Drawing up and solving chemical equations. Drawing up chemical formulas of salts
Lesson in 8th grade
Subject: " Compilation chemical formulas by valence."
Goals:
consolidate the ability to determine valency using the formulas of compounds;
introduce the concept of “binary compounds”;
teach how to make names of binary compounds using their formulas;
teach how to compose formulas for compounds based on the valency of elements.
(You will learn :
what substances are called binary;
how to correctly formulate the name of a binary compound;
how valence is used to clarify the names of substances;
how to make up their formulas based on the names of binary compounds.
Remember :
what is valence;
how to determine valency, knowing the formula of a substance.)
During the classes.
Organizing time. Checking homework.
What is the valence of chemical elements?
Why is the valence of hydrogen taken as unity?
Which chemical elements have constant valence?
What chemical elements have variable valence?
New topic.
In the last lesson we learned how to determine the valency of chemical elements using the formulas of substances. Determine the valency of the elements in these compounds.
(independently, then check with the whole class)
Na 2 O SO 3 Fe 2 O 3 Ag 2 OCaH 2 H 2 S
In all these compounds we knew the valence of one element. What if there is no chemical element with a known valence? PSHE will come to the rescue (8 groups, metals and non-metals).
Rules for determining valence:
The valence of metals in group A is equal to the group number.
Nonmetals exhibit two valences: the maximum, equal to the group number, and the minimum, equal to 8 - the group number.
Let's look again at the series of connections written on the board. What do these connections have in common?
(complex substances; consist of two chemical elements)
Compounds formed by atoms of two chemical elements are calledbinary . Give another example of a binary compound that you encounter every day (water ).
Now we will learn how to give names to binary compounds. In chemistry, for naming substances and drawing up formulas, special rules, which are called nomenclature. Only for a small number of substances are the so-called trivial names (i.e., historically established) retained. We will become familiar with the rules of chemical nomenclature gradually as we become familiar with the classification of substances.
Compilation of names of binary compounds (Appendix 1):
We name the chemical element whose sign in the formula is in second place. We use its Latin name. Select the root and add the suffix – id.
Presentation, slide 2.
Give names to the substances shown on the board.(Together).
Let's create the nomenclature names for carbon dioxide and carbon monoxide:
carbon dioxide - CO 2 – carbon monoxide;
carbon monoxide – CO – carbon monoxide.
It turned out that different substances have the same names. But this cannot happen. What do we do?
Valence will help here. Determine the valence of carbon in these compounds. Write down: carbon monoxide (IV), carbon monoxide (II).
Knowing the valency of elements, we can create formulas for substances. Let's create the formula for nitric oxide (V). To do this, you need to perform the following steps (Appendix 2, presentation, slide 3):
Find NOC.
Divide the NOC by the valency of the elements.
Presentation, slide 4.
Using the algorithm, create the formula for aluminum oxide.
Lesson summary.
Determine the valence of chromium atoms in the compounds:
CrO 3
CrO
Cr 2 O 3
Give them names.
Check: presentation, slide 6.
§12, questions 4-7 p. 37 (written), task 2 p. 37.
Appendix 1. Compilation of names of binary compounds:
We name the chemical element whose sign in the formula is in second place. We use its Latin name. Select the root and add the suffix –id to it.
We name the chemical element whose sign comes first in the formula of the substance. We use the Russian name in the genitive case.
CaO – ok eid calcium
NaCl – chlorine eid sodium
PbS – sulf eid lead
Chemical element sign
Latin name
Sa
calcium
ok igenium
oxygen
sodium
chlorine mind
chlorine
lead
sulf ur
sulfur
Appendix 2. Drawing up chemical formulas of binary compounds by their names.
Nitric oxide ( V )
Write down the signs of chemical elements.N O
V II
Indicate the valence of elements.N O
10
Find NOC.
Divide the NOC by the valency of the elements. [N] 10: V=2 [O] 10: II= 5
Place indexes (bottom right).N 2 O 5
Lesson type. Combined.
Teaching methods. Partially searchable.
Goals. Didactic: consolidate the concept of “valency”, skills in determining valency using the formula and the Periodic Table.
Psychological: arouse interest in the subject, develop the ability to reason logically, and correctly express one’s thoughts.
Educational: develop the ability to work collectively, evaluate the answers of your comrades.
Equipment. Kits for building models of molecules of various substances, anagram tablets for chemical warm-up, efficiency target
DURING THE CLASSES
1. Indicative-motivational stage
Chemical warm-up
Anagrams are words in which the order of the letters has been changed. Try to solve some of the chemical anagrams. Rearrange the letters in each word and get the name of the chemical element. Pay attention to the hint.
“Odovrod” – this element has the smallest relative atomic mass.
“Mailinu” – this element is called “winged” metal.
“Dikosolr” is part of the air.
“Tsalkiy” - without it our bones would be weak and fragile.
“Ozegel” - this element is part of the blood and is involved in the transfer of oxygen.
Teacher. If you can easily guess the anagram words, tell yourself: “I’m great!”
2. Updating knowledge
Catch a mistake (Guys look for a mistake, work in pairs, argue, confer. Having come to an opinion, they offer their own reasoned answer)
The word “valency” (from the Latin “valentia”) arose in the middle of the 19th century, during the period of completion of the chemical-analytical stage of the development of chemistry. “Valence is the ability of atoms of one element to attach a certain number of atoms of another element.” One atom of another monovalent element (HF, NaCl) is combined with one atom of a monovalent element. Combine with an atom of a divalent element one atom of monovalent (H 2 O) or one divalent atom (CaO). This means that the valence of an element can be represented as a number that shows how many atoms of a monovalent element an atom of a given element can combine with.
