iodine was heated with an excess of phosphorus, and the reaction product was treated with a small amount of water. The gaseous reaction product was completely neutralized with sodium hydroxide solution and silver nitrate was added to the resulting solution. Write the equations of the described reactions.
8) The gas released when solid sodium chloride was heated with concentrated sulfuric acid was passed through a solution of potassium permanganate. The gaseous reaction product was taken up in cold sodium hydroxide solution. After adding hydroiodic acid to the resulting solution, a pungent odor appears and the solution acquires a dark color. Write the equations of the described reactions.
9) A gas was passed through a solution of sodium bromide, which is released during the interaction of hydrochloric acid with potassium permanganate. After completion of the reaction, the solution was evaporated, the residue was dissolved in water and subjected to electrolysis with graphite electrodes. The gaseous reaction products were mixed with each other and illuminated. The result was an explosion. Write the equations of the described reactions.
10) A solution of hydrochloric acid was carefully added to pyrolusite, and the released gas was passed into a beaker filled with a cold solution of caustic potash. After the end of the reaction, the beaker was covered with cardboard and left, while the beaker was illuminated Sun rays; after a while, a smoldering splinter was brought into the glass, which flared up brightly. Write the equations of the described reactions.
11) The substance released on the cathode and anode during the electrolysis of sodium iodide solution with graphite electrodes react with each other. The reaction product interacts with concentrated sulfuric acid with the release of gas, which was passed through a solution of potassium hydroxide. Write the equations of the described reactions.
12) Concentrated hydrochloric acid was added to lead (IV) oxide while heating. The escaping gas was passed through a heated solution of caustic potash. The solution was cooled, the oxygenated acid salt was filtered off and dried. When the resulting salt is heated with hydrochloric acid, a poisonous gas is released, and when it is heated in the presence of manganese dioxide, a gas that is part of the atmosphere is released. Write the equations of the described reactions.
13) Iodine was treated with concentrated nitric acid by heating. The reaction product was gently heated. The resulting oxide reacted with carbon monoxide. The isolated simple substance was dissolved in a warm solution of potassium hydroxide. Write the equations of the described reactions.
14) A solution of potassium iodide was treated with an excess of chlorine water, while at first the formation of a precipitate was observed, and then its complete dissolution. The resulting iodine-containing acid was isolated from the solution, dried and gently heated. the resulting oxide reacted with carbon monoxide. Write the equations of the described reactions.
15) Iodine was treated with chloric acid. The reaction product was gently heated. the reaction product was gently heated. The resulting oxide reacts with carbon monoxide to form two substances - simple and complex. A simple substance dissolves in a warm alkaline solution of sodium sulfite. Write the equations of the described reactions.
16) Potassium permanganate was treated with an excess of hydrochloric acid solution, a solution formed and gas was released. The solution was divided into two parts: potassium hydroxide was added to the first, and silver nitrate was added to the second. The evolved gas reacted The gas reacted with potassium hydroxide upon cooling. Write the equations of the described reactions.
17) Sodium chloride melt was subjected to electrolysis. The gas released at the anode reacted with hydrogen to form a new gaseous substance with a characteristic odor. It was dissolved in water and treated with the calculated amount of potassium permanganate, and a yellow-green gas was formed. This substance enters upon cooling with sodium hydroxide. Write the equations of the described reactions.
18) Potassium permanganate was treated with concentrated hydrochloric acid. The gas released in this case was collected, and a solution of potassium hydroxide was added dropwise to the reaction mass until the precipitation ceased. The collected gas was passed through a hot solution of potassium hydroxide, thus forming a mixture of two salts. The solution was evaporated, the solid residue was calcined in the presence of a catalyst, after which only salt remained in the solid residue. Write the equations of the described reactions.
Halogens.
