Chemical reaction equation examples 8. Chemical equations

The law of conservation of mass of substances, discovered by M. V. Lomonosov in 1748, states:

The material carriers of the mass of substances are atoms of chemical elements, of which both the substances that entered into the reaction (reagents) and the new substances formed as a result of it (reaction products) are composed. Since during chemical reactions atoms are not formed or destroyed, but only their rearrangement occurs, the validity of the law discovered by M. V. Lomonosov and later confirmed by A. Lavoisier becomes obvious.

The validity of the law of conservation of mass of substances can be easily verified by simple experiment. Let's put a little red phosphorus in the flask, close it with a stopper and weigh it on the scales (Fig. 96). Then carefully heat the flask. The fact that a chemical reaction has occurred can be determined by the appearance in the flask of thick white smoke consisting of phosphorus oxide (V), which was formed by the interaction of phosphorus with oxygen. When we weigh the flask with the products of this reaction again, we will make sure that the mass of substances in the flask has not changed, although phosphorus has been converted into its oxide.

Rice. 96.
Experimental verification of the law of conservation of mass of substances:
a - weighing the flask with phosphorus before the reaction; b - combustion of phosphorus in a closed flask; c - weighing the flask with the reaction product

We will draw the same conclusion when conducting another simple but very clear experiment. In a special vessel, pour separately hydrochloric acid and an alkali solution, for example sodium hydroxide (Fig. 97). Add a few drops of an indicator - phenolphthalein - to the alkali solution, causing the solution to turn crimson. We close the device with a stopper, balance it with weights on the scales, note the mass, and then salt the solutions. The crimson color will disappear because the acid and alkali have reacted with each other. The mass of the vessel with the resulting reaction products did not change.

Rice. 97.
Experiment confirming the law of conservation of mass of matter

A similar observation was made by the author of the law of conservation of mass of substances, M.V. Lomonosov, who conducted experiments in sealed glass vessels, “in order to investigate whether the weight of a metal comes from pure heat,” and discovered that “without the passage of external air, the weight of metals remains in one least."

Based on this law, they write chemical sentences, that is, they compose equations of chemical reactions using chemical words - formulas.

On the left side of the equation, write down the formulas of the substances that reacted, connecting them with a plus sign. On the right side of the equation, write down the formulas of the resulting substances, also connected by a plus sign. An arrow is placed between the parts of the equation. Then they find the coefficients - the numbers in front of the formulas of substances, so that the number of atoms of identical elements on the left and right sides of the equation is equal.

Let us write, for example, the equation for the reaction of hydrogen with oxygen. First, let's draw up a reaction diagram - indicate the formulas of the substances that enter into the reaction (hydrogen H 2 and oxygen O 2) and those formed as a result of it (water H 2 O), and connect them with an arrow:

H 2 + O 2 → H 2 O (Fig. 98, a).

Rice. 98.
Drawing up an equation for the reaction between hydrogen and oxygen

Since the number of oxygen atoms on the left side is twice as large as on the right, we write the coefficient 2 in front of the water formula:

H 2 + O 2 → 2H 2 O (Fig. 98, b).

But now there are four hydrogen atoms on the right side of the equation, and there are two left on the left. To equalize the number of hydrogen atoms, we write the coefficient 2 in front of its formula on the left side. Since we have equalized the number of atoms of each element on the left and right sides of the equation, we replace the arrow with an equal sign:

2H 2 + O 2 = 2H 2 O (Fig. 98, c).

Now, you probably understand why such a record is called an equation (Fig. 99).

Rice. 99.
The law of conservation of mass of substances using the example of a reaction whose equation is 2H 2 + O 2 = 2H 2 O

To draw up equations of chemical reactions, in addition to knowing the formulas of the reagents and reaction products, it is necessary to select the correct coefficients.

This can be done using simple rules,

1. Before the formula of a simple substance, you can write a fractional coefficient, which shows the amount of reacting and resulting substances.

