Implementation of transformations according to the scheme. Implementation of transformations according to the scheme Tasks for independent solution
Exercise 1.
Establish a genetic chain for obtaining carbonate: A calcium oxide; B calcium hydroxide; in calcium; G calcium carbonate. Write down the equations of the corresponding reactions.
Decision:
Scheme of transformations of substances (Compose equations of the corresponding reactions):
A B C D
CaO ----> Ca(OH) 2 ----> Ca ---> CaCO 3
Equations of the corresponding reactions:
A) 2Сa + O 2 \u003d 2CaO;
B) CaO + H 2 O \u003d Ca (OH) 2;
C) Ca (OH) 2 \u003d Ca + O 2 + H 2 O;
D) Ca + H 2 O + CO 2 \u003d CaCO 3 + H 2 or Ca + H 2 O + CO 2 \u003d CaCO 3 + H 2.
Task 2.
Write the reaction equations in molecular and ionic forms, with the help of which transformations can be carried out according to the scheme:
Fe --> Fe(SO4 )3 ---> FeCI3 --->Fe(NO3 )3 ---> Fe 2 O 3 .
For redox reactions, specify the oxidizing agent and reducing agent.
Decision:
chemical reaction equations:
one). 2Fe + 6H2 SO4 = Fe2 (SO4 )3 + 3SO2 + 6H2 O - In this reaction, iron is oxidized to the +3 oxidation state, i.e. iron is a reducing agent. In this case, sulfur (VI) is reduced to sulfur (IV) in the reaction. those. sulfur is an oxidizing agent.
2Fe 0 + 12Н 0 + SO 4 2- = 2Fe 3+ + 3 SO 4 2- + 3SO 2 + 6H2 O.
2). Fe 2 (SO 4) 3 + 3ВаCl2 --> 2FeCl3 + 3ВаSO4 ↓ - exchange reaction;
Full ionic form of the reaction equation:
2Fe 3+ + 3SO 4 2- + 3Ва 2+ + 6Cl - = 2Fe 3+ + 6Cl - + 3ВаSO 4 ↓;
3Ва 2+ + 3SO 4 2- = 3ВаSO 4 ↓.
3). FeCl3 + 3HNO3 --> Fe(NO3 )3 + 3HCl - exchange reaction;
Full ionic form of the reaction equation:
Fe 3+ + 3Cl - + 3Ag + + 3NO 3 - = Fe 3+ + 3NO 3 - + 3AgCl↓.
Abbreviated ionic reaction equation:
3Ag + + 3Cl - = 3AgCl↓.
4). 4Fe (NO3) 3 ---> Fe2 O3 + NO2 + O2 - the reaction of decomposition of iron (III) nitrate when it is heated. In this case, oxidation-reduction occurs. – In this reaction, oxygen is oxidized to the oxidation state 0, i.e. oxygen is a reducing agent. In this case, nitrogen (V) is reduced to nitrogen (IV) in the reaction. those. nitrogen is an oxidizing agent.
Tasks for independent solution.
1. Write the equation for the reaction between propylene and potassium permanganate in a neutral medium.
2. Write the equation for the reaction between butene-2 and potassium permanganate in an acidic medium.
3. Compare the ratio of all isomeric alcohols of the composition C 4 H 10 O to oxidizing agents. For butanol-1 and butanol-2, write the reaction equations with a solution of potassium dichromate in an acidic medium.
4. Write the equation for the reaction between ethyl alcohol and a solution of potassium dichromate in an acidic medium.
5. Write the equation for the reaction between ethylbenzene and potassium permanganate in an acidic medium.
6. Write the reaction equation between styrene and potassium permanganate in a neutral medium.
7. Write the equation for the reduction of 1,3-dimethylnitrobenzene with ammonium sulfide in a neutral medium (Zinin reaction).
8. When glucose is oxidized with bromine water, gluconic acid is formed, and when oxidized with concentrated nitric acid, glucaric acid is formed. Write down the equations of the corresponding reactions.
9. Give the reaction equations for the interaction of acetylene with potassium permanganate in acidic and neutral media to form oxalic acid and potassium oxalate, respectively.
10. Write the equation for the oxidation of formaldehyde when heated with copper (II) hydroxide.
11. Write the equation for the reaction between butanediol-1,4 and potassium permanganate in an acidic medium to form a dibasic carboxylic acid.