1) 2NaI 
2Na + I 2
at the cathode at the anode
I 2 + H 2 S = 2HI + S↓
2HI + 2FeCl 3 \u003d I 2 + 2FeCl 2 + 2HCl
I 2 + 6NaOH (gor.) \u003d NaIO 3 + 5NaI + 3H 2 O
2) 2NaI + 2H 2 O 
2H 2 + 2NaOH + I 2
At the cathode At the anode
8KI + 8H 2 SO 4 (conc.) = 4I 2 ↓ + H 2 S + 4K 2 SO 4 + 4H 2 O or
8KI + 9H 2 SO 4 (conc.) = 4I 2 ↓ + H 2 S + 8KHSO 4 + 4H 2 O
3H 2 S + 2K 2 CrO 4 + 2H 2 O = 2Cr(OH) 3 + 3S + 4KOH
3) Cl 2 + H 2 O ↔ HCl + HClO
HCl + NaOH = NaCl + H2O
HClO + NaOH = NaClO + H2O
NaClO + 2HCl \u003d NaCl + Cl 2 + H 2 O
4) H 2 SO 4 (conc.) + NaCl (solid) = NaHSO 4 + HCl
9H 2 SO 4 (conc.) + 8NaI (solid) \u003d 8NaHSO 4 + 4I 2 ↓ + H 2 S + 4H 2 O
NH 4 OH + HCl \u003d NH 4 Cl + H 2 O
NH 4 OH + H 2 S \u003d NH 4 HS + H 2 O
5) 2KClO 3 2KCl + 3O 2
4KClO 3 KCl + 3KClO 4
KClO 3 + 6HCl \u003d KCl + 3Cl 2 + 3H 2 O
KCl + AgNO 3 = AgCl↓ + KNO 3
6) 4HCl + MnO 2 = MnCl 2 + Cl 2 + 2H 2 O
3Cl 2 + 6KOH (gor.) = 5KCl + KClO 3 + 3H 2 O
2KClO 3 2KCl + 3O 2
H 2 SO 4 (conc.) + NaCl (solid) = NaHSO 4 + HCl
7) 3I 2 + 3P = 2PI 3
PI 3 + 3H 2 O \u003d H 3 PO 3 + 3HI
HI + NaOH = NaI + H 2 O
NaI + AgNO 3 = AgI↓ + NaNO 3
8) H 2 SO 4 (conc.) + NaCl (solid) = NaHSO 4 + HCl
16HCl + 2KMnO 4 = 5Cl 2 + 2KCl + 2MnCl 2 + 8H 2 O
Cl 2 + 2NaOH (cold) = NaCl + NaClO + H 2 O
NaClO + 2HI \u003d NaCl + I 2 + H 2 O
9) 16HCl + 2KMnO 4 = 5Cl 2 + 2KCl + 2MnCl 2 + 8H 2 O
Tasks C 2 (2013)
Reactions confirming the relationship of various classes of inorganic substances
Copper(II) oxide was heated in a current 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.
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.
The 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.
Metallic aluminum powder 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.
As a result of incomplete combustion of coal, a gas was obtained, in the flow of which iron(III) oxide 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.
A certain 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.
Sulfur was fused with iron. The reaction product was dissolved in water. The resulting gas was burned in an excess of oxygen. The combustion products were absorbed by an aqueous solution of iron(III) sulfate. Write the equations for the four described reactions.
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.
A solution of potassium iodide was treated with an excess of chlorine water, while first the formation of a precipitate was observed, 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.
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.
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.
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.
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.
Magnesium was dissolved in dilute nitric acid, and no gas evolution 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.
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.
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.
Chlorine reacted with hot potassium hydroxide solution. 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.
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.
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.