So, for the example discussed above:

H 2 + O 2 → H 2 O

the number of oxygen atoms on the right and left sides of the equation can be made equal using the coefficient 1/2, placing it in front of the oxygen formula:

H2 + 1/2O2 = H2O

But since the coefficient shows not only the amount of substance, but also the number of molecules (atoms), and it is impossible to take half a molecule, it is better to rewrite the above equation, doubling all the coefficients in it:

2H 2 + O 2 = 2H 2 O.

Let us give another example of composing the equation for the combustion reaction of ethane C 2 H 6 contained in natural gas. It is known that as a result of this process, carbon dioxide and water. Scheme of this reaction:

C 2 H 6 + O 2 → CO 2 + H 2 O.

Let's equalize the number of carbon and hydrogen atoms:

C 2 H 6 + O 2 → 2CO 2 + ZH 2 O.

Now there are 7 oxygen atoms on the right side of the reaction equation, and only 2 on the left. Let’s equalize the number of oxygen atoms by writing the coefficient 3.5 (7:2 = 3.5) before formula 02:

C 2 H 6 + 3.5O 2 = 2CO 2 + ZH 2 O.

And finally, we rewrite the resulting reaction equation, doubling the coefficients in front of the formulas of all reaction participants:

2C 2 H 6 + 7O 2 = 4CO 2 + 6H 2 O.

2. If the reaction scheme contains a salt formula, then first the number of ions forming the salt is equalized.

For example, the interaction of sulfuric acid and aluminum hydroxide is described by the following scheme:

H 2 SO 4 + Al (OH) 3 → Al 2 (SO 4) 3 + H 2 O.

The salt formed as a result of the reaction - aluminum sulfate Al 2 (SO 4) 3 - consists of aluminum ions Al3+ and sulfate ions. Let’s equalize their number by writing coefficients 3 and 2 before the formulas H 2 SO 4 and Al (OH) 3, respectively:

3H 2 SO 4 + 2Al(OH) 3 → Al 2 (SO 4) 3 + H 2 O.

To equalize the number of hydrogen and oxygen atoms, we use the third rule.

3. If the substances involved in the reaction contain hydrogen and oxygen, then the hydrogen atoms are equalized in the penultimate place, and the oxygen atoms in the last place.

Therefore, we equalize the number of hydrogen atoms. There are 12 hydrogen atoms on the left side of the reaction diagram, and only 2 on the right side, so before the water formula we write the coefficient 6:

3H 2 SO 4 + 2Al(OH) 3 → Al 2 (SO 4) 3 + 6H 2 O.

An indicator of the correctness of the arrangement of coefficients is the equality of the number of oxygen atoms in the left and right sides of the reaction equation - 24 oxygen atoms each. Therefore, we replace the arrow with an equal sign:

3H 2 SO 4 + 2Al(OH) 3 = Al 2 (SO 4) 3 + 6H 2 O.

4. If there are several salt formulas in the reaction scheme, then it is necessary to begin the equation with the ions that are part of the salt containing a larger number of them.

For example, the interaction of solutions of sodium phosphate and calcium nitrate is described by the following scheme:

Na 3 PO 4 + Ca(NO 3) 2 → Ca 3 (PO 4) 2 + NaNO 3.

Largest number ions contains one of the reaction products - calcium phosphate Ca 3 (PO 4) 2, therefore the ions that form this salt are equalized - Ca 2+ and:

2Na 3 PO 4 + 3Ca(NO 3) 2 → Ca 3 (PO 4) 2 + NaNO 3.

and finally, Na + and N0 - 3 ions:

2Na 3 PO 4 + 3Ca(NO 3) 2 → Ca 3 (PO 4) 2 + 6NaNO 3.

Key words and phrases

Chemical equations.
Rules for selecting coefficients in reaction equations.

