characteristics of lithium. Characteristics of lithium Characteristics of lithium according to plan 9
chemistry, grade 9. Give a description of lithium according to the plan. and got the best answer
Answer from
Lithium is an element of the 2nd period of the main subgroup of group I of the periodic system of D. I. Mendeleev, an element of IA or a subgroup of alkali metals.
The structure of the lithium atom can be reflected as follows: 3Li - 2ē, 1ē. Lithium atoms will exhibit strong reducing properties: they will easily give up their only outer electron and, as a result, will receive an oxidation state (s. o.) +1. These properties of lithium atoms will be less pronounced than those of sodium atoms, which is associated with an increase in the atomic radii: Rat (Li)< Rат (Na). Восстановительные свойства атомов лития выражены сильнее, чем у бериллия, что связано и с числом внешних электронов, и с расстоянием от ядра до внешнего уровня.
Lithium is a simple substance, is a metal, and, therefore, has a metallic crystal lattice and a metallic chemical bond. The charge of the lithium ion: not Li + 1 (as indicated by the s. o.), but Li +. Are common physical properties metals arising from their crystalline structure: electrical and thermal conductivity, malleability, ductility, metallic luster, etc.
Lithium forms an oxide with the formula Li2O - it is a salt-forming, basic oxide. This compound is formed due to the ionic chemical bond Li2 + O2-, interact with water, forming an alkali.
Lithium hydroxide has the formula LiOH. This base is alkali. Chemical properties: interaction with acids, acid oxides and salts.
In the subgroup of alkali metals is absent general formula"Volatile Hydrogen Compounds". These metals do not form volatile hydrogen compounds. Compounds of metals with hydrogen are binary compounds of the ionic type with the formula M+H-.
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Lithium? this is an element of the 2nd period of the main subgroup of group I of the periodic system of D. I. Mendeleev, an element of IA or a subgroup of alkali metals.
The structure of the lithium atom can be reflected as follows: 3Li? 2e, 1e. Lithium atoms will exhibit strong reducing properties: they will easily give up their only outer electron and, as a result, will receive an oxidation state (s. o.) +1. These properties of lithium atoms will be less pronounced than those of sodium atoms, which is associated with an increase in the atomic radii: Rat (Li)< Rат (Na). Восстановительные свойства атомов лития выражены сильнее, чем у бериллия, что связано и с числом внешних электронов, и с расстоянием от ядра до внешнего уровня.
Lithium? a simple substance is a metal, and, therefore, has a metallic crystal lattice and a metallic chemical bond. The charge of the lithium ion: not Li + 1 (as indicated by the s. o.), but Li +. General physical properties of metals arising from their crystalline structure: electrical and thermal conductivity, malleability, ductility, metallic luster, etc.
Lithium forms an oxide with the formula Li2O ? it is a salt-forming, basic oxide. This compound is formed due to the ionic chemical bond Li2 + O2-, interact with water, forming an alkali.
Lithium hydroxide has the formula LiOH. Is this the foundation? alkali. Chemical properties: interaction with acids, acid oxides and salts.
In the subgroup of alkali metals, there is no general formula "Volatile hydrogen compounds". These metals do not form volatile hydrogen compounds. Compounds of metals with hydrogen? binary compounds of the ionic type with the formula M+H-.
Characteristics of the chemical element-metal based on its position in the Periodic system of D. I. Mendeleev
Lesson goals. Give a plan for the general characteristics of a chemical element according to its position in the Periodic system and teach ninth graders to use it to draw up a characteristic of a metal element. Repeat on the basis of this information at the 8th grade course on the structure of the atom, on the types of chemical bonds, on the classification of inorganic substances and their properties in the light of TED and OVR, on the genetic relationship between classes of compounds. To acquaint students with solving problems on the share of the yield of the reaction product.
Equipment and reagents. Li, Li 2 O, LiOH; CaCO 3 and HNO 3 to get CO 2 ; solutions: CuSO 4 , NH 4 Cl, HCl, phenolphthalein; test tubes, a device for obtaining gases.
I. Plan for the characteristics of a chemical element according to its position in the Periodic system
Unlike the plan given in the textbook, it would obviously be logical to start general characteristics element precisely from the definition of its "coordinates", i.e., the position in the Periodic system. Students very often simply call this point of the plan: “the address of a chemical element”, i.e., indicate the serial number of the element, the period (its type is called: small or large) and the group (the type of subgroup is indicated: main or side). When fulfilling this point of the plan, the characteristics will be correct if the teacher also introduces new designations for the type of subgroup: A - for the main and C (B) - for the secondary, which is caused by the use of such symbols in tests and formulations of tickets for the final exams for the course of the main and secondary schools.
The textbook provides an abbreviated version of the characteristics of magnesium. Let us reveal in more detail the characteristics of another chemical element - metal - lithium.
II. Characteristics of the chemical element lithium according to its position in the Periodic system
1. Lithium is an element of the 2nd period of the main subgroup of group I of the Periodic Table of D. I. Mendeleev, an element of group IA or (if students remember the eighth grade course) of the alkali metal subgroup.
2. The structure of the lithium atom can be reflected as follows:
It will be correct if here the students characterize the first form of the existence of a chemical element - atoms.
Lithium atoms will exhibit strong reducing properties: they will easily give up their only external electron and, as a result, will receive an oxidation state (s.o.) + 1. These properties of lithium atoms will be less pronounced than those of sodium atoms, which is associated with an increase in the atomic radii:
The teacher can pay attention to the problem: why, in the electrochemical series of voltages, lithium is ahead of sodium. The thing is that a series of stresses characterizes not the properties of atoms, but the properties of metals - simple substances, i.e. the second form of existence chemical elements, for which it is not R at, and parameters of a different kind: the binding energy of the crystal lattice, standard electrode potentials, etc.
The reducing properties of lithium atoms are more pronounced than those of its neighbor in period, beryllium, which is related both to the number of external electrons and to R at.
3. Lithium is a simple substance, it is a metal, and therefore it has a metal crystal lattice and a metal chemical bond (the teacher repeats the definitions of these two concepts with the students), the formation of which can be reflected using the scheme:
The teacher pays attention to how the charge of the lithium ion is written: not Li +1 (so indicate s. o.), a Li + .