Choice questions:
1. The interaction of methane with chlorine is a reaction
1) connections 3) splitting off
2) substitution 4) exchange
2. Propanol-2 can be obtained from propene by the reaction
1) hydrogenation 3) hydration
2) hydrolysis 4) halogenation
3. The interaction of an alcoholic solution of alkali with 2-chlorobutane is the reaction
1) connections 3) splitting off
2) substitution 4) exchange
4. 3,3-dimethylbutanal is formed during oxidation
1) (CH 3) 3 C - CH 2 - CH 2 OH
2) CH 3 CH 2 C (CH 3) - CH 2 OH
3) CH 3 CH (CH 3) CH (CH 3) - CH 2 OH
4) CH 3 CH 2 CH (CH 3) - CH 2 OH
5. The reaction proceeds according to the ionic mechanism
1) C 2 H 6 + Cl 2 → C 2 H 5 Cl + HCl
2) C 3 H 6 + H 2 → C 3 H 8
3) C 3 H 6 + HCl → C 3 H 7 Cl
4) nCH 2 \u003d CH 2 → (-CH 2 -CH 2 -) n
One of the most common types of tasks in organic chemistry are those in which it is required to carry out transformations according to the proposed scheme. At the same time, in some cases it is necessary to indicate specific reagents and conditions for the occurrence of reactions leading to substances that make up the chain of transformations. In others, on the contrary, it is necessary to determine which substances are formed under the action of these reagents on the starting compounds.
Usually, in such cases, it is not required to indicate the fine technical details of the synthesis, the exact concentration of reagents, specific solvents, purification and isolation methods, etc. however, exemplary reaction conditions must be specified.
Most often, the essence of the task lies in the sequential solution of the following tasks:
construction (lengthening or shortening) of the carbon skeleton;
introduction of functional groups into aliphatic and aromatic compounds;
substitution of one functional group for another;
removal of functional groups;
change in the nature of functional groups.
The sequence of operations may be different, depending on the structure and nature of the initial and resulting compounds.
Present facts and their relationships visually. Write down, in as much detail as possible, the essence of the problem in the form of a diagram.
Look at the problem as broadly as possible, take into account even solutions that seem unthinkable. In the end, they may be the right ones and lead you to the right decision.
Use the trial and error method. If there is a limited set of options, try them all.
First level
Option 1
EXERCISE 1
Write the equations of possible reactions of interaction of sodium and zinc with substances: chlorine, water, hydrochloric acid.
2Nа + 2Н2О = 2NаОН + Н2
2Na + 2HCl = 2NaCl + H2
2Nа + Cl2 = 2NаСl
Zn + Cl2 = ZnCl2
Zn + H2O = ZnO + H2.
Zn + 2HCl = ZnCl2 + H2
TASK 2
Fe →FeCl2 →Fe(N03)2 →Fe(OH)2 →FeO.
Fe + 2HCl = FeCl2 + H2
FeCl2 + 2AgNO3 = 2AgCl + Fe(NO3)2
Fe(NO3)2 + 2NaOH = Fe(OH)2 + 2NaNO3
Fe2++ 2NO3-+ 2Na++2OH- =2Na++2NO3-+ Fe(OH)2
Fe2++ 2OH- \u003d Fe (OH) 2
Fe (OH) 2 \u003d FeO + H2O
TASK 3
When interacting 12 g of magnesium with an excess of hydrochloric acid 10 liters of hydrogen (n.a.) were released. Calculate volume fraction the yield of the reaction product.
Mg + 2HCl = MgCl2 + H2
n (Mg) = 12/24 = 0.5 mol
V (H2) \u003d 10 / 22.4 \u003d 0.446 mol
n (Mg) = n (H2) according to the equation (theor.)
yield = 0.446 / 0.5 = 0.89 = 89%
Option 2
EXERCISE 1
Write the equations of possible reactions of lithium and copper with substances: chlorine, water, hydrochloric acid.
Consider one of the recorded reactions in the light of OVR.
2Li + Cl2 = 2LiCl
2Li + 2H2O = 2LiOH + H2
2Li + 2HCl = 2LiCl + H2
Cu + Cl2 = CuCl2
Cu + H2O = not reacting
Cu + HCl = no reaction
2Li+CL2= 2LiCL
Li- e---> Li+ reducing agent
CL2+2e----> 2CL- oxidant
TASK 2
Write reaction equations that can be used to carry out transformations according to the scheme
Ca →CaO →Ca(OH)2 →Ca(N03)2 → Ca3(PO4)2.