Work with computer

Refer to the electronic application. Study the lesson material and complete the assigned tasks.
Find email addresses on the Internet that can serve as additional sources that reveal the content of keywords and phrases in the paragraph. Offer your help to the teacher in preparing a new lesson - send a message by keywords and phrases in the next paragraph.

Questions and tasks

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 assemble iron 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 up potassium 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 oxide(Fe 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 #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 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 figured out the structure various substances, you can begin to compile 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 conditional 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 big number The “2” 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 it down correctly chemical equation, necessary 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, the formation of sediment, the 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 the case of reaction (6), the soluble salts AgNO 3 and KCl, when both solutions are merged, form an insoluble precipitate of the 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 a chemical reaction belongs to, its recording (chemical equation) must correspond to the condition that the number of atoms before and after the reaction is equal.

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 valences 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 into 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 for 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 number chemical bonds, which one or another 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 valences 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 are taken only even values, and for odd ones they can have both even and odd values (Table 2).

Of course, there are exceptions to the valence 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 a compound simple substances, for example, when interacting with oxygen ( oxidation reactions). Let's say you need to display an oxidation reaction aluminum. But let us remember that metals are designated by single atoms (Al), and non-metals in the gaseous state are designated by the indices “2” - (O 2). First, let's write the general reaction scheme:

Al + О 2 →AlО

At this stage it is not yet known which correct writing should be 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

Chemical compound composition 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

Already calling in a 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 expected 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, we 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 during the reaction?

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 first necessary to establish on the left side the oxidation states of the particles receiving in this chemical process. 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 create a complete equation, equalizing its left and right sides 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 the correct form of writing:

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 it 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.

blog.site, when copying material in full or in part, a link to the original source is required.

Detailed outline of the lesson “Equations of chemical reactions”.

Textbook: O.S. Gabrielyan.

Class: 8

Lesson topic: Equations of chemical reactions.

Lesson type: improvement of knowledge and skills.

Educational objectives: consolidate knowledge of composing chemical equations and arranging coefficients;

Educational tasks: continue the formation of a natural science worldview, an idea of the individual and the whole when becoming familiar with chemical equations.

Developmental tasks: continue to develop the ability to observe, analyze, explain, and draw conclusions.

Teaching methods: verbal (explanation and story of the teacher), verbal - visual (explanation using notes on the blackboard).

Equipment: chalkboard, table of D.I. Mendeleev.

During the classes:

1. Organizational moment (2-4 min.)

Hello guys, have a seat. Today in the lesson we will consolidate the equations of chemical reactions, their writing and the distribution of coefficients.

2. Improving knowledge and skills (20 – 35 min.)

Let's write down:

Algorithm for composing a chemical reaction equation.

Write down the formulas (formula) of the starting substances, connecting them with a “+” sign (this is the left side of the equation).

Place an arrow.

Write down the formula (formula) of the reaction products after the arrow (this is the right side of the equation).

Arrange the coefficients so that the number of atoms of identical elements on the left and right sides of the equation are equal.

Replace the arrow with an equal sign.

Using an algorithm for composing equations of chemical reactions, let's write down the equation of the interaction of iron (II) sulfide with oxygen as a result of a chemical reaction, iron (III) oxide and sulfur oxide (IV) are formed.

We write according to the algorithm:

We define the starting substances - these are the substances that reacted: these are iron (II) sulfide and oxygen, and now we write the formulas of these substances on the left side of the equation:

FeS 2 + O 2

Place the arrow:

FeS 2 + O 2 →

3. We define the reaction products - these are substances that were obtained as a result of a chemical reaction: these are iron oxide (III) and sulfur oxide (IV), and now we write the formulas of these substances on the right side of the equation:

FeS 2 + O 2 → Fe 2 O 3 + SO 2

4. We arrange the coefficients so that the number of atoms of identical elements on the left and right sides of the equation are equal:

2FeS 2 + 5.5O 2 → Fe 2 O 3 + 4SO 2

Now there are 11 oxygen atoms on the right side of the reaction equation, and only 2 on the left. Let’s equalize the number of oxygen atoms by writing a coefficient of 5.5 in front of the O 2 formula (11:2 = 5.5).