In the course of this characteristic, the general physical properties of metals, arising from their crystalline structure, are also repeated: electrical and thermal conductivity, malleability, plasticity, metallic luster, etc.
4. Lithium forms an oxide with the formula Li 2 ABOUT.
The teacher repeats the composition and classification of oxides with the student, as a result of which the students themselves formulate that Li 2
0 is a salt-forming, basic oxide. This compound is formed due to an ionic chemical bond (why?; the teacher asks to write down the scheme for the formation of this bond:
) and, like all basic oxides, reacts with acids to form salt and water, and with acidic oxides and water to form alkali. Students name the type of the corresponding reactions, write down their equations, and the reaction with acids is also considered in ionic form.
5. Lithium hydroxide has the formula LiOH. It's a base, an alkali.
The teacher repeats with the students two blocks of theoretical information based on the material of the previous year: the structure and properties of LiOH.
Structure. Students themselves name the type of connection between Li + and he - - ionic, they say that Li + is a simple ion, and OH - - difficult. Then the teacher asks to determine the type of bond between the oxygen and hydrogen atoms in the hydroxide ion. The guys easily call it: a covalent polar bond. And then the teacher emphasizes that the presence of different types of bonds in one substance is an argument in favor of the assertion that division chemical bonds into different types relatively, all connections are of the same nature.
Chemical properties: interaction with acids, acid oxides and salts - are considered in the light of TED and are illustrated by reaction equations in ionic and molecular forms (preferably in that order).
6. To characterize a hydrogen compound (it can only be given in a strong class), it is better to use a problematic situation: why is there no general formula in the horizontal column "Volatile hydrogen compounds" in the subgroup of alkali metals?
Students reasonably answer that this is obvious, since these metals do not form volatile hydrogen compounds. The teacher in response asks: what compounds can these metals give with hydrogen? To this, students quite often answer that, probably, binary compounds of the ionic type with the formula M + H - . Then the teacher can complete this part of the characterization by substantiating the conclusion that hydrogen quite legitimately occupies a dual position in the Periodic system: both in group IA and in group VIIA.
III. Solving problems on finding the share of the yield of the reaction product from the theoretically possible
The first part of the lesson is devoted to the application of theoretical knowledge from the eighth grade course to describe the properties of a particular chemical element. This, so to speak, is the qualitative side of a repetitive-generalizing lesson, an introduction to the course of the chemistry of elements.
The quantitative side of such a lesson can be represented by calculations related to such a generalizing concept as "the fraction of the yield of the reaction product from the theoretically possible".
The teacher recalls that the concept of “share” is universal in nature - it shows with what part of the whole calculations are being made - and recalls the varieties of this concept that students operated on last year: the fraction of an element in a compound, the mass or volume fraction of a component in a mixture substances.
Now, the teacher continues, let's get acquainted with the fraction of the yield of the reaction product from the theoretically possible and proposes to solve the problem:
"Find the volume of carbon dioxide (n.a.) that can be obtained by reacting 250 g of limestone containing 20% impurities with an excess of nitric acid."
Students can easily cope with the task by repeating the algorithm for solving calculations according to chemical equations:
The teacher poses a problem: is it possible in fact (in practice) to obtain the calculated theoretical volume? After all, the technology for obtaining chemical products often leaves much to be desired. And it demonstrates the interaction of a piece of marble with acid, as well as the collection of CO 2 into the flask. Students can easily guess that the collected volume of the product will always be less than the calculated one: part of it will be lost until the teacher closes the device with a cork, part will evaporate until the end of the gas outlet tube is lowered into the flask, etc.
The teacher summarizes that the ratio of the volume (or mass) of the product obtained is the practical output to the volume (or mass) calculated theoretically and is called the yield fraction - ω out.or W:
Then the teacher asks to find the volume of CO 2 for the considered problem, if its output is 75% of the theoretically possible:
The reverse task is proposed for the house:
“In the interaction of 800 mg of a 30% solution of sodium hydroxide (sodium hydroxide) with an excess of a solution of copper sulfate (copper (I) sulfate), 196 mg of a precipitate was obtained. What is its output as a percentage of the theoretically possible?
IV. Metal genetic series
At the end of the lesson, students recall the signs of the genetic series of the metal:
1) the same chemical element - metal;
2) different forms of existence of this chemical element: a simple substance and compounds - oxides, bases, salts;
3) interconversions of substances of different classes.
As a result, students write down the genetic series of lithium:
which the teacher suggests at home to illustrate with reaction equations in ionic (where it takes place) and molecular forms, as well as to analyze all redox reactions.
First level
Option 1
1. The reaction equation for the neutralization of sodium hydroxide with hydrochloric acid is given:
NaOH + HCl = NaCl + H20 + Q.
thermal effect;
participation of a catalyst;
direction.
Consider this chemical reaction from the point of view of the theory of electrolytic dissociation. Write down the full and abbreviated ionic equations.
NaOH + HCl = NaCl + H2O + Q
Starting materials: 1 mol of sodium hydroxide solid (1 sodium atom, 1 hydrogen atom, 1 oxygen atom), 1 mol of hydrochloric acid (1 hydrogen atom, 1 chlorine atom).
Reaction products: 1 mol of sodium chloride solid (1 sodium atom, 1 chlorine atom), 1 mol of water (1 oxygen atom, 2 hydrogen atom).
The reaction is exothermic
The starting materials and products are in solution.
without catalyst
irreversible reaction
Na+ + OH- + H+ + Cl- = Na+ + Cl- + H2O
OH- + H+ = H2O
2. Give a description of the chemical element magnesium according to the plan:
the position of the element in the PSCE;
the structure of the atom;
Magnesium -- Mg
Ordinal number Z=12; mass number A = 24, nuclear charge + 12, number of protons = 12, neutrons (N = A-Z = 12) 24 - 12 = 12 neutrons, electrons = 12, period - 3, energy levels - 3,
The structure of the electron shell: 12 M g 2e; 8e; 2e.
12 M g)))
2 8 2
+2 oxidation state;
The reduction properties of magnesium are more pronounced than those of beryllium, but weaker than those of calcium, which is associated with an increase in the radii of the atoms Be - M g - Ca;
Magnesium ion M g 2+
MgO - magnesium oxide is the main oxide and exhibits all the characteristic properties of oxides. Magnesium forms hydroxide Mg (OH) 2, which exhibits all the characteristic properties of bases.