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1) CuO + CO \u003d Cu + CO 2 2) Cu + Cl 2 \u003d CuCl 2 3) 2CuCl 2 + 2KI \u003d 2CuCl ↓ + I 2 + 2KCl 4) CuCl 2 + 2AgNO 3 \u003d 2AgCl ↓ + Cu (NO 3) 2 |
1) Cu (NO 3) 2 2CuO + 4NO 2 + O 2 2) CuO + 2H 2 SO 4 \u003d CuSO 4 + SO 2 + 2H 2 O 3) CuSO 4 + H 2 O \u003d Cu ↓ + H 2 SO 4 + O 2 (electronic 4) Cu + 4HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O |
1) 2Fe + 3Cl 2 = 2FeCl 3 2) FeCl 3 + 3NaOH \u003d Fe (OH) 3 ↓ + 3NaCl 4) Fe 2 O 3 + 6HI \u003d 2FeI 2 + I 2 + 3H 2 O |
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1) 2Al + 3I 2 \u003d 2AlI 3 2) AlI 3 + 3NaOH \u003d Al (OH) 3 + 3NaI 3) Al (OH) 3 + 3HCl \u003d AlCl 3 + 3H 2 O 4) 2AlCl 3 + 3Na 2 CO 3 + 3H 2 O \u003d 2Al (OH) 3 + 3CO 2 + 6NaCl |
2) Fe 2 O 3 + CO \u003d Fe + CO 2 3) 2Fe + 6H 2 SO 4 \u003d Fe 2 (SO 4) 3 + 3SO 2 + 6H 2 O 4) Fe 2 (SO 4) 3 + 4H 2 O \u003d 2Fe + H 2 + 3H 2 SO 4 + O 2 (electrolysis) |
1) ZnS + 2HNO 3 \u003d Zn (NO 3) 2 + H 2 S 2) 2ZnS + 3O 2 \u003d 2ZnO + 2SO 2 3) 2H 2 S + SO 2 \u003d 3S ↓ + 2H 2 O 4) S + 6HNO 3 \u003d H 2 SO 4 + 6NO 2 + 2H 2 O |
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2) FeS + 2H 2 O \u003d Fe (OH) 2 + H 2 S 3) 2H 2 S + 3O 2 2SO 2 + 2H 2 O 4) Fe 2 (SO 4) 3 + SO 2 + 2H 2 O \u003d 2FeSO 4 + 2H 2 SO 4 |
1) 2Fe + 3Cl 2 = 2FeCl 3 2) 2FeCl 3 + 3Na 2 CO 3 \u003d 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 |
1) 2KI + Cl 2 \u003d 2KCl + I 2 2) I 2 + 5Cl 2 + 6H 2 O \u003d 10HCl + 2HIO 3 3)2HIO 3 I 2 O 5 + H 2 O 4) I 2 O 5 + 5CO \u003d I 2 + 5CO 2 |
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1) Cr 2 S 3 + 3H 2 SO 4 \u003d Cr 2 (SO 4) 3 + 3H 2 S 2) Cr 2 (SO 4) 4 + 6NH 3 + 6H 2 O \u003d 2Cr (OH) 3 ↓ + 3 (NH 4) 2 SO 4 3) H 2 S + Pb (NO 3) 2 \u003d PbS ↓ + 2HNO 3 4) PbS + 4H 2 O 2 \u003d PbSO 4 + 4H 2 O |
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 +KOH=K 4) 3K + AlCl 3 \u003d 3KCl + Al (OH) 3 ↓ |
1) KNO 3 + Pb KNO 2 + PbO 2) 2KNO 2 + 2H 2 SO 4 + 2KI \u003d 2K 2 SO 4 + 2NO + I 2 + 2H 2 O 3) I 2 + 10HNO 3 2HIO 3 + 10NO 2 + 4H 2 O 4) 10NO 2 + P \u003d 2P 2 O 5 + 10NO |
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1) 3Cu + 8HNO 3 \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O 4) (OH) 2 + 3H 2 SO 4 \u003d CuSO 4 + 2 (NH 4) 2 SO 4 + 2H 2 O |
1) 4Mg + 10HNO 3 \u003d 4Mg (NO 3) 2 + NH 4 NO 3 + 3H 2 O 2) Mg (NO 3) 2 + 2KOH \u003d Mg (OH) 2 ↓ + 2KNO 3 3) NH 4 NO 3 + KOHKNO 3 + NH 3 + H 2 O 4) 4NH 3 + 3O 2 \u003d 2N 2 + 6H 2 O |
1) KNO 2 + NH 4 Cl KCl + N 2 + 2H 2 O 2) 3Mg + N 2 \u003d Mg 3 N 2 3) Mg 3 N 2 + 8HCl \u003d 3MgCl 2 + 2NH 4 Cl 4) 2MgCl 2 + 2Na 2 CO 3 + H 2 O \u003d (MgOH) 2 CO 3 ↓ + CO 2 + 4NaCl |
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1) Al 2 O 3 + 2NaOH 2NaAlO 2 + H 2 O 2) NaAlO 2 + NH 4 Cl + H 2 O \u003d NaCl + Al (OH) 3 ↓ + NH 3 3) 2NH 3 + H 2 SO 4 \u003d (NH 4) 2 SO 4 4) (NH 4) 2 SO 4 NH 3 + NH 4 HSO 4 |
1) 3Cl 2 + 6KOH6KCl + KClO 3 + 3H 2 O 2) 6HCl + KClO 3 \u003d KCl + 3Cl 2 + 3H 2 O 3) 2Fe + 3Cl 2 \u003d 2FeCl 3 4) 2FeCl 3 + Fe3FeCl 2 |
1) 3Cu + 4HNO 3 \u003d 3Cu (NO 3) 2 + 2NO 2 + 4H 2 O 2) Cu (NO 3) 2 + 2NH 3 H 2 O \u003d Cu (OH) 2 + 2NH 4 NO 3 3) Cu (OH) 2 + 4NH 3 H 2 O \u003d (OH) 2 + 4H 2 O 4) (OH) 2 + 6HCl \u003d CuCl 2 + 4NH 4 Cl + 2H 2 O |