And finally, we rewrite the resulting reaction equation, doubling the coefficients in front of the formulas of all reaction participants:

4FeS 2 + 11O 2 → 2Fe 2 O 3 + 8SO 2

5. Replace the arrow with an equal sign:

4FeS 2 + 11O 2 = 2Fe 2 O 3 + 8SO 2.

Next we will work in workbooks (from Chemistry. 8th grade: workbook for the textbook by O.S. Gabrielyan “Chemistry. 8th grade” / O.S. Gabrielyan, A.V. Yashukova. - 12th ed., additional – M.: Bustard, 2010. – 192 pp.: ill.) open pp. 98 – 99 and perform exercise. 5 – 7 in writing. (The teacher collects notebooks for some students to check).

3. Control of knowledge and skills (10 -20 min).

1. Place coefficients in the equations of chemical reactions:

P + Cl 2 = PCl 5

CH 4 = C + H 2

Na + S = Na 2 S

HCl + Mg = MgCl 2 + H 2

ZnSO 4 + KOH = K 2 SO 4 + Zn(OH) 2

BaCl 2 + Na 2 SO 4 = BaSO 4 + NaCl

AlCl 3 + NaOH = NaCl + Al(OH) 3

H 2 SO 4 + Al = Al 2 (SO 4) 3 + H 2

P 2 O 5 + Na 2 O = Na 3 PO 4

Al 2 (SO 4) 3 + Ba(NO 3) 2 = Al(NO 3) 3 + BaSO 4

Write down the equations of chemical reactions according to the following schemes and arrange the coefficients:

a) sodium + chlorine → sodium chloride

b) carbonic acid → water + carbon dioxide

c) iron (III) hydroxide → iron (III) oxide + water

d) aluminum + oxygen → aluminum oxide (III)

e) sodium oxide + water → sodium hydroxide

e) potassium oxide + phosphorus oxide → potassium phosphate.

4. Lesson summary (1-3 min.)

Today in class we learned how to write equations for chemical reactions and assign coefficients in the equations.

5. Homework(1-3 min).

§27, ex. 2 (written).

Target: teach students how to write chemical equations. Teach them to equalize using coefficients based on knowledge of the law of conservation of mass of matter M.V. Lomonosov.

Tasks:

Educational:
- continue the study of physical and chemical phenomena with the introduction of the concept of “chemical reaction”,
- introduce the concept of “chemical equation”;
- teach students how to write chemical equations and balance equations using coefficients.
Developmental:
- continue to develop creative potential students’ personalities through creating a situation of problem-based learning, observation, and conducting experiments on chemical reactions.
Educational:
- develop the ability to work in a team or group.

Equipment: tabular material, reference books, algorithms, a set of tasks.

BEFORE:“Burning sparklers”: matches, dry fuel, iron sheet/TB when working with fire.

DURING THE CLASSES

I. Organizational moment

Determining the purpose of the lesson.

II. Repetition

1) On the board is a set of physical and chemical phenomena: evaporation of water; filtration; rusting; wood burning; souring of milk; melting ice; eruption; dissolving sugar in water.

Exercise:

Give an explanation of each phenomenon, name the practical application of this phenomenon in human life.

2) Task:

A drop of water is drawn on the board. Create a complete diagram of the transformation of water from one state of aggregation to another. What is this phenomenon called in nature and what is its significance in the life of our planet and all living things?

III. D/O “Burning sparklers”

1. What happens to magnesium, which forms the basis of sparklers?
2. What was the main reason for this phenomenon?
3. What type of chemical reaction is this?
4. Try to diagrammatically depict the chemical reaction that you observed in this experiment.