3. Write the equations for the reactions of magnesium oxide and hydroxide with hydrochloric acid in molecular and ionic form.
MgO+2HCl=MgCl₂ + H₂O
MgO+2H+=Mg2+ + H₂O
Mg(OH)2+2HCl= MgCl₂ + 2H₂O
Mg(OH)2+2H+= Mg2+ + 2H₂O
Option 2
1. The scheme of aluminum combustion reaction is given.
Al + 02 → A1203 + Q.
Describe the following responses:
the number and composition of the starting materials and reaction products;
thermal effect;
aggregate state of substances;
participation of a catalyst;
change in the oxidation states of elements;
direction.
0 0 +3 –2
Al + O2 = Al2O3+Q
4Al + 3O2 = 2Al2O3
Aluminum is a reducing agent and oxygen is an oxidizing agent.
Starting materials: 4 moles of aluminum, 3 moles of oxygen (3 molecules of 2 oxygen atoms). Reaction product: 2 moles of aluminum oxide (2 aluminum atoms, 3 oxygen atoms in one molecule).
The reaction is exothermic.
Aluminum - solid, oxygen - g., aluminum oxide - solid.
Without the participation of a catalyst
Irreversible.
2. Give a description of the chemical element sodium according to the plan:
the position of the element in the PSCE;
the structure of the atom;
formulas of oxide and hydroxide, their character.
Sodium -- Na
11 Na)))
2 8 1
+1 oxidation state;
Sodium ion Na+
3. Write the equations for the reactions of sodium oxide and hydroxide with a solution of sulfuric acid in molecular and ionic form.
2NaOH+H2SO4=2H2O+Na2SO4
2OH-+2H+=2H2O
Na2O+H2SO4=H2O+Na2SO4
Na2O+2H+=H2O+2Na+
Option 3
1. The reaction scheme for obtaining sulfur oxide (VI) from sulfur oxide (IV) is given.
S02 + 02 S03 + Q.
Write an equation for this reaction by placing the coefficients in it using the electronic balance method. Specify the oxidizing agent and reducing agent.
Describe the following responses:
the number and composition of the starting materials and reaction products;
thermal effect;
aggregate state of substances;
participation of a catalyst;
change in the oxidation states of elements;
direction.
2S+4O2 + O02 = 2S+6O-23+ Q
S+4 -2e →S+6 reducing agent
O02 +4e→2O-2 oxidant
The initial substances are 2 mol of sulfur oxide 4 (in one molecule 1 sulfur atom, 2 oxygen atoms) and 1 mol of oxygen (in one molecule 2 oxygen atoms).
The reaction product is 2 mol of sulfur oxide 6 (one molecule contains 1 sulfur atom, 3 oxygen atoms)
The reaction is exothermic.
Sulfur oxide 4 and oxygen - gases, Sulfur oxide (VI) liquid
with catalyst
Reversible.
2. Give a description of the chemical element lithium according to the plan:
the structure of the atom;
formulas of oxide and hydroxide, their character.
lithium Li
Ordinal number Z=3; mass number A \u003d 7, nuclear charge + 3, number of protons \u003d 3, neutrons (N \u003d A-Z \u003d 4) 7 - 3 \u003d 4 neutrons, electrons \u003d 3, period - 2, energy levels - 2
The structure of the electron shell: 3 Li 2e; 1e.
3Li))
2 1
+1 oxidation state;
The reducing properties of lithium are less pronounced than those of sodium and potassium, which is associated with an increase in the atomic radii;
Lithium ion Li+
Li 2O - lithium oxide is the main oxide and exhibits all the characteristic properties of oxides. Lithium Li forms hydroxide Li OH (alkali), which exhibits all the characteristic properties of bases.
3. Write the equations for the reactions of lithium oxide and hydroxide with sulfuric acid in molecular and ionic form.
2 LiOH+H2SO4=2H2O+ Li2SO4
2OH-+2H+=2H2O
Li2O+H2SO4=H2O+Li2SO4
Li2O+2H+=H2O+2Li +
Option 4
1. The equation for the reaction of zinc with hydrochloric acid is given:
Zn + 2HCl = ZnCl2 + H2 + Q.
Describe the following responses:
the number and composition of the starting materials and reaction products;
thermal effect;
aggregate state of the substances involved in the reaction;
participation of a catalyst;
change in the oxidation states of chemical elements;
direction.
Consider this chemical reaction from the point of view of the theory of electrolytic dissociation: write down the full and reduced ionic equations.
2HCl+Zn=ZnCl2+H2 + Q
Starting materials: 1 mol zinc, 2 mol hydrochloric acid (1 hydrogen atom, 1 chlorine atom per molecule). Reaction products: 1 mol of zinc chloride (1 zinc atom, 2 chlorine atoms in FE), 1 mol of hydrogen (2 hydrogen atoms).
exothermic reaction
Zinc - TV., Hydrochloric acid - well., Zinc chloride TV. (solution), hydrogen - g.
without catalyst
With a change in oxidation states
irreversible
2H++2Cl-+Zn0=Zn2++2Cl-+H20
2H++Zn0=Zn2++H20
2. Give a description of the chemical element calcium according to the plan:
the position of the element in the Periodic system;
the structure of the atom;
formulas of higher oxide and hydroxide, their character.
Calcium Ca
Ordinal number Z=20; mass number A \u003d 40, nuclear charge + 20, number of protons \u003d 20, neutrons (N \u003d A-Z \u003d 20) 40 - 20 \u003d 20 neutrons, electrons \u003d 20, period - 4, energy levels - 4,
The structure of the electron shell: 20 M g 2e; 8e; 8e; 2e.
20 Ca))))
2 8 8 2
+2 oxidation state;
The reducing properties of calcium are more pronounced than those of magnesium, but weaker than those of strontium, which is associated with an increase in the atomic radii
Calcium ion Ca 2+
CaO - calcium oxide is the main oxide and exhibits all the characteristic properties of oxides. Calcium forms hydroxide Ca (OH) 2, which exhibits all the characteristic properties of bases.