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19 Document C2 Reactions, confirming relationship various classes inorganic substances Given aqueous solutions ... Write the equations of four possible reactions. Are given substances: hydrobromic acid, ... precipitate during reactions substance yellow color burned on... Summary report of the chairmen of the subject commissions of the Astrakhan region on academic subjects of the state final certification on educational programs of secondary general educationReport4 21.9 40.6 C2 Reactions, confirming relationship various classes inorganic substances 54.1 23.9 9.9 7.7 4.3 С3 Reactions, confirming relationship organic compounds 56.8 10 ... Calendar-thematic planning of classes in preparation for the exam in chemistry in the 2013-2014 academic yearCalendar-thematic planning25) 27.03.2014 Reactions, confirming relationship various classes inorganic substances. Solution of exercises C-2 ... simple substances. Chemical properties complex substances. Relationship various classes inorganic substances. Reactions ion exchange... Calendar-thematic lesson plan for preparing for the exam in chemistry for graduates of schools in Barnaul and the Altai Territory in 2015. Lesson numberCalendar-thematic plan... (on the example of aluminum and zinc compounds). Relationship inorganic substances. Reactions, confirming relationship various classes inorganic substances(37 (C2)). February 28, 2015 ... |
Chemistry Tutor
ACTIVITY 10
10th grade(first year of study)
Continuation. For the beginning, see No. 22/2005; 1, 2, 3, 5, 6, 8, 9, 11/2006
Redox reactions
Plan
1. Redox reactions (ORD), the degree of oxidation.
2. Oxidation process, the most important reducing agents.
3. The recovery process, the most important oxidizing agents.
4. Redox duality.
5. Main types of OVR (intermolecular, intramolecular, disproportionation).
6. The value of OVR.
7. Methods for compiling OVR equations (electronic and electron-ion balance).
All chemical reactions On the basis of changes in the oxidation states of the atoms involved in them, they can be divided into two types: OVR (occurring with a change in oxidation states) and non-ORD.
Oxidation state is the conditional charge of an atom in a molecule, calculated on the assumption that only ionic bonds exist in the molecule.
R e u l e d e n t o d i n g o c i d i n d
The oxidation state of atoms of simple substances is zero.
The sum of the oxidation states of atoms in a complex substance (in a molecule) is zero.
The oxidation state of alkali metal atoms is +1.
The oxidation state of alkaline earth metal atoms is +2.
The oxidation state of boron, aluminum atoms is +3.
The oxidation state of hydrogen atoms is +1 (in hydrides of alkali and alkaline earth metals -1).
The oxidation state of oxygen atoms is –2 (in peroxides –1).
Any OVR is a set of processes of recoil and attachment of electrons.
The process of donating electrons is called oxidation. Particles (atoms, molecules or ions) that donate electrons are called reducing agents. As a result of oxidation, the oxidation state of the reducing agent increases. Reducing agents can be particles in lower or intermediate oxidation states. The most important reducing agents are: all metals in the form of simple substances, especially active ones; C, CO, NH 3 , PH 3 , CH 4 , SiH 4 , H 2 S and sulfides, metal halides and halides, metal hydrides, metal nitrides and phosphides.