– I propose to try to draw up a diagram of this reaction:

Mg + air = other substance

– How did we know that a different substance was obtained? (According to signs of a chemical reaction: change in color, appearance of odor.)
– What gas is in the air that supports combustion? (Oxygen – O)

IV. New material

A chemical reaction can be written using a chemical equation.
You can recall the concept of “equation”, which is given in mathematics. What is the essence of the equation itself? Some things are equalized, some parts.
Let's try to give a definition of a "chemical equation", you can look at the diagram and try to give a definition:

A chemical equation is a conventional notation of a chemical reaction using chemical symbols, formulas and coefficients.
Chemical equations are written on the basis of the Law of Conservation of Mass of Matter, discovered by M.V. Lomonosov in 1756, which states (textbook p. 96): “The mass of substances that entered into a reaction is equal to the mass of substances resulting from it.”
– We must learn to equalize chemical equations using coefficients.
– In order to learn how to write chemical equations well, we need to remember:
– What is a coefficient?
– What is an index?
Don’t forget the “Creating Chemical Formulas” algorithm.

I suggest step-by-step algorithm drawing up a chemical equation:

V. Drawing up a chemical equation

1. I write down the equation of the reacting substances on the left side: Al + O 2

2. I put the “=” sign and write the resulting substances on the right side of the equation - reaction products: Al + O 2 = Al 2 O 3

3. I start equalizing with the chemical element that is greater or with oxygen, then I make a construction:

Al + O 2 = Al 2 O 3
2 /6 3

oxygen entered “2”, but it turned out “3”, their number is not equal.

4. I am looking for the LCM (least common multiple) of two digits “2” and “3” - this is “6”

5. I divide the LCM “6” by the number “2” and “3” and set it as coefficients in front of the formulas.

Al + 3O 2 = 2Al 2 O 3
6 = 6

6. I begin to equalize the following chemical elements - Al, I reason in the same way. Al “1” entered, but “4” turned out, I’m looking for NOC

Al + 3O 2 = 2Al 2 O 3
1 /4 4
4 = 4
4 Al + 3O 2 = 2Al 2 O 3

The coefficient “1” is not written in the equations, but is taken into account when drawing up the equation.

7. I read the entire entry of the chemical equation.

Such long reasoning allows you to quickly learn how to equalize in chemical equations, given that the correct composition of reaction equations for chemistry is of great importance: solving problems, writing chemical reactions.

VI. Reinforcement task

Phosphorus + oxygen = phosphorus oxide (V)
Sulfuric acid + aluminum = aluminum sulfate + hydrogen
Water = hydrogen + oxygen

– One strong student is working on the board.

Zn + O 2 = ZnO;
H 2 + O 2 = H 2 O;
Ba + O 2 = BaO;
S + O 2 = SO 2;
Na + O 2 = Na 2 O 2;
Fe + O 2 = Fe 3 O 4

– Arrange the coefficients in the equations of chemical reactions.

Chemical equations vary in type, but we'll look at that in the next lesson.

VII. Summing up the lesson

Conclusion. Grading.

VIII. Homework:§ 27, ex. 2, p. 100.

Additional material: R.t.s. 90-91, exercise 2 – individually.

LESSON PLAN ON THE TOPIC: “CHEMICAL REACTION EQUATIONS.”

Lesson type: learning new material

Textbook : Gabrielyan O.S. "Chemistry. 8th grade”, From “Bustard”

Lesson number for planning is No. 35, in the topic “Changes that occur with substances” - No. 2.

Tasks:

1. Educational:1) form a concept about the equations of chemical reactions; 2) begin to develop the ability to compose equations of chemical reactions.

2. Developmental: 1) develop in students the ability to observe and analyze what they see; 2) to develop self-control skills in mastering the studied material; 3) develop the cognitive interest and emotions of students, introducing into the content of the lesson an element of novelty of knowledge, its connection with other subjects, with life; 4) activate students' thinking through conversation and experiment.

3. Educating: 1) apply the acquired knowledge in the following lessons (types of chemical reactions); 2) help prevent fatigue in schoolchildren during the lesson, using techniques for maintaining performance, such as the use of various types of work and demonstrations of experiments.