3. Write the equations for the reactions of calcium oxide and hydroxide with nitric acid in molecular and ionic form.
CaO + 2HNO3 \u003d Ca (NO3) ₂ + H ₂ O
CaO + 2H + \u003d Ca 2+ + H₂O
Ca(OH)2+2HNO3= Ca(NO3)₂ + 2H₂O
Ca (OH) 2 + 2H + \u003d Ca 2+ + 2H₂O
Second level
Option 1
1. The reaction equation for the production of nitric oxide (II) is given:
N2 + 02 2NO - Q.
N20 + O20 2N+2O-2 - Q
N20 - 2 * 2e \u003d 2N + 2 reducing agent
O20 + 2 * 2e \u003d 2O-2 oxidizing agent
Starting materials: nitrogen 1 mol, 2 N atoms, oxygen 1 mol (2 O atoms).
Reaction product: 2 mol of nitric oxide 2 (in the molecule 1 nitrogen atom and 1 oxygen atom).
The starting materials and reaction products are gases.
The reaction is endothermic.
Reversible.
Without catalyst.
With a change in oxidation states.
6 C))
2 4
+4 oxidation state;
3. Make formulas for higher carbon monoxide and hydroxide, indicate their nature.
CO2 + H2O ↔ H2CO3
CO2 + H2O ↔ 2H+ + CO32-
Na2O + CO2 → Na2CO3
Na2O + CO2 → 2Na+ + CO32-
2NaOH + CO2 → Na2CO3 + H2O
OH- + CO2 → CO32- + H2O
Ca(OH)2 + CO2 → CaCO3 ↓+ H2O
H2CO3 + Ca = CaCO3 + H2
2H+ +CO32- + Ca = CaCO3 ↓+ H2
H2CO3 + CaO = CaCO3 ↓+ H2O
H2CO3 + 2NaOH = Na2CO3 + 2H2O
2H+ +OH- = 2H2O
Option 2
1. The reaction equation for the synthesis of ammonia is given:
N2 + 3H2 2NH3 + Q.
Give a description of the reaction according to all the classification features you have studied.
Consider this reaction in terms of OVR. Specify the oxidizing agent and reducing agent.
3H2 + N2 2NH3 + Q
N20 +2*3е→2N-3 oxidant
H20 -2*1e→2H+1 reducing agent
Starting substances: 1 mol of nitrogen (a molecule of 2 nitrogen atoms), 3 mol of hydrogen (a molecule of 2 hydrogen atoms). The reaction product is ammonia, 2 mol. Molecule of 1 nitrogen atom and 2 hydrogen atoms. The starting materials are the products of the reaction - gases.
Reaction:
exothermic.
Redox.
Straight.
catalytic.
Reversible.
2. Give a description of the chemical element sulfur according to its position in the Periodic system.
Sulfur - S
Serial number Z=16 and mass number A= 32, nuclear charge + 16, number of protons = 16, neutrons (N= A-Z= 12) 32 - 16=16 neutrons, electrons = 16, period - 3, energy levels - 3
16S)))
The structure of the electron shell: 16 S 2e; 8e; 6e.
16S)))
2 8 6
Oxidation state - (-2) and (+ 2; +4; +6)
The oxidizing properties of sulfur are more pronounced than those of selenium, but weaker than those of oxygen, which is associated with an increase in the atomic radii from oxygen to selenium
SO 3 - sulfur oxide is an acidic oxide and exhibits all the characteristic properties of oxides.
Sulfur forms hydroxide H2SO4, which exhibits all the characteristic properties of acids.
Sulfur from hydrogen compounds forms H2S.
3. Make formulas for higher oxide and sulfur hydroxide, indicate their nature. Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
SO3 + H2O → H2SO4
2NaOH + SO3 → Na2SO4 + H2O
2OH- + SO3 → SO42- + H2O
Na2O + SO3 → Na2SO4
Na2O + SO3 → 2Na+ +SO42-
Zn0 + H2+1SO4(razb) → Zn+2SO4 + H20
Zn0 + 2H+ → Zn2+ + H20
CuO + H2SO4 → CuSO4 + H2O
CuO + 2H+ → Cu2+ + H2O
H2SO4 + 2NaOH → Na2SO4 + 2H2O (neutralization reaction)
H+ + OH- → H2O
H2SO4 + Cu(OH)2 → CuSO4 + 2H2O
2H+ + Cu(OH)2 → Cu2+ + 2H2O
BaCl2 + H2SO4 → BaSO4↓ + 2HCl
Ba2+ + SO42- → BaSO4↓
MgCO3 + H2SO4 → MgSO4 + H2O + CO2
MgCO3 + 2H+ → Mg2+ + H2O + CO2¬
Option 3
1. The equation for the reaction of copper (II) chloride with sodium hydroxide is given:
CuCl2 + 2NaOH = Cu(OH)2↓ + 2NaCl.
Give a description of the reaction according to all the classification features you have studied.
Consider the reaction from the point of view of TED: write down the full and reduced ionic equations.
CuCl2 + 2NaOH = Cu(OH)2↓ + 2NaCl
Cu2+ + 2OH- = Cu(OH)2↓
Starting materials: 1 mol of copper chloride (1 copper atom, 2 chlorine atoms), 2 mol of sodium hydroxide (1 sodium atom, 1 oxygen atom, 1 hydrogen atom in FE).
Reaction products: 1 mol of copper hydroxide (1 copper atom, 2 oxygen atoms, 2 hydrogen atoms), 2 mol sodium chloride (1 sodium atom, 1 chlorine atom in FE).
The reaction products and starting materials are solid dissolved. Cu(OH)2 is a solid precipitate.
Reaction:
exothermic
No change in oxidation states
Straight
Without the participation of a catalyst
Irreversible.
2. Give a description of the chemical element phosphorus according to its position in the Periodic system of D. I. Mendeleev.
Characteristic P (phosphorus)
Atomic mass \u003d 31. The charge of the nucleus of the atom is P + 15, t. There are 15 protons in the nucleus. Scheme:
15R 2e) 8e) 5e)
3. Make formulas for higher oxide and phosphorus hydroxide, indicate their nature. Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
P2O5 + 3H2O = 2H3PO4
P2O5 + 3H2O = 6H+ +2PO43-
3CaO + P2O5 = Ca3(PO4)2
6H++ 3CO3 2-= 3H2O + 3CO2
3NaOH + H3PO4 = Na3PO4 + 3H2O
3OH- + 3H+= 3H2O
Option 4
1. The equation for the reaction of potassium carbonate with hydrochloric acid is given:
K2CO3 + 2HCl = 2KCl + CO2 + H20.