Consider transformation 3 in the light of TED.
2Ca+O2=2CaO
CaO+2NaOH=Ca(OH)2+Na2O
2HNO3 + Ca(OH)2 = Ca(NO3)2 + 2H2O
2H+ +2NO3- + Ca(OH)2 =Ca2+ +2NO3- + 2H2O
2H+ + Ca(OH)2 =Ca2+ +2H2O
3Ca(NO3)2 + 2Н3PO4 = Ca3(PO4)2 + 6HNO3
TASK 3
During the thermal decomposition of 10 g of limestone, 1.68 liters of carbon dioxide (n.a.) were obtained. Calculate the volume fraction of the yield of the reaction product.
CaCO3 = CaO+CO2
n (CaCO3) = 10/100 = 0.1 mol
V (CO2) = 1.68 / 22.4 = 0.075 mol
n (CaCO3) = n (CO2) according to the equation (theor.)
yield = 0.075/0.1 = 0.75 = 75%
Option 3
EXERCISE 1
Write the equations of possible reactions of calcium and iron with substances: chlorine, water, hydrochloric acid.
Consider one of the recorded reactions in the light of OVR.
Ca + Cl2 = CaCl2
Ca + 2HCl = CaCl2 + H2
Ca + 2H2O = Ca(OH)2 + H2
2Fe + 3Cl2 = 2FeCl3
Fe + 2HCl = FeCl2 + H2
TASK 2
Write the reaction equations with which you can carry out transformations according to the scheme:
Al →Al203 →AlCl3 →Al(OH)3 →Al(N03)3.
Consider transformation 3 in the light of TED.
4Al + 3O2 = 2Al2O3
Al2O3 + 6HCl = 2AlCl3 + 3H2O
AlCl3 + 3NH3 + 3H2O = Al(OH)3 + 3NH4Cl
Al3+ +3Cl- + 3NH3 + 3H2O = Al(OH)3 + 3NH4+ +Cl-
Al3+ + 3NH3 + 3H2O = Al(OH)3 + 3NH4+
Al(OH)3+3HNO3=Al(NO3)3+3H2O.
TASK 3
In the interaction of 23 g of sodium with water, 8.96 liters of hydrogen (n.a.) were obtained. Find the volume fraction of the yield of the reaction product (in %).
2Na + 2H2O = 2NaOH + H2
n (Na) \u003d 23/23 \u003d 1 mol
V (H2) \u003d 8.96 / 22.4 \u003d 0.4 mol
n (Na) : n (H2) = 2:1
n (H2) \u003d 1/ 2 \u003d 0.5 mol according to the equation (theor.)
yield = 0.4/0.5 = 0.8 = 80%
Option 4
EXERCISE 1
Write the equations of possible reactions of potassium and magnesium metals with chlorine, water, hydrochloric acid.
Consider one of these reactions in the light of OVR.
2K + 2H2O = 2KOH + H2.
2K + CI2 = 2KCI
2K + 2HCI = 2KCI + H2
Mg + 2HCl = MgCl2 + H2
Mg + Cl2 = MgCl2
Mg 0 -2e → Mg 2+ oxidizes, reducing agent
Cl20+2e→2Cl- is reduced, oxidizer
Mg+2H2O=Mg(OH)2+H2
TASK 2
Write the reaction equations for these transitions:
Li →Li20 →LiOH →Li2SO4.
Consider transformation 3 in the light of TED.
4Li + O2 = 2Li2O
Li2O + H2O = 2LiOH
2LiOH + H2SO4 = Li2SO4 + 2H2O
2Li+ +2OH- + 2Н+ +SO42- = 2Li+ +SO42- + 2Н2О
2OH- + 2H+ - = 2H2O
TASK 3
When 60 g of calcium interacted with water, 30 liters of hydrogen (n.a.) were released. Find the volume fraction of the yield of the reaction product.
Ca + 2H2O \u003d Ca (OH) 2 + H2
n (Ca) \u003d 60/40 \u003d 1.5 mol
V (H2) \u003d 30 / 22.4 \u003d 1.34 mol
n (Ca) \u003d n (H2) \u003d according to the equation (theor.)
yield = 1.34/1.5 = 0.89 = 89%
Second level
Option 1
EXERCISE 1
Write the equations of possible reactions of magnesium and aluminum with substances: oxygen, bromine, dilute sulfuric acid.