TARGET: To form a concept about the equations of chemical reactions as a conventional notation reflecting the transformations of substances. Begin to develop students’ ability to write equations for chemical reactions.

DURING THE CLASSES.

1.Organization of the beginning of the lesson (2 min.).

Topic of today's lesson:"Equations of chemical reactions."

Task: Today we will get acquainted with the conventional notation of chemical reactions - equations. Let's learn how to write equations for chemical reactions and how to place coefficients in them.

2.Checking homework (5 min.).

Let us repeat with you what phenomena are called physical?

Physical phenomena are those in which the size, shape of bodies and state of aggregation substances, but their composition remains constant.

What phenomena are called chemical?

Phenomena as a result of which other substances are formed from one substance are called chemical phenomena, or chemical reactions.

What signs of chemical reactions do you know?

Color change
Odor appears
Formation of sediment
Dissolution of sediment
Gas release
The release or absorption of heat, sometimes light is released.

Now, try to guess what phenomena these verses are talking about.

3. Preparation for mastering new knowledge (5-7 min.).

Now I will conduct several experiments, and you and I will try to draw up a diagram of the observed transformation.

Experiment 1. Combustion of magnesium.

What are you observing? Let's draw up a diagram of the observed phenomenon.

Magnesium + oxygen → magnesium oxide

Initial substances reaction product

Mg+O 2 → MgO

This conditional notation is called a reaction scheme. On the left side of the diagram the starting substances are written (i.e., those substances that were taken for interaction), on the right side are the reaction products (i.e., those substances that were formed as a result of the interaction).

Experience 2. Producing carbon dioxide

Place a piece of chalk in a test tube and pour in 1-2 ml of hydrochloric acid solution. What are we observing? What's happening? What are the signs of these reactions?

Let us draw up a diagram of the observed transformation using chemical formulas:

calcium carbonate + hydrochloric acid →

starting materials

CaCO 3 + HCl→

Calcium chloride + water + carbon dioxide

reaction products

CaCl 2 + H 2 O + CO 2

4.Assimilation of new material (10-15 min).

Formation of the concept of “coefficients and the ability to arrange coefficients in the equation of a chemical reaction.

Now we will learn about the law of conservation of mass of substances, which was discovered by M.V. Lomonosov in 1756.

Law of conservation of mass of substances (The mass of substances that entered into a reaction is equal to the mass of substances resulting from it).

The material carriers of the mass of substances are atoms of chemical elements, because They are not formed or destroyed during chemical reactions, but their rearrangement occurs, then the validity of this law becomes obvious.

The number of atoms of one element on the left side of the equation must be equal to the number of atoms of that element on the right side of the equation.

Task 1 (for groups). Determine the number of atoms of each chemical element involved in the reaction. 1. Calculate the number of atoms:

a) hydrogen: 8NH3, NaOH, 6NaOH, 2NaOH, H3PO4, 2H2SO4, 3H2S04, 8H2SO4;

6) oxygen: C0 2, 3C0 2, 2C0 2, 6CO, H 2 SO 4, 5H 2 SO 4, 4H 2 S0 4, HN0 3.

2. Calculate the number of atoms: a) hydrogen:

1) NaOH + HCl 2)CH 4 +H 2 0 3)2Na+H 2

b) oxygen:

1) 2СО + 0 2 2) С0 2 + 2Н.О. 3)4NO 2 + 2H 2 O + O 2

Algorithm for arranging coefficients in equations of chemical reactions (source:Borovskikh T. A. Workbook in chemistry: 8th grade: to the textbook by G. E. Rudzitis, F. G. Feldman “Chemistry. 8th grade", M. "Exam", 2011)