The process of adding electrons is called recovery. The particles that accept electrons are called oxidizers. As a result of reduction, the oxidation state of the oxidizing agent decreases. Oxidizing agents can be particles in higher or intermediate oxidation states. The most important oxidizing agents: simple non-metal substances with high electronegativity (F 2, Cl 2, O 2), potassium permanganate, chromates and dichromates, nitric acid and nitrates, concentrated sulfuric acid, perchloric acid and perchlorates.
There are three types of redox reactions.
Intermolecular OVR - an oxidizing agent and a reducing agent are part of various substances, for example:
Intramolecular OVR The oxidizing agent and reducing agent are in the same substance. These can be different elements, for example:
or one chemical element in different oxidation states, for example:
Disproportionation (self-oxidation-self-healing)- The oxidizing and reducing agent is the same element that is in an intermediate oxidation state, for example:
OVR are of great importance, since most of the reactions occurring in nature are of this type (the process of photosynthesis, combustion). In addition, OVR are actively used by man in his practical activities (metal reduction, ammonia synthesis):
The electronic balance method ( electronic circuits) or the electron-ion balance method.
Electronic balance method:
Electron-ion balance method:
Test on the topic "Redox reactions"
1. Potassium dichromate was treated with sulfur dioxide in a sulfuric acid solution, and then with an aqueous solution of potassium sulfide. The final substance X is:
a) potassium chromate; b) chromium(III) oxide;
c) chromium(III) hydroxide; d) chromium(III) sulfide.
2. What reaction product between potassium permanganate and hydrobromic acid can react with hydrogen sulfide?
a) Bromine; b) manganese(II) bromide;
c) manganese dioxide; d) potassium hydroxide.
3. Iron(II) iodide is oxidized with nitric acid to form iodine and nitrogen monoxide. What is the ratio of the coefficient at the oxidizing agent to the coefficient at the reducing agent in the equation for this reaction?
a) 4: 1; b) 8: 3; at 11; d) 2:3.
4. The oxidation state of the carbon atom in the hydrocarbonate ion is:
a) +2; b) -2; c) +4; d) +5.
5. Potassium permanganate in a neutral medium is reduced to:
a) manganese; b) manganese(II) oxide;
c) manganese(IV) oxide; d) potassium manganate.
6. The sum of the coefficients in the equation for the reaction of manganese dioxide with concentrated hydrochloric acid is:
a) 14; b) 10; at 6; d) 9.
7. Of the listed compounds, only the oxidizing ability is shown by:
A) sulfuric acid; b) sulfurous acid;
c) hydrosulphuric acid; d) potassium sulfate.
8. Of the listed compounds, redox duality is exhibited by:
a) hydrogen peroxide; b) sodium peroxide;
c) sodium sulfite; d) sodium sulfide.
9. Of the following types of reactions, redox reactions are:
a) neutralization; b) recovery;
c) disproportionation; d) exchange.
10. The oxidation state of the carbon atom does not numerically coincide with its valence in the substance:
a) carbon tetrachloride; b) ethane;
c) calcium carbide; d) carbon monoxide.
Key to the test
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| V | A | A | V | V | G | a, g | a B C | b, c | b, c |
Redox exercises
(electronic and electron-ionic balance)
Task 1. Compile OVR equations using the electronic balance method, determine the type of OVR.
1. Zinc + Potassium Dichromate + Sulfuric Acid = Zinc Sulfate + Chromium(III) Sulfate + Potassium Sulfate + Water.
Electronic balance:
2. Tin(II) sulfate + potassium permanganate + sulfuric acid = tin(IV) sulfate + manganese sulfate + potassium sulfate + water.
3. Sodium iodide + potassium permanganate + potassium hydroxide = iodine + potassium manganate + sodium hydroxide.
4. Sulfur + potassium chlorate + water = chlorine + potassium sulfate + sulfuric acid.
5. Potassium iodide + potassium permanganate + sulfuric acid = manganese(II) sulfate + iodine + potassium sulfate + water.
6. Iron(II) sulfate + potassium dichromate + sulfuric acid = iron(III) sulfate + chromium(III) sulfate + potassium sulfate + water.