Give a description of the reaction according to all the classification features you have studied.
Consider this reaction from the point of view of TED: write down the full and reduced ionic equations.
K2CO3 + 2HCl = 2KCl + H2O + CO2
2K+ +CO32- + 2H+ + 2Cl-= 2K+ 2Cl-+ H2O + CO2
CO32- + 2H+= H2O + CO2
Starting materials: 1 mol of potassium carbonate (2 potassium atoms, 1 carbon atom, 3 oxygen atoms) solid, 2 mol of hydrochloric acid (1 hydrogen atom, 1 chlorine atom in a molecule) liquid.
Reaction products: 2 mol of potassium chloride (in FE 1 potassium atom, 1 chlorine atom) solid, 1 mol of water (2 volumes of hydrogen, 1 oxygen atom) liquid, 1 mol of carbon dioxide (1 carbon atom, 2 oxygen atoms) - gas.
Reaction:
exothermic.
No change in oxidation states.
Straight.
Without the participation of a catalyst.
Irreversible.
2. Give a description of the chemical element nitrogen according to its position in the Periodic system.
Nitrogen N - non-metal, period II (small), group V, main subgroup.
Atomic mass = 14, nuclear charge - +7, number of energy levels = 2
p=7, e=7,n=Ar-p=14-7=7.
The structure of the electron shell: 7 N 2e; 5e
7 N))
2 5
+5 oxidation state;
The oxidizing properties are more pronounced than those of carbon, but weaker than those of oxygen, which is associated with an increase in the charge of the nucleus.
N2O5 nitric oxide is an acidic oxide and exhibits all the characteristic properties of oxides. Nitrogen forms the acid HNO3, which exhibits all the characteristic properties of acids.
Volatile hydrogen compound - NH3
3. Make the formulas of the higher nitrogen oxide and hydroxide, indicate their nature.
Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
N2O5 + H2O = 2НNO3
N2O5 + H2O = 2H+ + NO3-
N2O5 + BaO = Ba(NO3)2
N2O5 + BaO = Ba2+ +2NO3-
N2O5 + 2KOH (solution) = 2KNO3 + H2O
N2O5 + 2K+ +2OH- = 2K+ +NO32- + H2O
N2O5 + 2OH- = NO32- + H2O
K2O + 2HNO3 → 2KNO3 + H2O
K2O + 2H+ + 2NO3- → 2K+ + 2NO3- + H2O
K2O + 2H+ → 2K+ + H2O
HNO3 + NaOH → NaNO3 + H2O
H+ + NO3- + Na+ + OH- → Na+ + NO3- + H2O
H+ + OH- → H2O
2HNO3 + Na2CO3 → 2NaNO3 + H2O + CO2¬
2H+ + 2NO3- + 2Na+ + СO32- → 2Na+ + 2NO3- + H2O + CO2¬
2H+ + CO32- → H2O + CO2¬
S0 + 6HNO3(conc) → H2S+6O4 + 6NO2 + 2H2O
B0 + 3HNO3 → H3B+3O3 + 3NO2
3P0 + 5HNO3 + 2H2O → 5NO + 3H3P+5O4
From razb.
4Zn + 9HNO3 = NH3 + 4Zn(NO3)2 + 3H2O
4Zn + 9H+ + 9NO3- = NH3 + 4Zn2+ + 8NO3- + 3H2O
3Cu + 8HNO3 = 2NO + 3Cu(NO3)2+ 4H2O
3Cu + 8H+ +8NO3-= 2NO + 3Cu2+ +6NO3-+ 4H2O
conc.
Zn + 4HNO3 = 2NO2 + 2H2O + Zn(NO3)2
Zn + 4H+ +4NO3-= 2NO2 + 2H2O + Zn2+ +2NO3-
Cu + 4HNO3 = 2NO2 + 2H2O + Cu(NO3)2
Cu + 4H+ +4NO3- = 2NO2 + 2H2O + Cu2+ +2NO3-
Third level
Option 1
1. The equation for the reaction of obtaining nitric acid is given:
4N02 + 02 + 2H20 = 4HN03 + Q.
Give a description of the reaction according to all the classification features you have studied.
4N+4O2 + О02 + 2H2O ↔ 4HN+5O-23
N+4 -1e = N+5 reducing agent
O20 +4e = 2O-2 oxidizer
Reaction:
exothermic.
With a change in the oxidation state (OVR).
Without the participation of a catalyst.
Straight.
Reversible.
Source substances: 4 mol of nitric oxide 4 (1 nitrogen atom, 2 oxygen atoms in a molecule) - gas, 1 mol of oxygen (2 oxygen atoms in a molecule) - gas, 2 mol of water (1 oxygen atom, 2 hydrogen atoms in a molecule) - liquid
The reaction product - 4 moles of nitric acid (1 nitrogen atom, 1 hydrogen atom, 3 oxygen atoms in a molecule) - is a liquid.
2. Give a description of the chemical element magnesium according to its position in the Periodic system.
Magnesium - serial number in the Periodic system Z = 12 and mass number A = 24. Nuclear charge +12 (number of protons). The number of neutrons in the nucleus N \u003d A - Z \u003d 12. The number of electrons \u003d 12.
The element magnesium is in the 3rd period of the Periodic Table. The structure of the electron shell:
12 Mg)))
2 8 2
+2 oxidation state.
The reduction properties of magnesium are more pronounced than those of beryllium, but weaker than those of calcium (group IIA elements), which is associated with an increase in the atomic radii upon transition from Be to Mg and Ca.
Magnesium oxide MgO is a basic oxide and exhibits all the typical properties of basic oxides. Magnesium hydroxide corresponds to the base Mg(OH)2, which exhibits all the characteristic properties of bases.
3. Make formulas for magnesium oxide and hydroxide, indicate their nature.
Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
Magnesium oxide MgO is the basic oxide, the base Mg(OH)2 exhibits all the characteristic properties of bases.