2 Mg + O2 = 2 MgO
Mg + Br 2 = Mg Br2
Mg 0 -2e → Mg 2+ is oxidized, reducing agent
Mg + H2SO4 = MgSO4 + H2
Mg + 2H+ = Mg2+ + H2
4Al + 3O2 = 2Al2O3,
2Al + 3H2SO4 = Al2(SO4)3 + 3H2
2Al + 3Br2 = 2AlBr3
TASK 2
Ca →X1 →Ca(OH)2 →X2 →Ca(HC03)2.
2Ca+O2=2CaO
CaO+H2O=Ca(OH)2
Ca(OH)2 + CO2 = CaCO3 ↓ + H2O
CaCO3 + CO2 + H2O ↔ Ca(HCO3)2
TASK 3
When 12 g of magnesium interacted with an excess of hydrochloric acid, 10 liters of hydrogen (n.a.) were released.
Mg + 2HCl = MgCl2 + H2
n (Mg) \u003d 12/24 \u003d 0.5 mol
n (H2) = n (Mg) according to the equation (theor.)
ŋ = V (H2) (ex.) / V (H2) (theor.) = n (H2) (ex.) / n (H2) (theor.) = 0.44 / 0.5 = 0.89 = 89%
Option 2
EXERCISE 1
Write the equations of possible reactions of copper and magnesium with substances: oxygen, iodine, dilute sulfuric acid.
2Cu + O2 = 2CuO
2Cu + I2 = 2CuI
Cu + H2SO4 = not available
2 Mg + O2 = 2 MgO
Mg 0 -2e → Mg 2+ is oxidized, reducing agent
O20 + 4e → 2 O 2- is reduced, oxidizing agent
Mg + H2SO4 = MgSO4 + H2
Mg + 2H+ = Mg2+ + H2
Mg + I 2 = Mg I 2
Mg 0 -2e → Mg 2+ is oxidized, reducing agent
I 20+2e→2 I - is reduced, oxidizer
TASK 2
Determine the formulas of substances X1 and X2 in the chain of transformations:
Zn → X1 → ZnSO4 → X2 → ZnO.
Write reaction equations that can be used to carry out transformations according to this scheme.
2Zn + O2 = 2ZnO
ZnO+H2SO4 --->ZnSO4 + H2O
ZnSO4 + 2NaOH (dec.) = Zn(OH)2↓ + Na2SO4
Zn(OH)2= ZnO + H2O
TASK 3
Thermal decomposition of 10 g of calcium carbonate produced 1.68 liters of carbon dioxide (n.a.). Calculate the volume fraction of the yield of the reaction product.
CaCO3 → CaO + CO2
n (CaCO3) \u003d 10/100 \u003d 0.1 mol
n (CO2) = n (CaCO3) according to the equation (theor.)
n (CO2) \u003d 1.68 / 22.4 \u003d 0.075 mol (ex.)
ŋ = V (CO2) (ex.) / V (CO2) (theor.) = n (CO2) (ex.) / n (CO2) (theor.) = 0.075 / 0.1 = 0.75 = 75%
Option 3
EXERCISE 1
Write the equations of possible reactions of iron and zinc with substances: oxygen, chlorine, dilute sulfuric acid.
Consider two of these reactions in the light of OVR and one in the light of TED.
3Fe + 2O2 = Fe2O3 FeO
Fe + H2SO4 = FeSO4 + H2
2Fe + 3Cl2 = 2FeCl3
2Zn + O2 = 2ZnO
Zn 0 -2e → Zn 2+ oxidizes, reducing agent
O20 + 4e → 2 O 2- is reduced, oxidizing agent
Zn + Cl2 = ZnCl2
Zn0 -2e→Zn2+ oxidizes, reducing agent
Cl20+2e→2Cl- is reduced, oxidizer
Zn + H2SO4 = ZnSO4 + H2
Zn + 2Н+ = Zn 2+ + Н2
TASK 2
Determine the formulas of substances X1 and X2 in the chain of transformations:
Fe →X1 →Fe(OH)2 →X2 →Fe.