7. Ammonium nitrate \u003d nitric oxide (I) + water.
8. Phosphorus + Nitric acid\u003d phosphoric acid + nitric oxide (IV) + water.
9. Nitrous acid \u003d nitric acid + nitric oxide (II) + water.
10. Potassium chlorate + hydrochloric acid = chlorine + potassium chloride + water.
11. Ammonium dichromate \u003d nitrogen + chromium (III) oxide + water.
12. Potassium hydroxide + chlorine = potassium chloride + potassium chlorate + water.
13. Sulfur(IV) oxide + bromine + water = sulfuric acid + hydrobromic acid.
14. Sulfur(IV) oxide + hydrogen sulfide = sulfur + water.
15. Sodium sulfite = sodium sulfide + sodium sulfate.
16. Potassium permanganate + hydrochloric acid = manganese(II) chloride + chlorine + potassium chloride + water.
17. Acetylene + oxygen = carbon dioxide + water.
18. Potassium nitrite + potassium permanganate + sulfuric acid = potassium nitrate + manganese(II) sulfate + potassium sulfate + water.
19. Silicon + Potassium Hydroxide + Water = Potassium Silicate + Hydrogen.
20. Platinum + nitric acid + hydrochloric acid = platinum(IV) chloride + nitric oxide(II) + water.
21. Arsenic Sulfide + Nitric Acid = Arsenic Acid + sulphur dioxide+ nitrogen dioxide + water.
22. Potassium permanganate \u003d potassium manganate + manganese (IV) oxide + oxygen.
23.
Copper(I) Sulfide + Oxygen + Calcium Carbonate = Copper(II) Oxide + Calcium Sulfite +
+ carbon dioxide.
24.
Iron(II) chloride + potassium permanganate + hydrochloric acid = iron(III) chloride + chlorine +
+ manganese(II) chloride + potassium chloride + water.
25. Iron(II) sulfite + potassium permanganate + sulfuric acid = iron(III) sulfate + manganese(II) sulfate + potassium sulfate + water.
Answers to the exercises of task 1
When using the half-reaction method (electron-ionic balance), it should be borne in mind that in aqueous solutions, the binding of excess oxygen and the addition of oxygen by the reducing agent occur differently in acidic, neutral, and alkaline media. In acidic solutions, excess oxygen is bound by protons to form water molecules, and in neutral and alkaline solutions, by water molecules to form hydroxide ions. The addition of oxygen by the reducing agent is carried out in acidic and neutral environments due to water molecules with the formation of hydrogen ions, and in an alkaline environment - due to hydroxide ions with the formation of water molecules.
Neutral environment:
Alkaline environment:
oxidizing agent + H 2 O \u003d ... + OH -,
reducing agent + OH - \u003d ... + H 2 O.
Acid environment:
oxidizing agent + H + \u003d ... + H 2 O,
reducing agent + H 2 O \u003d ... + H +.
Task 2. Using the method of electron-ion balance, compose the equations of the OVR occurring in a certain medium.
1. Sodium sulfite + potassium permanganate + water = ...................... .
2. Iron(II) hydroxide + oxygen + water = ..............................
3. Sodium bromide + potassium permanganate + water = .......... .
4. Hydrogen sulfide + bromine + water = sulfuric acid + ...................... .
5. Silver(I) nitrate + phosphine + water = silver + phosphoric acid + ..............................
In alkaline environment
1. Sodium sulfite + potassium permanganate + potassium hydroxide = ...................... .
2. Potassium bromide + chlorine + potassium hydroxide = potassium bromate + ...................... .
3. Manganese(II) sulfate + potassium chlorate + potassium hydroxide = potassium manganate + ...................... .
4. Chromium(III) chloride + bromine + potassium hydroxide = potassium chromate + ...................... .
5. Manganese(IV) oxide + potassium chlorate + potassium hydroxide = potassium manganate + ...................... .
In acidic environment
1. Sodium sulfite + potassium permanganate + sulfuric acid = ...................... .
2. Potassium nitrite + potassium iodide + sulfuric acid = nitric oxide (II) + ...................... .
3. Potassium permanganate + nitric(II) oxide + sulfuric acid = nitric(IV) oxide + ...................... .
4. Potassium iodide + potassium bromate + hydrochloric acid = ...................... .
5. Manganese(II) Nitrate + Lead(IV) Oxide + Nitric Acid = Permanganic Acid +
+ ...................... .