There are elements that have constant valency:
monovalent (I) - H, Li, Na, Rb, Cs, F, I
divalent (II) - Be, Mg, Ca, Sr, Ba, Zn, Cd K
trivalent (III) - B, Al, O
Tic Tac Toe: (Connect the elements with a straight line, the criterion for the correct answer is the constant valency of the selected elements)
1 option
Option 2
3. Learning new knowledge
Task 1: the general formula for combining hydrogen with any element is given
Knowing that the valency of hydrogen is I, determine the valence of the element.
The guys work in pairs, if necessary, they unite in groups of four, argue, and confer. Having come to an opinion, they offer their own reasoned answer. As a result, we get scheme No. 1
scheme 1
Assignment for consolidation:
- determine the valences of elements in compounds with hydrogen: PH 3, HF, H 2 S, CaH 2,
- name the connections.
Task 2: in the same way, you can determine the valence of elements in compounds with oxygen, knowing that oxygen is divalent. For example:
The guys work in pairs, if necessary, they unite in groups of four, argue, and confer. Having come to an opinion, they offer their own reasoned answer. As a result, we get scheme No. 2
scheme 2
Assignment for consolidation:
- Determine the valences of elements in compounds with oxygen:
- What are binary compounds containing oxygen called?
NO 2, N 2 O 5, SO 2, SO 3, Cl 2 O 7.
Task 3: what do you need to know in order to determine the valence of elements in a binary compound? (valency of one of the elements)
Determine the valence of atoms in a compound
The guys work in pairs, if necessary, they unite in groups of four, argue, and confer. Having come to an opinion, they offer their own reasoned answer. As a result, we get scheme No. 3
Teacher: which of the following diagrams
scheme 2
most fully reflects the rule for determining valency using the formula? (Scheme 3, because it reflects the general case, and Schemes 1 and 2 are only particular)
4. Consolidation of the studied material.
Independent work
The text of the work is written in advance on the board. Two students solve the problem on the back of the board, the rest in their notebooks.
Task 4. Check whether the formulas of the following compounds are written correctly: Na 2 S, KBr, Al 2 O 3, Mg 3 N 2, MgO.
5. Generalization and systematization of knowledge.
Creative work in groups
Task 5. Using kits for making models of molecules of various substances, create formulas and models of molecules for the following compounds:
1st group – copper and oxygen,
2nd group – zinc and chlorine,
3rd group – potassium and iodine,
4th group – magnesium and sulfur.
After finishing the work, one student from the group reports on the completed task and, together with the class, analyzes the errors.
Task 6. Write the formulas for compounds of metals with non-metals: calcium with oxygen, aluminum with chlorine, sodium with phosphorus. Name these connections.
After completing the work, students exchange notebooks and mutual checking takes place.
Task 7. Write down the procedure for drawing up formulas of substances, analyzing the proposed example
Procedure |
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6. Reflection
You have the opportunity to self-assess your activities in class. You are presented with an “Efficiency Target”.
Mark your knowledge on new topic, marking the corresponding sector in the figure with shading. Submit your notes.
7. Homework. According to the textbook “Chemistry-8” (UMK Kuznetsova N.E. and others) § 14, exercise 1-71 is mandatory (additional from 1-72 to 1-74).
Reports about the French scientist J.L. Proust and English scientist J. Dalton.
Literature
- Kuznetsova N.E. and others. Chemistry: Textbook for 8th grade students of general education institutions. - M.: Ventana-Graf, 2010. - 320 pp.: ill.
- Kuznetsova N.E., Shatalov M.A. Teaching chemistry based on interdisciplinary integration: grades 8-9: Educational manual. - M.: Ventana-Graf, 2004. - 352 p.
- Emelyanova E.O., Iodko A.G. Organization of cognitive activity of students in chemistry lessons in grades 8-9. Supporting notes With practical tasks, tests: In 2 parts.
- Part I. – M.: School press, 2002.- 144 p. Kuznetsova L.M. New technology
teaching chemistry in grade 8. - Obninsk: Title, 1999. - 208 p.: ill.
.
Chemistry is the science of substances, their properties and transformations
That is, if nothing happens to the substances around us, then this does not apply to chemistry. But what does “nothing happens” mean? If a thunderstorm suddenly caught us in the field, and we were all wet, as they say, “to the skin,” then isn’t this a transformation: after all, the clothes were dry, but they became wet. If, for example, you take an iron nail, file it, and then assembleiron filings () Fe , then isn’t this also a transformation: there was a nail - it became powder. But if you then assemble the device and carry out obtaining oxygen (O 2) : heat uppotassium permanganate(KMpO 4)
and collect oxygen in a test tube, and then place these red-hot iron filings into it, then they will flare up with a bright flame and after combustion will turn into a brown powder. And this is also a transformation. So where is the chemistry? Despite the fact that in these examples the shape (iron nail) and the condition of the clothing (dry, wet) change, these are not transformations. The fact is that the nail itself was a substance (iron), and remained so, despite its different shape, and our clothes absorbed the water from the rain and then evaporated it into the atmosphere. The water itself has not changed. So what are transformations from a chemical point of view? From a chemical point of view, transformations are those phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It doesn’t matter what shape it took after filing, but after collecting from it iron filings placed in an oxygen atmosphere - it turned into(iron filings ( 2 O 3 ) . So, something has changed after all? Yes, it has changed. There was a substance called a nail, but under the influence of oxygen a new substance was formed - element oxide gland. Molecular equation This transformation can be represented by the following chemical symbols:
4Fe + 3O 2 = 2Fe 2 O 3 (1)
For someone uninitiated in chemistry, questions immediately arise. What is "molecular equation", what is Fe? Why are the numbers “4”, “3”, “2”? What are the little numbers “2” and “3” in the formula Fe 2 O 3? This means it’s time to sort everything out in order.