MgO + H2O = Mg(OH)2
MgO + CO2 = MgCO3
MgO + CO2 = Mg2+ + CO32-
MgO + H2SO4 = MgSO4 + H2O
MgO + 2H+ = Mg2+ + H2O
Mg(OH)2 + 2HCl = MgCl2 + 2H2O
Mg(OH)2 + 2H+ = Mg2+ + 2H2O
Mg(OH)2 + CO2 = Mg2+ +CO32- + H2O
3Mg(OH)2 + 2FeCl3 = 2Fe(OH)3 + 3MgCl2
3Mg(OH)2 + 2Fe3+ = 2Fe(OH)3 + 3Mg2+
Mg(OH)2 + 2NH4Cl = MgCl2 + 2NH3 + 2H2O
Mg(OH)2 + 2NH4+= Mg2+ + 2NH3 + 2H2O
MgSO4 + 2NaOH = Mg(OH)2 + Na2SO4
Mg2+ + 2OH- = Mg(OH)2
Option 2
1. The equation for the reaction of iron with chlorine is given:
2Fe + 3Cl2 = 2FeCl3 + Q.
Give a description chemical reaction according to all the classification features studied by you.
Consider the reaction in terms of redox processes. Specify the oxidizing agent and reducing agent.
2Fe + 3Cl2 = 2FeCl3 + Q
2
3 Fe - 3e - = Fe + III,
Cl2 + 2e– = 2Cl–I
2Fe – 6e– = 2Fe+III,
3Cl2 + 6e– = 6Cl–I.
Fe – 3e– = Fe+III reducing agent
Cl2 + 2e– = 2Cl–I oxidant
exothermic
OVR
Straight
irreversible
non-catalytic
Starting substances: 2 mol of iron - solid, 2 mol of chlorine (a molecule of 2 atoms) - gas
Product: 2 mol of iron chloride (from 1 iron atom, 2 chlorine atoms in FE) - tv.
2. Give a description of the chemical element sodium according to its position in the Periodic system of D. I. Mendeleev.
Sodium -- Na
Ordinal number Z=11; mass number A \u003d 23, nuclear charge + 11, number of protons \u003d 11, neutrons (N \u003d A-Z \u003d 11) 23 - 11 \u003d 12 neutrons, electrons \u003d 11, period - 3, energy levels - 3,
The structure of the electron shell: 11 Na 2е; 8e; 1e.
11 Na)))
2 8 1
+1 oxidation state;
The reducing properties of sodium are more pronounced than those of lithium, but weaker than those of potassium, which is associated with an increase in the atomic radii;
Sodium ion Na+
Na 2O - sodium oxide is the main oxide and exhibits all the characteristic properties of oxides. Sodium forms hydroxide NaOH (alkali), which exhibits all the characteristic properties of bases.
3. Make formulas for sodium oxide and hydroxide, indicate their nature. Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
2NaOH+H2SO4=2H2O+Na2SO4
2OH-+2H+=2H2O
2NaOH + CO2 ---> Na2CO3 + H2O
2OH(-) + CO2 ---> CO3(2-) + H2O
2NaOH + SO2 ---> Na2SO3 + H2O
2OH(-) + SO2 ---> SO3(2-) + H2O
NaOH+ Al(OH)3 ---> Na
OH(-) + Al(OH)3 ---> Al(OH)4 (-)
Na2O+H2SO4=H2O+Na2SO4
Na2O+2H+=H2O+2Na+
Na2O + H2O ---> 2NaOH
Na2O + H2O ---> 2Na+ +2OH-
Na2O + 2HCl ----> 2NaCl + H2O
Na2O + 2H+ ----> 2Na+ + H2O
Na2O + CO2 ---> Na2CO3
Na2O + CO2 ---> 2Na++CO32-
Na2O + SO2 ---> Na2SO3
Na2O + SO2 ---> 2Na++SO32-
Option 3
1. The reaction equation for the decomposition of potassium nitrate is given:
2KN03 = 2KN02 + O2 - Q.
Give a description of the reaction according to all the classification features you have studied.
Consider the reaction in terms of redox processes. Specify the oxidizing agent and reducing agent.
2KNO3 = 2KNO2 + O2- Q
oxidizing agent: N5+ + 2e− = N=3+|2| recovery
reducing agent: O2− − 4e− = O20 |1| oxidation
Starting substances: 2 mol of potassium nitrate (in FE 1 potassium atom, 1 nitrogen atom, 3 oxygen atoms) - TV.
The reaction products - 2 mol of potassium nitrite (in FE 1 potassium atom, 1 nitrogen atom, 2 oxygen atoms) - solid, 1 mol of oxygen (2 oxygen atoms) - gas.
Endothermic
OVR
Straight
irreversible
non-catalytic
2. Give a description of the chemical element carbon by its position in the Periodic system.
Carbon C is a chemical element of group IV of the periodic system of Mendeleev: atomic number 6, atomic mass 12.011.
Ordinal number Z=6; mass number A \u003d 12, nuclear charge + 6 number of protons \u003d 6, neutrons (N \u003d A-Z \u003d 6) 12 - 6 \u003d 6 neutrons, electrons \u003d 6, period - 2, energy levels - 2,
The structure of the electron shell: 6 C 2e; 4th
6 C))
2 4
+4 oxidation state;
The oxidizing properties of carbon are more pronounced than those of boron, but weaker than those of nitrogen, which is associated with an increase in the nuclear charge.
CO2 acid oxide, H2CO3 acid.
3. Make formulas for carbon monoxide and hydroxide, indicate their nature.
Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
CO2 carbon monoxide is an acidic oxide and exhibits all the characteristic properties of oxides. Carbon forms the acid H2CO3, which exhibits all the characteristic properties of acids.
CO2 + H2O ↔ H2CO3
CO2 + H2O ↔ 2H+ + CO32-
Na2O + CO2 → Na2CO3
Na2O + CO2 → 2Na+ + CO32-
2NaOH + CO2 → Na2CO3 + H2O
OH- + CO2 → CO32- + H2O
Ca(OH)2 + CO2 → CaCO3 ↓+ H2O
Ca2+ +2OH- + CO2 → CaCO3 ↓+ H2O
H2CO3 + Ca = CaCO3 + H2
2H+ +CO32- + Ca = CaCO3 ↓+ H2
H2CO3 + CaO = CaCO3 ↓+ H2O
2H+ +CO32- + CaO = CaCO3 ↓+ H2O
H2CO3 + 2NaOH = Na2CO3 + 2H2O
2H+ + CO32- + 2Na+ +OH- = 2Na++CO32- + 2H2O
2H+ +OH- = 2H2O
Ca(OH)2 + H2CO3 → CaCO3 ↓+ 2H2O
Ca2+ +2OH- + 2H+ +CO32- → CaCO3 ↓+ 2H2O
Option 4
1. The reaction equation for the formation of iron hydroxide (III) is given:
4Fe(OH)2 + 2H20 + 02 = 4Fe(OH)3.