Answers to the exercises of task 2
N e t r a l en viroment
Task 3. Using the method of electron-ion balance, compose the OVR equations.
1. Manganese(II) hydroxide + chlorine + potassium hydroxide = manganese(IV) oxide + ...................... .
Electron-ion balance:
2. Manganese(IV) oxide + oxygen + potassium hydroxide = potassium manganate + ....................... .
3. Iron(II) sulfate + bromine + sulfuric acid = ...................... .
4. Potassium iodide + iron(III) sulfate = ....................... .
5. Hydrogen bromide + potassium permanganate = ..............................
6. Hydrogen chloride + chromium(VI) oxide = chromium(III) chloride + ...................... .
7. Ammonia + bromine = ...................... .
8. Copper(I) oxide + nitric acid = nitric(II) oxide + ...................... .
9. Potassium sulfide + potassium manganate + water = sulfur + ...................... .
10. Nitric oxide (IV) + potassium permanganate + water = ...................... .
11. Potassium iodide + potassium dichromate + sulfuric acid = ..............................
12. Lead(II) sulfide + hydrogen peroxide = ..............................
13. Hypochlorous acid + hydrogen peroxide = hydrochloric acid + ...................... .
14. Potassium iodide + hydrogen peroxide = .............................. .
15. Potassium permanganate + hydrogen peroxide = manganese(IV) oxide + .................................... .
16. Potassium iodide + potassium nitrite + acetic acid = nitric oxide (II) + .............................. .
17. Potassium permanganate + potassium nitrite + sulfuric acid = ..................................
18. Sulfurous acid + chlorine + water = sulfuric acid + ...................... .
19. Sulfurous acid + hydrogen sulfide = sulfur + .............................. .
dissolving water. The solution obtained after passing gases through water had an acidic reaction. When this solution was treated with silver nitrate, 14.35 g of a white precipitate precipitated. Determine the quantitative and qualitative composition of the initial mixture of gases. Solution.
The gas that burns to form water is hydrogen, which is slightly soluble in water. React in sunlight with an explosion hydrogen with oxygen, hydrogen with chlorine. Obviously, there was chlorine in the mixture with hydrogen, because. the resulting HC1 is highly soluble in water and gives a white precipitate with AgNO3.
Thus, the mixture consists of gases H2 and C1:
1 mol 1 mol
HC1 + AgN03 -» AgCl 4- HN03.
x mol 14.35
When processing 1 mol of HC1, 1 mol of AgCl is formed, and when processing x mol, 14.35 g or 0.1 mol. Mr(AgCl) = 108 + 2 4- 35.5 = 143.5, M(AgCl) = 143.5 g/mol,
v= - = = 0.1 mol,
x = 0.1 mol of HC1 was contained in the solution. 1 mol 1 mol 2 mol H2 4-C12 2HC1 x mol y mol 0.1 mol
x \u003d y \u003d 0.05 mol (1.12 l) of hydrogen and chlorine reacted to form 0.1 mol
HC1. The mixture contained 1.12 liters of chlorine, and hydrogen 1.12 liters + 1.12 liters (excess) = 2.24 liters.
Example 6 A laboratory has a mixture of sodium chloride and iodide. 104.25 g of this mixture was dissolved in water and an excess of chlorine was passed through the resulting solution, then the solution was evaporated to dryness and the residue was calcined to constant weight at 300 °C.
The mass of dry matter turned out to be 58.5 g. Determine the composition of the initial mixture in percent.
Mr(NaCl) = 23 + 35.5 = 58.5, M(NaCl) = 58.5 g/mol, Mr(Nal) = 127 + 23 = 150 M(Nal) = 150 g/mol.
In the initial mixture: the mass of NaCl - x g, the mass of Nal - (104.25 - x) g.
When passing through a solution of chloride and sodium iodide, iodine is displaced by them. When passing the dry residue, the iodine evaporated. Thus, only NaCl can be a dry matter.