Fe - 1 --> FeCl2 - 2 --> Fe(OH)2 - 3 --> FeO -4-> Fe
1. Fe + 2HCl --> FeCl2 + H2
2. FeCl2 + 2NaOH --> Fe(OH)2 + 2NaCl
Fe2+ + 2OH- --> Fe(OH)2
3. Fe (OH) 2 - t --> FeO + H2O
4. FeO + C = Fe + CO
TASK 3
In the interaction of 23 g of sodium with water, 8.96 liters of hydrogen (n.a.) were obtained. Find the volume fraction of the yield of the reaction product.
2Na + 2H2O = 2NaOH + H2
n (Na) \u003d 23/23 \u003d 1 mol
2n (H2) = n (Na) according to the equation (theor.)
n (H2) = 0.5 mol theor.
Option 4
EXERCISE 1
Write the equations of possible chemical reactions of beryllium and iron with substances: oxygen, bromine, dilute sulfuric acid.
Consider two of these reactions in the light of OVR and one in the light of TED.
3Fe + 2O2 = Fe2O3 FeO
Fe + H2SO4 = FeSO4 + H2
2Fe + 3Br 2 = 2FeBr3
2 Be + O2 = 2 BeO
Be0 -2e → Be 2+ is oxidized, reducing agent
O20 + 4e → 2 O 2- is reduced, oxidizing agent
Be + Br 2 = Be Br2
Be 0 -2e → Be 2+ oxidizes, reducing agent
Br 20+2e→2 Br - is reduced, oxidizer
Be + H2SO4 = Be SO4 + H2
Be + 2H+ = Be 2+ + H2
TASK 2
Determine the formulas of substances X1 and X2 in the chain of transformations:
Fe → X1 → Fe(OH)3 → X2 → Fe.
Write reaction equations that can be used to carry out transformations according to this scheme.
Fe 1→FeCl3 2→Fe(OH)3 3→Fe2O3 4→ Fe
1. 2Fe + 3Cl2 t → 2FeCl3
2. FeCl3+ 3NaOH → Fe(OH)3 ↓ + 3NaCl
3. 2Fe(OH)3 t → Fe2O3 + H2O
4. 2Fe2O3 + 3Ct → 4Fe + 3CO2
TASK 3
When 60 g of calcium interacted with water, 30 liters of hydrogen (n.a.) were released. Find the volume fraction of the yield of the reaction product (in %).
Ca + 2H2O \u003d Ca (OH) 2 + H2
n (Ca) \u003d 60/40 \u003d 1.5 mol
n (H2) = n (Ca) according to the equation (theor.)
n (H2) \u003d 30 / 22.4 \u003d 1.34 mol (ex.)
ŋ = V (H2) (ex.) / V (H2) (theor.) = n (H2) (ex.) / n (H2) (theor.) = 1.34 / 1.5 = 0.89 = 89%
Third level
Option 1
EXERCISE 1
Write equations of possible reactions of magnesium, aluminum and silver with non-metal, water, acid, salt solution.
1. Mg + 2HCl = MgCl2 + H2
Mg + 2H+ = Mg2+ + H2
2H+ + 2e= H2 oxidant
Mg + Cl2 = MgCl2
Mg+2H2O=Mg(OH)2+H2
Mg + CuCl2 = MgCl2 + Cu
Mg0 + Cu2+ = Mg2+ + Cu0
Mg0 - 2e = Mg2+ reducing agent
Cu2+ + 2e= Cu0 oxidant
2. 2Al + 6HCl = 2AlCl3 + 3H2
2Al + 6H+ = 2Al3+ + 3H02
2H+ + 2e= H2 oxidant
2Al + 3Cl2 = 2AlCl3
2Al + 6H2O = 2Al(OH)3 + 3H2
3HgCI2 + 2Al = 2AICI3 + 3Hg
3Hg2+ + 2Al0 = 2AI3+ + 3Hg0
Al0 - 3e \u003d AI3 + oxidizing agent
3. Ag + 2HCl = not reacting
2Ag + S = Ag2S
Ag+ H2O = not reacting
Ag + FeCl3 = AgCl + FeCl2
Ag0 + Fe3+= Ag++ Fe2+
Ag0 - 1e= Ag+ reducing agent
Fe3++1е= Fe2+oxidizer
TASK 2
Write reaction equations that can be used to carry out transformations according to this scheme.