Signs of chemical elements.
Despite the fact that chemistry begins to be studied in the 8th grade, and some even earlier, many people know the great Russian chemist D.I. Mendeleev. And of course, his famous “Periodic Table of Chemical Elements”. Otherwise, more simply, it is called the “Periodical Table”.
In this table, the elements are arranged in the appropriate order. To date, about 120 of them are known. The names of many elements have been known to us for a long time. These are: iron, aluminum, oxygen, carbon, gold, silicon. Previously, we used these words without thinking, identifying them with objects: an iron bolt, an aluminum wire, oxygen in the atmosphere, Golden ring etc. etc. But in fact, all these substances (bolt, wire, ring) consist of their corresponding elements. The whole paradox is that the element cannot be touched or picked up. How so? They are in the periodic table, but you can’t take them! Yes exactly. A chemical element is an abstract (that is, abstract) concept, and is used in chemistry, as well as in other sciences, for calculations, drawing up equations, and solving problems. Each element differs from the other in that it has its own characteristic electronic configuration of an atom. The number of protons in the nucleus of an atom is equal to the number of electrons in its orbitals. For example, hydrogen is element No. 1. Its atom consists of 1 proton and 1 electron. Helium is element number 2. Its atom consists of 2 protons and 2 electrons. Lithium is element #3. Its atom consists of 3 protons and 3 electrons. Darmstadtium – element No. 110. Its atom consists of 110 protons and 110 electrons.
Each element is indicated by a specific symbol, with Latin letters, and has a certain reading translated from Latin. For example, hydrogen has the symbol "N", read as "hydrogenium" or "ash". Silicon has the symbol "Si" read as "silicium". Mercury has a symbol "Hg" and is read as "hydrargyrum". And so on. All these notations can be found in any 8th grade chemistry textbook. The main thing for us now is to understand that when composing chemical equations, it is necessary to operate with the indicated symbols of the elements.
Simple and complex substances.
Denoting various substances with single symbols of chemical elements (Hg mercury, Fe iron, Cu copper, Zn zinc, Al aluminum) we essentially denote simple substances, that is, substances consisting of atoms of the same type (containing the same number of protons and neutrons in an atom). For example, if the substances iron and sulfur interact, then the equation will take the following writing form:
Fe + S = FeS (2)
Simple substances include metals (Ba, K, Na, Mg, Ag), as well as non-metals (S, P, Si, Cl 2, N 2, O 2, H 2). Moreover, one should pay attention
Special attention to the fact that all metals are designated by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals are either simple symbols: C, S, P or may have different indices that indicate their molecular structure: H 2, Cl 2, O 2, J 2, P 4, S 8. In the future this will have a very great importance when writing equations. It is not at all difficult to guess that complex substances are substances formed from atoms of different types, for example,
1). Oxides:
aluminium oxide Al 2 O 3, 
sodium oxide Na2O,
copper oxide CuO,
zinc oxide ZnO,
titanium oxide Ti2O3,
carbon monoxide or carbon monoxide (+2) CO,
sulfur oxide (+6) SO 3
2). Reasons:
iron hydroxide(+3) Fe(OH) 3,
copper hydroxide Cu(OH)2,
potassium hydroxide or alkali potassium KOH,
sodium hydroxide NaOH.
3). Acids:
hydrochloric acid HCl,
sulfurous acid H2SO3,
Nitric acid HNO3
4). Salts:
sodium thiosulfate Na 2 S 2 O 3 ,
sodium sulfate or Glauber's salt Na2SO4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl2
5). Organic matter:
sodium acetate CH 3 COONa,
methane CH 4,
acetylene C 2 H 2,
glucose C 6 H 12 O 6
Finally, after we have figured out the structure of various substances, we can begin to write chemical equations.
Chemical equation.
The word “equation” itself is derived from the word “equalize”, i.e. divide something into equal parts. In mathematics, equations constitute almost the very essence of this science. For example, you can give a simple equation in which the left and right sides will be equal to “2”:
40: (9 + 11) = (50 x 2) : (80 – 30);
And in chemical equations the same principle: the left and right sides of the equation must correspond to the same numbers of atoms and elements participating in them. Or, if an ionic equation is given, then in it number of particles must also meet this requirement. A chemical equation is a conventional representation of a chemical reaction using chemical formulas and mathematical symbols. A chemical equation inherently reflects one or another chemical reaction, that is, the process of interaction of substances, during which new substances arise. For example, it is necessary write a molecular equation reactions in which they take part barium chloride BaCl 2 and sulfuric acid H 2 SO 4. As a result of this reaction, an insoluble precipitate is formed - barium sulfate BaSO 4 and hydrochloric acid HCl:
BaCl 2 + H 2 SO 4 = BaSO 4 + 2HCl (3)
First of all, it is necessary to understand that the large number “2” standing in front of the substance HCl is called a coefficient, and the small numbers “2”, “4” under the formulas BaCl 2, H 2 SO 4, BaSO 4 are called indices. Both coefficients and indices in chemical equations act as multipliers, not summands. To write a chemical equation correctly, you need assign coefficients in the reaction equation. Now let's start counting the atoms of the elements on the left and right sides of the equation. On the left side of the equation: the substance BaCl 2 contains 1 barium atom (Ba), 2 chlorine atoms (Cl). In the substance H 2 SO 4: 2 hydrogen atoms (H), 1 sulfur atom (S) and 4 oxygen atoms (O). On the right side of the equation: in the BaSO 4 substance there is 1 barium atom (Ba), 1 sulfur atom (S) and 4 oxygen atoms (O), in the HCl substance: 1 hydrogen atom (H) and 1 chlorine atom (Cl). It follows that on the right side of the equation the number of hydrogen and chlorine atoms is half as much as on the left side. Therefore, before the HCl formula on the right side of the equation, it is necessary to put the coefficient “2”. If we now add up the numbers of atoms of the elements participating in this reaction, both on the left and on the right, we obtain the following balance:
In both sides of the equation, the numbers of atoms of the elements participating in the reaction are equal, therefore it is composed correctly.