Order of operations	example
1. Determine the number of atoms	A1 + O 2 → A1 2 O 3 A1-1 atom A1-2 atoms O-2 atoms 0-3 atoms
	O-2 atoms on the left O-3 atoms on the right
3. Find least common multiple(LCM) number of atoms of this element on the left and right side of the equation	LCM = 6
4. Split the LOC with the left parts of the equation, getcoefficient for left parts of the equation	6:2 = 3 Al + ZO 2 → Al 2 O 3
5. Split the LOC on the right parts of the equation, getcoefficient for right parts of the equation	6:3 = 2 A1 + ZO 2 → 2 A1 2 O 3
	A1 + ZO 2 → 2 A1 2 O 3 A1 - 1 atom A1 - 4 atoms LCM = 4 4:1=4 4:4=1 4A1 + ZO 2 → 2 A1 2 O 3

5.Primary test of knowledge acquisition (8-10 min.). Formation

There are two oxygen atoms on the left side of the diagram, and one on the right. The number of atoms must be equalized using coefficients. The number of atoms must be equalized using coefficients . Let's summarize the lesson:

1)2Mg+O 2 →2MgO

2) CaCO 3 + 2HCl → CaCl 2 + H 2 O + CO 2

Task 2

Mg + N 2 → Mg 3 N 2;

Al + S → A1 2 S 3 ;

A1 + C → A1 4 C 3 ;

Ca + P → Ca 3 P 2 ;

C + H 2 → CH 4;

Ca + C → CaC 2;

Fe + O 2 → Fe 3 O 4;

Si + Mg → Mg 2 Si;

Na + S → Na 2 S;

CaO + C → CaC 2 + CO;

Ca + N 2 → Ca 3 N 2;

Si + C1 2 → SiCl 4 ;

Ag + S → Ag 2 S;

Exercise (reserve)3.

H 2 + C1 2 → HC1;

N 2 + O 2 → NO;

CO 2 + C → CO;

HI → H 2 + 1 2;

Mg + HC1 → MgCl 2 + H 2;

Zn+ HC1 → ZnCl 2 + H 2 ;

Br 2 + KI → KBr+ I 2 ;

KC1O 3 + S → KC1+ SO 2;

C1 2 + KBr → KC1 + Br 2;

SiO 2 + C → Si + CO;

SiO 2 + C → SiC + CO;

Mg + SiO 2 → Mg 2 Si + MgO

6. Summing up (2 min.).

So, today we became acquainted with the concept“equation of chemical reactions”, learned to place coefficients in these equations based on the law of conservation of mass.

What is a chemical reaction equation?

What is written on the right side of the equation? And on the left?

What does the "+" sign mean in an equation?

Why are coefficients placed in chemical reaction equations?

7.Homework.§ 27, ex. 1.3(pis.).

Lesson grades.

Handout:

Algorithm for arranging coefficients in chemical reaction equations

Order of operations	example
1. Determine the number of atomseach element on the left and right sides of the reaction diagram	A1 + O 2 → A1 2 O 3 A1-1 atom A1-2 atoms O-2 atoms 0-3 atoms
2. Among the elements with different numbers atoms on the left and right sides of the diagramchoose the one whose number of atoms is greater	O-2 atoms on the left O-3 atoms on the right
3. Find least common multiple(LCM) number of atoms of this element in the left parts of the equation and the number of atoms of that element on the right side of the equation	LCM = 6
4. Split the NOC by the number of atoms of this element in left parts of the equation, getcoefficient for left parts of the equation	6:2 = 3 Al + ZO 2 → Al 2 O 3
5. Split the NOC by the number of atoms of this element on the right parts of the equation, getcoefficient for right parts of the equation	6:3 = 2 A1 + ZO 2 → 2 A1 2 O 3
6. If the set coefficient has changed the number of atoms of another element, then repeat steps 3, 4, 5 again.	A1 + ZO 2 → 2 A1 2 O 3 A1 - 1 atom A1 - 4 atoms LCM = 4 4:1=4 4:4=1 4A1 + ZO 2 → 2 A1 2 O 3