Give a description of the reaction according to all the classification features you have studied.
Consider the reaction in terms of redox processes. Specify the oxidizing agent and reducing agent.
4Fe(OH)2 + O2 + 2H2O = 4Fe(OH)3↓
Fe2+ -1е→ Fe3+ reducing agent
O20 + 4е → 2O2- oxidant
Initial substances: 4 mol of iron hydroxide 2 (in FE 1 iron atom, 2 oxygen atoms, 2 hydrogen atoms) - solid, 1 mol of oxygen (2 oxygen atoms) - gas, 2 mol of water (2 hydrogen atoms, 1 oxygen atom in molecule) - f.
The reaction product is 4 mol of iron hydroxide 3 (in FE 1 iron atom, 3 oxygen atoms, 3 hydrogen atoms) - TV.
exothermic
OVR
Straight
irreversible
Non-catalytic.
2. Give a description of the chemical element phosphorus according to its position in the Periodic system.
Characteristic P (phosphorus)
The element with serial number 15 is in the 3rd period of the 5th group, the main subgroup.
Atomic mass \u003d 31. The charge of the nucleus of the atom is P + 15, t. There are 15 protons in the nucleus.
Scheme 15P 2e) 8e) 5e)
There are 16 neutrons in the nucleus of an atom. There are 15 electrons in an atom, since their number is equal to the number of protons and the serial number. There are 3 electron layers in the phosphorus atom, since P is in the 3rd period. There are 5 electrons on the last layer, since phosphorus is in group 5. The last layer is not completed. P-non-metal, because in the chemical. reactions with metals takes 3 electrons to complete the layer. Its oxide is Р2О5-acid. He is mutual. with H2O, bases and basic oxides. Its hydroxide is H3PO4-acid. She interacts. with metals standing up to H (hydrogen), with basic oxides, bases.
3. Make formulas for phosphorus oxide and hydroxide, indicate their nature.
Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
P2O5 + 3H2O = 2H3PO4
P2O5 + 3H2O = 6H+ +2PO43-
3CaO + P2O5 = Ca3(PO4)2
3Ca(OH)2 + P2O5 = Ca3(PO4)2 + 3H2O.
3Mg + 2H3PO4 = Mg3(PO4)2↓ + 3H2
3Mg + 6H++ 2PO43- = Mg3(PO4)2↓ + 3H2
2H3PO4+3Na2CO3 = 2Na3PO4 + 3H2O + 3CO2
6H++ 3CO3 2-= 3H2O + 3CO2
3NaOH + H3PO4 = Na3PO4 + 3H2O
3OH- + 3H+= 3H2O
The chemical element Lithium gained fame thanks to the discovery of Johann August Arfvedson in 1817 as part of an aluminosilicate, petalite. Then "flammable alkali" was found in other minerals of natural origin. It is a white metal with a silvery sheen that can be cut with a knife. In the periodic table, it occupies the third place and is designated Li (from the Latin Lithium).
Brief description of the chemical element Lithium
Ordinal (atomic) number of the element in periodic system chemical elements of Mendeleev is equal to three. Under normal conditions, metallic Li has the lowest density of all known metals. In addition, it leads the alkali metal family in terms of melting and boiling points.
Historical facts
The first metal sample was obtained by Sir Humphrey Davy in the process of decomposition of lithium hydroxide melt by electric current. Together with the first result of the electrolysis of lithium, Leopold Gmelin, experimenting with lithium-containing salts, noted the coloring of the flame in a dark carmine color.
Chemical properties of lithium
Lithium exhibits "capricious" properties when mixed with sodium, does not react at all with melts of potassium, rubidium and cesium. At room temperature, lithium does not interact with dry air or hydrogen. Unlike other alkali metals, it cannot be stored in kerosene. For this purpose, Sherwood oil, paraffin, gas gasoline or mineral oil are used in sealed tin containers.
At temperatures above 100, but below 300 degrees Celsius, a protective oxide film is formed on the surface of lithium, which prevents further interaction of the chemical. Element with environment even in humid air. The metal form of the element burns when it comes into contact with a wet surface of the skin or mucous membrane.
Application of lithium
The element itself and its compounds are widely used for the production of glass, as a coating for porcelain. Ferrous and non-ferrous metallurgy use lithium to impart strength and ductility to alloys, in the manufacture of lubricants. The textile industry uses this element as a fabric bleach, the food industry as a preservative, and the pharmaceutical industry successfully uses it in cosmetic preparations.
Liquid lithium has found its application in nuclear reactors, radioactive tritium is obtained using the isotope of lithium-6. The alkali metal has found wide application in the chemical industry as a catalyst for many processes, a component of alloys from which cold cathodes are made, as well as anodes of current sources.
Lithium fluoride in the form of single crystals is used to create high-precision lasers with an efficiency of 80%. Various compounds with lithium are used in flaw detection, pyrotechnics, radio electronics, and optoelectronics.
Lithium salts are a psychotropic substance, the positive effect of which on the mental state of a person was confirmed only in the middle of the 20th century. Lithium carbonate has been successfully used to treat people with bipolar disorder, manic depression, and suicidal tendencies.
This explains the low crime rate in areas where lithium is found to a large extent in drinking water. The mechanism of action of the element is still poorly understood, but there are suggestions that a positive effect is achieved by the regulatory function of the activity of some of the enzymes involved in the transfer of sodium and potassium ions to the brain. The balance of Na and K is directly responsible for the state of the psyche. So it has been proven that in people prone to depression, there is an excess content of sodium in the cells, and lithium evens out the ion picture.
The property of lithium to reduce depression and the risk of suicide is reflected in the work of Nirvana and Evanescence. Their discography includes psychedelic songs called Lithium.