Chemical equation and chemical reactions
As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are those phenomena during which the transformation of one substance into another occurs. Among their diversity, two main types can be distinguished:
1). Compound reactions
2). Decomposition reactions.
The overwhelming majority of chemical reactions belong to addition reactions, since changes in its composition can rarely occur with an individual substance if it is not exposed to external influences (dissolution, heating, exposure to light). Nothing characterizes a chemical phenomenon or reaction better than the changes that occur during the interaction of two or more substances. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, color changes, sediment formation, release of gaseous products, and noise.
For clarity, we present several equations reflecting the processes of compound reactions, during which we obtain sodium chloride(NaCl), zinc chloride(ZnCl2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)
Cl 2 + 2Nа = 2NaCl (4)
CuCl 2 + Zn = ZnCl 2 + Cu (5)
AgNO 3 + KCl = AgCl + 2KNO 3 (6)
3HCl + Al(OH) 3 = AlCl 3 + 3H 2 O (7)
Among the reactions of the compound, special mention should be made of the following: : substitution (5), exchange (6), and how special case exchange reactions - reaction neutralization (7).
Substitution reactions include those in which atoms of a simple substance replace atoms of one of the elements in a complex substance. In example (5), zinc atoms replace copper atoms from the CuCl 2 solution, while zinc passes into the soluble salt ZnCl 2, and copper is released from the solution in the metallic state.
Exchange reactions include those reactions in which two complex substances exchange their components. In case of reaction (6) soluble salts AgNO 3 and KCl, when both solutions are merged, form an insoluble precipitate of AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are added to the NO 3 anions, and silver cations Ag + are added to the Cl - anions.
A special, special case of exchange reactions is the neutralization reaction. Neutralization reactions include those reactions in which acids react with bases, resulting in the formation of salt and water. In example (7), hydrochloric acid HCl reacts with the base Al(OH) 3 to form the salt AlCl 3 and water. In this case, aluminum cations Al 3+ from the base are exchanged with Cl - anions from the acid. What happens in the end neutralization of hydrochloric acid.
Decomposition reactions include those in which two or more new simple or complex substances, but of a simpler composition, are formed from one complex substance. Examples of reactions include those in the process of which 1) decomposes. Potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate (K 2 MnO 4) is formed, manganese oxide(MnO 2) and oxygen (O 2); 3). Calcium carbonate or marble; in the process are formed carbonicgas(CO2) and calcium oxide(CaO)
2KNO 3 = 2KNO 2 + O 2 (8)
2KMnO 4 = K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 = CaO + CO 2 (10)
In reaction (8), one complex and one simple substance are formed from a complex substance. In reaction (9) there are two complex and one simple. In reaction (10) there are two complex substances, but simpler in composition
All classes of complex substances are subject to decomposition:
1). Oxides: silver oxide 2Ag 2 O = 4Ag + O 2 (11)
2). Hydroxides: iron hydroxide 2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O (12)
3). Acids: sulfuric acid H 2 SO 4 = SO 3 + H 2 O (13)
4). Salts: calcium carbonate CaCO 3 = CaO + CO 2 (14)
5). Organic matter: alcoholic fermentation of glucose
C 6 H 12 O 6 = 2C 2 H 5 OH + 2CO 2 (15)
According to another classification, all chemical reactions can be divided into two types: reactions that release heat are called exothermic, and reactions that occur with the absorption of heat - endothermic. The criterion for such processes is thermal effect of the reaction. As a rule, exothermic reactions include oxidation reactions, i.e. interaction with oxygen, for example methane combustion:
CH 4 + 2O 2 = CO 2 + 2H 2 O + Q (16)
and to endothermic reactions - decomposition reactions already given above (11) - (15). The Q sign at the end of the equation indicates whether heat is released (+Q) or absorbed (-Q) during the reaction:
CaCO 3 = CaO+CO 2 - Q (17)
You can also consider all chemical reactions according to the type of change in the degree of oxidation of the elements involved in their transformations. For example, in reaction (17), the elements participating in it do not change their oxidation states:
Ca +2 C +4 O 3 -2 = Ca +2 O -2 +C +4 O 2 -2 (18)
And in reaction (16), the elements change their oxidation states:
2Mg 0 + O 2 0 = 2Mg +2 O -2
Reactions of this type are redox . They will be considered separately. To compose equations for reactions of this type, you must use half-reaction method and apply electronic balance equation.
After bringing various types chemical reactions, you can proceed to the principle of compiling chemical equations, otherwise, selecting the coefficients on the left and right sides.
Mechanisms for composing chemical equations.
Whatever type this or that chemical reaction, its recording (chemical equation) must correspond to the condition of equality of the number of atoms before the reaction and after the reaction.
There are equations (17) that do not require equalization, i.e. placement of coefficients. But in most cases, as in examples (3), (7), (15), it is necessary to take actions aimed at equalizing the left and right sides of the equation. What principles should be followed in such cases? Is there any system for selecting odds? There is, and not only one. These systems include:
1). Selection of coefficients according to given formulas.
2). Compilation by valence of reacting substances.