In the resulting substance: the mass of NaCl of the original x g, the mass of the obtained (58.5-x):
2 150 g 2 58.5 g
2NaI + C12 -> 2NaCl + 12
(104.25 - x)g (58.5 - x)g
2 150 (58.5 - x) = 2 58.5 (104.25 x)
x = - = 29.25 (g),
those. NaCl in the mixture was 29.25 g, and Nal - 104.25 - 29.25 = 75 (g).
Find the composition of the mixture (in percent):
w(Nal) = 100% = 71.9%,
©(NaCl) = 100% - 71.9% = 28.1%.
Example 7 68.3 g of a mixture of nitrate, iodide and potassium chloride are dissolved in water and treated with chlorine water. As a result, 25.4 g of iodine was released (neglect the solubility of which in water). The same solution was treated with silver nitrate. 75.7 g of sediment fell out. Determine the composition of the initial mixture.
Chlorine does not interact with potassium nitrate and potassium chloride:
2KI + C12 -» 2KS1 + 12,
2 mol - 332 g 1 mol - 254 g
Mg (K1) \u003d 127 + 39 - 166,
x = = 33.2 g (KI was in the mixture).
v(KI) - - = = 0.2 mol.
1 mol 1 mol
KI + AgN03 = Agl + KN03.
0.2 mol x mol
x = = 0.2 mol.
Mr(Agl) = 108 + 127 = 235,
m(Agl) = Mv = 235 0.2 = 47 (r),
then AgCl will be
75.7 g - 47 g = 28.7 g.
74.5 g 143.5 g
KCl + AgN03 = AgCl + KN03
X \u003d 1 L_ \u003d 14.9 (KCl).
Therefore, the mixture contained: 68.3 - 33.2 - 14.9 = 20.2 g KN03.
Example 8. To neutralize 34.5 g of oleum, 74.5 ml of a 40% potassium hydroxide solution is consumed. How many moles of sulfur oxide (VI) account for 1 mole of sulfuric acid?
100% sulfuric acid dissolves sulfur oxide (VI) in any ratio. The composition expressed by the formula H2S04*xS03 is called oleum. Let's calculate how much potassium hydroxide is needed to neutralize H2SO4:
1 mol 2 mol
H2S04 + 2KOH -> K2S04 + 2H20 xl mol y mol
y - 2*x1 mole of KOH is used to neutralize SO3 in oleum. Let's calculate how much KOH is needed to neutralize 1 mol of SO3:
1 mol 2 mol
S03 4- 2KOH -> K2SO4 + H20 x2 mol z mol
z - 2 x2 mol of KOH goes to neutralize SOg in oleum. 74.5 ml of a 40% KOH solution is used to neutralize the oleum, i.e. 42 g or 0.75 mol KOH.
Therefore, 2 xl + 2x 2 \u003d 0.75,
98 xl + 80 x2 = 34.5 g,
xl = 0.25 mol H2SO4,
x2 = 0.125 mol SO3.
Example 9 There is a mixture of calcium carbonate, zinc sulfide and sodium chloride. If 40 g of this mixture is treated with an excess of hydrochloric acid, 6.72 liters of gases will be released, the interaction of which with an excess of sulfur oxide (IV) releases 9.6 g of sediment. Determine the composition of the mixture.
When exposed to a mixture of excess hydrochloric acid, carbon monoxide (IV) and hydrogen sulfide could be released. Only hydrogen sulfide interacts with sulfur oxide (IV), therefore, according to the amount of precipitate, its volume can be calculated:
CaC03 + 2HC1 -> CaC12 + H20 + C02t(l)
100 g - 1 mol 22.4 l - 1 mol
ZnS + 2HC1 -> ZnCl2 + H2St (2)
97 g - 1 mol 22.4 l - 1 mol
44.8 l - 2 mol 3 mol
2H2S + S02 -» 3S + 2H20 (3)
xl l 9.6 g (0.3 mol)
xl = 4.48 L (0.2 mol) H2S; from equations (2 - 3) it can be seen that ZnS was 0.2 mol (19.4 g):
2H2S + S02 -> 3S + 2H20.
Obviously, the carbon monoxide (IV) in the mixture was:
6.72 l - 4.48 l \u003d 2.24 l (CO2).
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