2Be + O2 = 2BeO
BeO + 2HCl = BeCl2 + H2O
BeCl2 + 2NaOH = Be(OH)2 + 2NaCl
Be(OH)2 + 2NaOH= Na2BeO2 + 2H2O
TASK 3
When 12 g of technical magnesium containing 5% impurities interacted with an excess of hydrochloric acid, 10 liters of hydrogen (n.a.) were released. Calculate the volume fraction of the yield of the reaction product.
Mg + 2HCl = MgCl2 + H2
n(Mg) \u003d 12-12 * 0.05 / 24 \u003d 11.4 / 24 \u003d 0.475 mol
n (H2) (theor.) \u003d n (Mg) \u003d 0.475 mol
n (H2) \u003d 10 / 22.4 \u003d 0.44 mol (ex.)
ŋ = V (H2) (ex.) / V (H2) (theor.) = n (H2) (ex.) / n (H2) (theor.) = 0.44 / 0.475 = 0.92 = 92%
Option 2
EXERCISE 1
Write the equations of possible reactions of lithium, copper, barium, aluminum with substances: non-metal, water, acid, salt solution.
Consider the reactions of metals with acid and salt solutions from the point of view of OVR and TED.
1. 2Li + Сl2 = 2LiСl
2Li + 2H2O = 2LiOH + H2
2Li + 2HCl = 2LiCl + H2
2Li0 + 2Н+ = 2Li+ + Н02
Li0 – 1e = Li+ reducing agent
2H+ + 2e= H02 oxidizer
2Li + CuСl 2= 2LiС1 + Cu
2Li0 + Cu2+ = 2Li+ + Cu0
Li0 – 1e = Li+ reducing agent
Cu2+ + 2e= Cu0 oxidant
2. 2Al + 6HCl = 2AlCl3 + 3H2
2Al + 6H+ = 2Al3+ + 3H02
Al 0 - 3e = Al3+ reducing agent
2H+ + 2e= H2 oxidant
2Al + 3Cl2 = 2AlCl3
2Al + 6H2O = 2Al(OH)3 + 3H2
3HgCI2 + 2Al = 2AICI3 + 3Hg
3Hg2+ + 2Al0 = 2AI3+ + 3Hg0
Hg2+ + 2e = Hg0 reducing agent
Al0 - 3e \u003d AI3 + oxidizing agent
3. Ba + 2HCl = BaCl2 + H2
Ba + 2H+ = Ba 2+ + H2
2H+ + 2e= H2 oxidant
Ba + Cl2 = BaCl2
Ba + 2H2O \u003d Ba (OH) 2 + H2
Ba + CuCl2 = BaCl2 + Cu
Ba + Cu2+ = Ba 2+ + Cu0
Va 0 - 2e \u003d Va 2+ reducing agent
Cu2+ + 2e= Cu0 oxidant
4. Cu + Cl2 = CuCl2
Cu + H2O = not reacting
Cu + HCl = no reaction
Cu + 2FeCl3 = CuCl2 + 2FeCl2.
Cu 0 + Fe3+= Cu2 ++ Fe2+
Fe3++1е= Fe2+oxidizer
TASK 2
Determine the formulas of substances X1, X2 and X3 in the chain of transformations:
Write reaction equations that can be used to carry out transformations according to this scheme.
Fe + 2HСl → FeCl2 + H2
FeCl2+ 2NaOH → Fe(OH)2 + 2NaCl
4Fe(OH)2 + O2 + 2H2O = 4Fe(OH)3↓
2Fe(OH)3 = Fe2O3 + 3H2O
TASK 3
During thermal decomposition of 20 g of limestone containing 10% non-carbonate impurities, 3.23 liters of carbon dioxide (n.a.) were obtained.
Calculate the volume fraction of the yield of the reaction product (in %).
CaCO3 = CaO+CO2
n (CaCO3) \u003d 20-20 * 0.1 / 100 \u003d 18/100 \u003d 0.18 mol
V (CO2) = 3.23 / 22.4 = 0.145 mol (pract.)
n (CaCO3) \u003d n (CO2) according to the equation (theor.) \u003d 0.18 mol
yield = 0.145/0.18 = 0.801 = 80.1%
Option 3
EXERCISE 1
Write equations of possible reactions of calcium, iron and zinc with non-metal, water, acid, salt solution.