3). Arrangement of reacting substances according to oxidation states.
In the first case, it is assumed that we know the formulas of the reacting substances both before and after the reaction. For example, given the following equation:
N 2 + O 2 →N 2 O 3 (19)
It is generally accepted that until equality is established between the atoms of the elements before and after the reaction, the equal sign (=) is not placed in the equation, but is replaced by an arrow (→). Now let's get down to the actual adjustment. On the left side of the equation there are 2 nitrogen atoms (N 2) and two oxygen atoms (O 2), and on the right side there are two nitrogen atoms (N 2) and three oxygen atoms (O 3). There is no need to equalize it in terms of the number of nitrogen atoms, but in terms of oxygen it is necessary to achieve equality, since before the reaction there were two atoms involved, and after the reaction there were three atoms. Let's make the following diagram:
before reaction after reaction
O 2 O 3
Let's determine the smallest multiple between the given numbers of atoms, it will be “6”.
O 2 O 3
\ 6 /
Let's divide this number on the left side of the oxygen equation by “2”. We get the number “3” and put it in the equation to be solved:
N 2 + 3O 2 →N 2 O 3
We also divide the number “6” for the right side of the equation by “3”. We get the number “2”, and also put it in the equation to be solved:
N 2 + 3O 2 → 2N 2 O 3
The numbers of oxygen atoms on both the left and right sides of the equation became equal, respectively, 6 atoms each:
But the number of nitrogen atoms on both sides of the equation will not correspond to each other:
The left one has two atoms, the right one has four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation, setting the coefficient to “2”:
Thus, equality in nitrogen is observed and, in general, the equation takes the form:
2N 2 + 3О 2 → 2N 2 О 3
Now in the equation you can put an equal sign instead of an arrow:
2N 2 + 3О 2 = 2N 2 О 3 (20)
Let's give another example. The following reaction equation is given:
P + Cl 2 → PCl 5
On the left side of the equation there is 1 phosphorus atom (P) and two chlorine atoms (Cl 2), and on the right side there is one phosphorus atom (P) and five oxygen atoms (Cl 5). There is no need to equalize it in terms of the number of phosphorus atoms, but in terms of chlorine it is necessary to achieve equality, since before the reaction there were two atoms involved, and after the reaction there were five atoms. Let's make the following diagram:
before reaction after reaction
Cl 2 Cl 5
Let's determine the smallest multiple between the given numbers of atoms, it will be “10”.
Cl 2 Cl 5
\ 10 /
Divide this number on the left side of the chlorine equation by “2”. Let’s get the number “5” and put it into the equation to be solved:
P + 5Cl 2 → PCl 5
We also divide the number “10” for the right side of the equation by “5”. We get the number “2”, and also put it in the equation to be solved:
P + 5Cl 2 → 2РCl 5
The numbers of chlorine atoms on both the left and right sides of the equation became equal, respectively, 10 atoms each:
But the number of phosphorus atoms on both sides of the equation will not correspond to each other:
Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation by setting the coefficient “2”:
Thus, equality in phosphorus is observed and, in general, the equation takes the form:
2Р + 5Cl 2 = 2РCl 5 (21)
When composing equations by valencies must be given valency determination and set values for the most famous elements. Valence is one of the previously used concepts, currently in a number of school programs not used. But with its help it is easier to explain the principles of drawing up equations of chemical reactions. Valence is understood as the number of chemical bonds that an atom can form with another or other atoms . Valency does not have a sign (+ or -) and is indicated by Roman numerals, usually above the symbols of chemical elements, for example:
Where do these values come from? How to use them when writing chemical equations? Numeric values valencies of elements coincide with their group number Periodic table chemical elements by D.I. Mendeleev (Table 1).
For other elements valence values may have other values, but never greater than the number of the group in which they are located. Moreover, for even group numbers (IV and VI), the valences of elements take only even values, and for odd ones they can have both even and odd values (Table 2).
Of course, there are exceptions to the valency values for some elements, but in each specific case these points are usually specified. Now let's consider general principle compiling chemical equations based on given valences for certain elements. Most often, this method is acceptable in the case of drawing up equations of chemical reactions of compounds of simple substances, for example, when interacting with oxygen ( oxidation reactions). Let's say you need to display an oxidation reaction aluminum. But let us recall that metals are designated by single atoms (Al), and non-metals in the gaseous state are designated by the indices “2” - (O 2). First we'll write general scheme reactions:
Al + О 2 →AlО
At this stage, it is not yet known what the correct spelling should be for aluminum oxide. And it is precisely at this stage that knowledge of the valences of elements will come to our aid. For aluminum and oxygen, let’s put them above the expected formula of this oxide:
III II
Al O
After that, “cross”-on-“cross” for these element symbols we will put the corresponding indices at the bottom:
III II
Al 2 O 3
Composition of a chemical compound Al 2 O 3 determined. The further diagram of the reaction equation will take the form:
Al+ O 2 →Al 2 O 3
All that remains is to equalize its left and right parts. Let us proceed in the same way as in the case of composing equation (19). Let's equalize the numbers of oxygen atoms by finding the smallest multiple:
before reaction after reaction
O 2 O 3
\ 6 /
Let's divide this number on the left side of the oxygen equation by “2”. Let’s get the number “3” and put it into the equation being solved. We also divide the number “6” for the right side of the equation by “3”. We get the number “2”, and also put it in the equation to be solved:
Al + 3O 2 → 2Al 2 O 3
To achieve equality in aluminum, it is necessary to adjust its quantity on the left side of the equation by setting the coefficient to “4”:
4Al + 3O 2 → 2Al 2 O 3
Thus, equality for aluminum and oxygen is observed and, in general, the equation will take its final form:
4Al + 3O 2 = 2Al 2 O 3 (22)
Using the valence method, you can predict what substance is formed during a chemical reaction and what its formula will look like. Let’s assume that the compound reacted with nitrogen and hydrogen with the corresponding valences III and I. Let’s write the general reaction scheme:
N 2 + N 2 → NH
For nitrogen and hydrogen, let’s put the valencies above the expected formula of this compound:
As before, “cross”-on-“cross” for these element symbols, let’s put the corresponding indices below:
III I
NH 3
The further diagram of the reaction equation will take the form:
N 2 + N 2 → NH 3
Equating in a well-known way, through the smallest multiple for hydrogen equal to “6”, we obtain the required coefficients and the equation as a whole:
N 2 + 3H 2 = 2NH 3 (23)
When composing equations according to oxidation states reactants, it is necessary to recall that the oxidation state of a particular element is the number of electrons accepted or given up during a chemical reaction. Oxidation state in compounds Basically, it numerically coincides with the valence values of the element. But they differ in sign. For example, for hydrogen, the valence is I, and the oxidation state is (+1) or (-1). For oxygen, the valence is II, and the oxidation state is -2. For nitrogen, the valences are I, II, III, IV, V, and the oxidation states are (-3), (+1), (+2), (+3), (+4), (+5), etc. . The oxidation states of the elements most often used in equations are given in Table 3.