The role of lithium in activating dormant bone marrow cells is the hope of modern medicine in the fight against blood cancer. It has been experimentally proven that lithium has a beneficial effect on areas affected by genital herpes. The use of Li in the complex treatment of hypertension and diabetes was positively noted. Undoubtedly effective in the prevention of sclerosis and diseases of the cardiovascular system.
Being present in lubricants, lithium makes it possible to explore Antarctica at critically low temperatures. Without this element, the technique will simply fail. It is considered as a component of solid rocket fuel, because the result of burning 1 kg of solid Li is more than ten thousand kilocalories, which is almost five times more than the result of burning 1 kg of kerosene.
4Li + O 2 = 2Li 2 O (1);
2Na + O 2 = Na 2 O 2 (2).
Find the total amount of oxygen substance:
n(O 2) \u003d V (O 2) / V m;
n(O 2) \u003d 3.92 / 22.4 \u003d 0.175 mol.
Let x mol of oxygen be used for the oxidation of lithium, then (0.175 - x) mol of oxygen participated in the oxidation of sodium.
Let's denote the amount of lithium substance as "a", and sodium - "b", then, according to the reaction equations written above:
b \u003d 2 × (0.175 - x) \u003d 0.35 - 2x.
Let us find the masses of lithium and sodium (the values of the relative atomic masses taken from the Periodic Table of D.I. Mendeleev, rounded up to integers - Ar(Li) = 7 a.m.u.; Ar(Na) = 23 amu):
m(Li) = 4x × 7 = 28x (r);
m (Na) \u003d (0.35 - 2x) × 23 \u003d 8.05 - 46x (g).
Considering that the mass of the mixture of lithium and sodium was 7.6 g, we can write the equation:
28x + (8.05 - 46x) = 7.6;
(-18) × x \u003d - (0.45);
Consequently, the amount of oxygen substance consumed for the oxidation of lithium is 0.025 mol, and sodium - (0.175 - 0.025) = 0.15 mol.
According to equation (1) n(O 2) :n(Li 2 O) = 1: 2, i.e.
n (Li 2 O) \u003d 2 × n (O 2) \u003d 2 × 0.025 \u003d 0.05 mol.
According to equation (2) n(O 2) :n(Na 2 O 2) = 1: 1, i.e. n (Na 2 O 2) \u003d n (O 2) \u003d 0.15 mol.
Let us write the reaction equations for the dissolution of lithium and sodium oxidation products in sulfuric acid:
Li 2 O + H 2 SO 4 = Li 2 SO 4 + H 2 O (3);
2Na 2 O 2 + 2H 2 SO 4 \u003d 2Na 2 SO 4 + 2H 2 O + O 2 (4).
Calculate the mass of sulfuric acid in solution:
m solute (H 2 SO 4) = m solution (H 2 SO 4) ×w(H 2 SO 4) / 100%;
m solute (H 2 SO 4) \u003d 80 × 24.5 / 100% \u003d 19.6 g.
The amount of sulfuric acid substance will be equal to (molar mass - 98 g / mol):
n (H 2 SO 4) \u003d m (H 2 SO 4) / M (H 2 SO 4);
n (H 2 SO 4) \u003d 19.6 / 98 \u003d 0.2 mol.
Let us determine the number of moles of reaction products (3) and (4). According to equation (3) n(Li 2 O) :n(Li 2 SO 4) = 1: 1, i.e. n (Li 2 O) \u003d n (Li 2 SO 4) \u003d 0.05 mol. According to equation (4) n(Na 2 O 2) :n(Na 2 SO 4) = 2: 2, i.e. n (Na 2 O 2) \u003d n (Na 2 SO 4) \u003d 0.15 mol.
Let's find the masses of sulfates formed (M (Li 2 SO 4) \u003d 110 g / mol; M (Na 2 SO 4) \u003d 142 g / mol):
m (Li 2 SO 4) \u003d 0.05 × 110 \u003d 5.5 (g);
m (Na 2 SO 4) \u003d 0.15 × 142 \u003d 21.03 (g).
To calculate the mass fractions of the obtained substances, it is necessary to find the mass of the solution. It includes sulfuric acid, lithium oxide and sodium peroxide. It is necessary to take into account the mass of oxygen that is released from the reaction mixture during reaction (4). Let's determine the masses of lithium oxide and sodium peroxide (M (Li 2 O) \u003d 30 g / mol, M (Na 2 O 2) \u003d 78 g / mol):
m(Li 2 O) = 0.05 x 30 = 1.5 (g);
m (Na 2 O 2) \u003d 0.15 × 78 \u003d 11.7 (g).
According to equation (4) n(O 2) :n(Na 2 O 2) = 1: 2, i.e.
n(O 2) \u003d ½ × n (Na 2 O 2) \u003d ½ × 0.15 \u003d 0.075 mol.
Then the mass of oxygen will be equal to (M (O 2) \u003d 32 g / mol):
m (O 2) \u003d 0.075 × 32 \u003d 2.4 (g).
In order to find the mass of the final solution, it is necessary to determine whether sulfuric acid remained in the solution. According to equation (3) n(Li 2 O): n(H 2 SO 4) = 1: 1, i.e. n (H 2 SO 4) \u003d n (Li 2 O) \u003d 0.05 mol. According to equation (4) n(Na 2 O 2) :n(H 2 SO 4) = 2: 2, i.e. n (H 2 SO 4) \u003d n (Na 2 O 2) \u003d 0.15 mol. Thus, (0.05 + 0.15) \u003d 0.2 mol of sulfuric acid entered into the reaction, i.e. she reacted completely.
Calculate the mass of the solution:
m solution = m(Li 2 SO 4) + m(Na 2 SO 4) - m(O 2);
m solution = 5.5 + 21.03 - 2.4 = 24.13 g.
Then, the mass fractions of sodium and lithium sulfates in the solution will be equal:
w(Li 2 SO 4) = m(Li 2 SO 4) /m solution × 100%;
w(Li 2 SO 4) = 5.5/24.13×100% = 22.79%.
w(Na 2 SO 4) = m(Na 2 SO 4) / m solution × 100%;
w(Na 2 SO 4) = 21.03 / 24.13 × 100% = 87.15%.
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