Consider the reactions of metals with acid and salt solutions from the point of view of OVR and TED.
1. Ca + 2HCl = CaCl2 + H2
Ca0 + 2H+ = Ca 2+ + H2
2H+ + 2e= H2 oxidant
Ca + Cl2 = CaCl2
Ca + 2H2O \u003d Ca (OH) 2 + H2
Ca + CuCl2 = CaCl2 + Cu
Ca + Cu2+ = Ca 2+ + Cu0
Ca 0 - 2e \u003d Ca 2+ reducing agent
Cu2+ + 2e= Cu0 oxidant
2.
2Fe + 3Cl2 t →2FeCl3
3. 2Zn + O2 = 2ZnO
Zn + H2O = ZnO + H2
Zn + H2SO4 = ZnSO4 + H2
Zn + 2Н+ = Zn 2+ + Н2
Zn 0 + 2H+ = Zn 2+ + H02
2H+ + 2e= H2 oxidant
Zn + CuSO4 = Cu + ZnSO4
Zn 0 + Cu2+ = Zn 2+ + Cu0
Zn 0 - 2e = Zn 2+ reducing agent
Cu2+ + 2e= Cu0 oxidant
TASK 2
Determine the formulas of substances X1, X2 and X3 in the chain of transformations:
Write reaction equations that can be used to carry out transformations according to this scheme.
2Ca+O2=2CaO
CaО+H2O=Ca(OH)2
Ca(OH)2+H2CO3=CaCO3+2H2O
CaCO3 + CO2 + H2O ↔ Ca(HCO3)2
Ca(HCO3)2 t →CaCO3↓ + CO2 + H2O
TASK 3
During the interaction of 24.15 g of technical sodium containing 5% impurities, 8.96 liters of hydrogen (n.a.) were obtained.
Calculate the volume fraction of the yield of the reaction product (in %).
2Na + 2H2O = 2NaOH + H2
n(Na) \u003d 24.15-24.15 * 0.05 / 23 \u003d 23 / 23 \u003d 1 mol
n (H2) (theor.) = 0.5 n (Mg) = 0.5 mol
n (H2) \u003d 8.96 / 22.4 \u003d 0.4 mol (ex.)
ŋ = V (H2) (ex.) / V (H2) (theor.) = n (H2) (ex.) / n (H2) (theor.) = 0.4 / 0.5 = 0.8 = 80%
Option 4
EXERCISE 1
Write the equations of possible reactions of beryllium, iron and copper with substances: non-metal, water, acid, salt solution.
Consider the reactions of metals with acid and salt solutions from the point of view of OVR and TED.
1. 2 Be + O2 = 2 BeO
Be + H2SO4 = BeSO4 + H2
Be + 2H+ = Be 2+ + H2
2. Cu + Cl2 = CuCl2
Cu + H2O = not reacting
Cu + HCl = no reaction
Cu + 2FeCl3 = CuCl2 + 2FeCl2.
Cu 0 + Fe3+= Cu2 ++ Fe2+
Cu 0 - 2e \u003d Cu + reducing agent
Fe3++1е= Fe2+oxidizer
3. 2Fe + 3Cl2 t → 2FeCl3
TASK 2
Determine the formulas of substances X1, X2 and X3 in the chain of transformations:
Write reaction equations that can be used to carry out transformations according to this scheme.
2Zn + O2 = 2ZnO
ZnO + 2HNO3 = Zn(NO3)2 + H2O
Zn(NO3)2 + 2NaOH = 2NaNO3 + Zn(OH)2
Zn(OH)2= ZnO + H2O
TASK 3
In the interaction of 60 g of technical calcium containing 2% impurities, with water, 30 liters of hydrogen (n.a.) were obtained. Calculate the volume fraction of the yield of the reaction product.
Ca + 2H2O \u003d Ca (OH) 2 + H2
n(Ca) \u003d 60-60 * 0.02 / 40 \u003d 58.8 / 40 \u003d 1.47 mol
n (H2) (theor.) \u003d n (Ca) \u003d 1.47 mol
n (H2) \u003d 30 / 22.4 \u003d 1.34 mol (ex.)
ŋ = V (H2) (ex.) / V (H2) (theor.) = n (H2) (ex.) / n (H2) (theor.) = 1.34 / 1.47 = 0.91 = 91%