In the case of compound reactions, the principle of compiling equations by oxidation states is the same as when compiling by valences. For example, let us give the equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write down the proposed equation:
Cl 2 + O 2 → ClO
Let us place the oxidation states of the corresponding atoms over the proposed compound ClO:
As in previous cases let us establish that the required compound formula will take the form:
7 -2
Cl 2 O 7
The reaction equation will take the following form:
Cl 2 + O 2 → Cl 2 O 7
Equating for oxygen, finding the smallest multiple between two and seven, equal to “14,” we ultimately establish the equality:
2Cl 2 + 7O 2 = 2Cl 2 O 7 (24)
A slightly different method must be used with oxidation states when composing exchange, neutralization, and substitution reactions. In some cases, it is difficult to find out: what compounds are formed during the interaction of complex substances?
How to find out: what will happen in the reaction process?
Indeed, how do you know what reaction products may arise during a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?
Ba(NO 3) 2 + K 2 SO 4 → ?
Maybe BaK 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, during this reaction the following compounds are formed: BaSO 4 and KNO 3. How is this known? And how to write the formulas of substances correctly? Let's start with what is most often overlooked: the very concept of “exchange reaction.” This means that in these reactions substances change their constituent parts with each other. Since exchange reactions are mostly carried out between bases, acids or salts, the parts with which they will be exchanged are metal cations (Na +, Mg 2+, Al 3+, Ca 2+, Cr 3+), H + ions or OH -, anions - acid residues, (Cl -, NO 3 2-, SO 3 2-, SO 4 2-, CO 3 2-, PO 4 3-). IN general view The exchange reaction can be given in the following notation:
Kt1An1 + Kt2An1 = Kt1An2 + Kt2An1 (25)
Where Kt1 and Kt2 are metal cations (1) and (2), and An1 and An2 are their corresponding anions (1) and (2). In this case, it is necessary to take into account that in compounds before and after the reaction, cations are always installed in first place, and anions are in second place. Therefore, if the reaction occurs potassium chloride And silver nitrate, both in dissolved state
KCl + AgNO 3 →
then in its process the substances KNO 3 and AgCl are formed and the corresponding equation will take the form:
KCl + AgNO 3 =KNO 3 + AgCl (26)
During neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):
HCl + KOH = KCl + H 2 O (27)
The oxidation states of metal cations and the charges of anions of acidic residues are indicated in the table of solubility of substances (acids, salts and bases in water). The horizontal line shows metal cations, and the vertical line shows the anions of acid residues.
Based on this, when drawing up an equation for an exchange reaction, it is necessary first to establish on the left side the oxidation states of the hosts in this chemical process particles. For example, you need to write an equation for the interaction between calcium chloride and sodium carbonate. Let’s create the initial diagram of this reaction:
CaCl + NaCO 3 →
Ca 2+ Cl - + Na + CO 3 2- →
Having performed the already known “cross”-on-“cross” action, we determine the real formulas of the starting substances:
CaCl 2 + Na 2 CO 3 →
Based on the principle of exchange of cations and anions (25), we will establish preliminary formulas for the substances formed during the reaction:
CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl
Let us place the corresponding charges above their cations and anions:
Ca 2+ CO 3 2- + Na + Cl -
Substance formulas written correctly, in accordance with the charges of cations and anions. Let's compose complete equation, equalizing its left and right parts for sodium and chlorine:
CaCl 2 + Na 2 CO 3 = CaCO 3 + 2NaCl (28)
As another example, here is the equation for the neutralization reaction between barium hydroxide and phosphoric acid:
VaON + NPO 4 →
Let us place the corresponding charges over the cations and anions:
Ba 2+ OH - + H + PO 4 3- →
Let's determine the real formulas of the starting substances:
Ba(OH) 2 + H 3 PO 4 →
Based on the principle of exchange of cations and anions (25), we will establish preliminary formulas for the substances formed during the reaction, taking into account that during an exchange reaction one of the substances must necessarily be water:
Ba(OH) 2 + H 3 PO 4 → Ba 2+ PO 4 3- + H 2 O
Let us determine the correct notation for the formula of the salt formed during the reaction:
Ba(OH) 2 + H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
Let's equalize the left side of the equation for barium:
3Ba (OH) 2 + H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
Since on the right side of the equation the orthophosphoric acid residue is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:
3Ba (OH) 2 + 2H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
It remains to match the number of hydrogen and oxygen atoms on the right side of water. Since on the left total hydrogen atoms is 12, then on the right it must also correspond to twelve, therefore before the formula of water it is necessary set the coefficient“6” (since the water molecule already has 2 hydrogen atoms). For oxygen, equality is also observed: on the left is 14 and on the right is 14. So, the equation has correct form entries:
3Ba (OH) 2 + 2H 3 PO 4 → Ba 3 (PO 4) 2 + 6H 2 O (29)
Possibility of chemical reactions
The world consists of a great variety of substances. The number of variants of chemical reactions between them is also incalculable. But can we, having written this or that equation on paper, say that a chemical reaction will correspond to it? There is a misconception that if it is correct set the odds in the equation, then it will be feasible in practice. For example, if we take sulfuric acid solution and put it in zinc, then you can observe the process of hydrogen evolution:
Zn+ H 2 SO 4 = ZnSO 4 + H 2 (30)
But if copper is dropped into the same solution, then the process of gas evolution will not be observed. The reaction is not feasible.
Cu+ H 2 SO 4 ≠
If concentrated sulfuric acid is taken, it will react with copper:
Cu + 2H 2 SO 4 = CuSO 4 + SO 2 + 2H 2 O (31)
In reaction (23) between the gases nitrogen and hydrogen, we observe thermodynamic equilibrium, those. how many molecules ammonia NH 3 is formed per unit time, the same amount of them will decompose back into nitrogen and hydrogen. Chemical equilibrium shift can be achieved by increasing pressure and decreasing temperature
N 2 + 3H 2 = 2NH 3
If you take potassium hydroxide solution and pour it on him sodium sulfate solution, then no changes will be observed, the reaction will not be feasible:
KOH + Na 2 SO 4 ≠
Sodium chloride solution when interacting with bromine, it will not form bromine, despite the fact that this reaction can be classified as a substitution reaction:
NaCl + Br 2 ≠
What are the reasons for such discrepancies? The point is that it is not enough just to correctly determine compound formulas, it is necessary to know the specifics of the interaction of metals with acids, skillfully use the table of solubility of substances, and know the rules of substitution in the activity series of metals and halogens. This article outlines only the most basic principles of how assign coefficients in reaction equations, How write molecular equations, How determine the composition of a chemical compound.
Chemistry, as a science, is extremely diverse and multifaceted. The above article reflects only a small part of the processes occurring in real world. Types, thermochemical equations, electrolysis, processes of organic synthesis and much, much more. But more on that in future articles.
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Lesson plan:
“The use of electronic educational resources in the work of a teacher.”
Place of work: MKOU "Khailinskaya" high school».
Position: teacher of chemistry and biology.
Subject: chemistry.
Basic textbook: G.E. Rudzitis, F.G. Feldman
Purpose of the lesson: to teach students to compose formulas of chemical compounds based on valency and oxidation state.
Tasks:
educational: teach how to compose formulas for binary compounds.
developing: to develop the ability to reason logically, correctly express one’s thoughts, to comprehend and understand it more deeply.
educational: develop independence, intelligence.
Lesson topic: Drawing up chemical formulas.
Lesson type: Lesson on studying and initially consolidating new knowledge.
Technical equipment: computer, multimedia projector
Lesson structure and flow:
Lesson steps:
Organization of the beginning of the lesson.
Motivation for learning activities.
The importance of understanding the study of this topic lies in its biological meaning.
The entire Universe, including planet Earth and all kingdoms (bacteria, fungi, protozoa, plants and animals) consist of the same chemical atoms and elements. Atoms, identical and different, combine to form inorganic and organic substances. All bodies and objects are made of substances. Let's express the composition of substances through formulas.
Preparing students to master, updating basic knowledge.
Students have already studied and become familiar with concepts such as:
The law of constancy of the composition of matter. (1799 – 1806 – J. Proust)
Each chemically pure substance, regardless of location and method of production, has the same permanent staff.
Based on the law of constancy of the composition of substances, chemical formulas can be derived.
2. A chemical formula is a conventional recording of the composition of a substance using chemical symbols and indices.
The subscript in chemical formulas denotes the number of atoms.
Al 2 index O 3 index iron filings ( Cl 3 – index
Valency is the property of an atom of a chemical element to attach or replace a certain number of atoms of another chemical element.
The valency of hydrogen is taken as unity.
The valence of oxygen is two.
The numerical value of valency is usually denoted by Roman numerals, which are placed above the signs of chemical elements.
The property of atoms of a given element to attract electrons from atoms of other elements in compounds is called electronegativity.
The oxidation state is the conditional charge of an element.
The oxidation state is determined by the number of electrons displaced from atoms of a less electronegative element to an atom of a more electronegative element.
Learning new material.
Chemical formulas are analogues of words, just as words are written using letters, so formulas are written using chemical symbols and signs. Chemical formulas reflect the composition of a substance.
The purpose of the lesson.
Lesson objectives.
Drawing up chemical formulas by valence.
To compose a chemical formula, you need to know the valency of the elements that form a given chemical compound. When compiling chemical formulas, you must follow the following procedure:
1. Write next to the chemical symbols of the elements that make up the compound:
K O Al Cl AlO
2. Their valency is indicated above the signs of chemical elements:
I II III I III II
K O Al Cl Al O
3. Determine the least common multiple of the numbers expressing the valence of both elements:
2 3 6
I II III I III II
K O Al Cl Al O
4. By dividing the least common multiple by the valence of the corresponding element, the indices are found
I II III I III II
K 2 O Al Cl 3 Al 2 O 3
5. In the names of substances formed by elements with variable valence, write in parentheses a number indicating the valence of the given element in this compound.
For example,
Cu O – copper oxide (II )
Cu 2 O – copper oxide (I )
iron filings ( Cl 2 – ferric chloride (II )
iron filings ( Cl 3 – ferric chloride(III )
6. Some elements in different compounds exhibit different valencies .
( see table)
Valency of some elements in chemical compounds.
Chemical elements.
With constant valence
O Be Mg Ca Ba Zn
Al B
With variable valence
I II
II III
Fe Co Ni
II IV
Sn Pb
III V
II III VI
II IV VI
When compiling chemical formulas by oxidation state, you need to know:
the degree of oxidation of the elements forming a given chemical compound;
their electronegativity, since the most electronegative element is placed last;
the sum of negative and positive oxidation states in a correctly composed formula is always zero.
Primary test of knowledge acquisition.
Rules for drawing up chemical formulas .
Primary consolidation of knowledge.
Let's talk about how to create a chemical equation, because they are the main elements of this discipline. Thanks to a deep understanding of all the patterns of interactions and substances, you can control them, apply them in various fields activities.
Theoretical features
Drawing up chemical equations is an important and responsible stage, considered in the eighth grade. secondary schools. What should precede at this stage? Before the teacher tells his students how to create a chemical equation, it is important to introduce schoolchildren to the term “valence” and teach them to determine this value for metals and non-metals using the periodic table of elements.
Compilation of binary formulas by valence
In order to understand how to create a chemical equation by valency, you first need to learn how to create formulas for compounds consisting of two elements using valence. We propose an algorithm that will help cope with the task. For example, you need to create a formula for sodium oxide.
First, it is important to take into account that the chemical element that is mentioned last in the name should be in first place in the formula. In our case, sodium will be written first in the formula, oxygen second. Let us recall that oxides are binary compounds in which the last (second) element must be oxygen with an oxidation state of -2 (valency 2). Next, using the periodic table, it is necessary to determine the valence of each of the two elements. To do this we use certain rules.
Since sodium is a metal that is located in the main subgroup of group 1, its valence is a constant value, it is equal to I.
Oxygen is a non-metal, since it is the last one in the oxide; to determine its valency, we subtract 6 from eight (the number of groups) (the group in which oxygen is located), we obtain that the valency of oxygen is II.
Between certain valences we find the least common multiple, then divide it by the valency of each of the elements to obtain their indices. We write down the finished formula Na 2 O.
Instructions for composing an equation
Now let's talk in more detail about how to write a chemical equation. First, let's look at the theoretical aspects, then move on to specific examples. So, composing chemical equations presupposes a certain procedure.
- 1st stage. After reading the proposed task, you need to determine which chemical substances must be present on the left side of the equation. A “+” sign is placed between the original components.
- 2nd stage. After the equal sign, you need to create a formula for the reaction product. When performing such actions, you will need the algorithm for composing formulas for binary compounds, which we discussed above.
- 3rd stage. Checking the number of atoms of each element before and after chemical interaction, if necessary, we put additional coefficients in front of the formulas.
Example of a combustion reaction
Let's try to figure out how to create a chemical equation for the combustion of magnesium using an algorithm. On the left side of the equation we write the sum of magnesium and oxygen. Do not forget that oxygen is a diatomic molecule, so it must be given an index of 2. After the equal sign, we compose the formula for the product obtained after the reaction. It will be in which magnesium is written first, and oxygen is written second in the formula. Next, using the table of chemical elements, we determine the valencies. Magnesium, which is in group 2 (the main subgroup), has a constant valency II; for oxygen, by subtracting 8 - 6 we also get valence II.
The process record will look like: Mg+O 2 =MgO.
In order for the equation to comply with the law of conservation of mass of substances, it is necessary to arrange the coefficients. First, we check the amount of oxygen before the reaction, after the process is completed. Since there were 2 oxygen atoms, but only one was formed, a coefficient of 2 must be added on the right side before the magnesium oxide formula. Next, we count the number of magnesium atoms before and after the process. As a result of the interaction, 2 magnesium was obtained, therefore, on the left side in front of the simple substance magnesium, a coefficient of 2 is also required.
The final type of reaction: 2Mg+O 2 =2MgO.
Example of a substitution reaction
Any chemistry abstract contains a description different types interactions.
Unlike a compound, in a substitution there will be two substances on both the left and right sides of the equation. Let's say we need to write the reaction of interaction between zinc and We use the standard writing algorithm. First, on the left side we write zinc and hydrochloric acid through the sum, and on the right side we write the formulas for the resulting reaction products. Since zinc is located before hydrogen in the electrochemical voltage series of metals, in this process it displaces molecular hydrogen from the acid and forms zinc chloride. As a result, we get the following entry: Zn+HCL=ZnCl 2 +H 2.
Now we move on to equalizing the number of atoms of each element. Since there was one atom on the left side of chlorine, and after the interaction there were two, it is necessary to put a factor of 2 in front of the formula of hydrochloric acid.
As a result, we obtain a ready-made reaction equation corresponding to the law of conservation of mass of substances: Zn+2HCL=ZnCl 2 +H 2 .
Conclusion
A typical chemistry note necessarily contains several chemical transformations. Not a single section of this science is limited to a simple verbal description of transformations, processes of dissolution, evaporation; everything is necessarily confirmed by equations. The specificity of chemistry lies in the fact that all processes that occur between different inorganic or organic substances, can be described using coefficients and indices.
How else does chemistry differ from other sciences? Chemical equations help not only to describe the transformations that occur, but also to carry out quantitative calculations based on them, thanks to which it is possible to carry out laboratory and industrial production of various substances.