A chemical equation is a record of a reaction using the symbols of the elements and the formulas of the compounds taking part in it. The relative amounts of reactants and products, expressed in moles, are indicated by numerical coefficients in the complete (balanced) reaction equation. These ratios are sometimes referred to as stoichiometric ratios. At present, there is an increasing tendency to include indications of the physical state of reactants and products in chemical equations. This is done using the following designations: (gas) or means a gaseous state, (-liquid, ) - solid, (-aqueous solution.

A chemical equation can be drawn up on the basis of experimentally established knowledge of the reactants and products of the reaction under study, as well as by measuring the relative amounts of each reactant and product that take part in the reaction.

Writing a chemical equation

Compiling a complete chemical equation involves the following four steps.

1st stage. Recording the reaction in verbal terms. For example,

2nd stage. Replacement of verbal names with formulas of reagents and products.

3rd stage. Equation balancing (determining its coefficients)

Such an equation is called balanced or stoichiometric. The need to balance the equation is dictated by the fact that in any reaction the law of conservation of matter must be fulfilled. In relation to the reaction we are considering as an example, this means that not a single atom of magnesium, carbon or oxygen can be formed in it or disappear. In other words, the number of atoms of each element on the left and right sides of a chemical equation must be the same.

4th stage. Indication of the physical state of each participant in the reaction.

Types of chemical equations

Consider the following complete equation:

This equation describes the entire reaction system as a whole. However, the reaction under consideration can also be represented in a simplified form using the ionic equation.

This equation does not include information about sulfate ions that are not listed in it because they do not take part in the reaction under consideration. Such ions are called observer ions.

The reaction between iron and copper (II) is an example of redox reactions (see Ch. 10). It can be conditionally divided into two reactions, one of which describes reduction, and the other, oxidation, occurring simultaneously in the overall reaction:

These two equations are called the half-reaction equations. They are especially often used in electrochemistry to describe processes occurring on electrodes (see Chap. 10).

Interpretation of chemical equations

Consider the following simple stoichiometric equation:

It can be interpreted in two ways. First, according to this equation, one mole of hydrogen molecules reacts with one mole of bromine molecules to form two moles of hydrogen bromide molecules. This interpretation of the chemical equation is sometimes called its molar interpretation.

However, this equation can also be interpreted in such a way that in the resulting reaction (see below) one hydrogen molecule reacts with one bromine molecule to form two hydrogen bromide molecules. Such an interpretation of a chemical equation is sometimes called its molecular interpretation.

Both molar and molecular interpretations are equally valid. However, it would be completely wrong to conclude on the basis of the equation of the reaction under consideration that one hydrogen molecule collides with one bromine molecule to form two hydrogen bromide molecules. The fact is that this reaction, like most others, is carried out in several successive stages. The totality of all these stages is usually called the reaction mechanism (see Chap. 9). In our example, the reaction includes the following steps:

Thus, the reaction under consideration is actually a chain reaction in which intermediates (intermediate reagents) called radicals participate (see Chapter 9). The mechanism of the reaction under consideration also includes other stages and side reactions. Thus, the stoichiometric equation only indicates the resulting reaction. It does not provide information about the reaction mechanism.

Calculations using chemical equations

Chemical equations are the starting point for a wide variety of chemical calculations. Here and later in the book, a number of examples of such calculations are given.

Calculation of the mass of reactants and products. We already know that a balanced chemical equation indicates the relative molar amounts of reactants and products involved in a reaction. These quantitative data allow the calculation of the masses of reactants and products.

Calculate the mass of silver chloride formed when an excess amount of sodium chloride solution is added to a solution containing 0.1 mol of silver in the form of ions

The first stage of all such calculations is to write the equation of the reaction under consideration: I

Since an excess amount of chloride ions is used in the reaction, it can be assumed that all the ions present in the solution are converted into. The reaction equation shows that one mole of ions is obtained from one mole. This allows you to calculate the mass of the formed as follows:

Hence,

Since g / mol, then

Determination of the concentration of solutions. Calculations based on stoichiometric equations underlie quantitative chemical analysis. As an example, consider the determination of the concentration of a solution from a known mass of the product formed in the reaction. This kind of quantitative chemical analysis is called gravimetric analysis.

An amount of potassium iodide solution was added to the nitrate solution, which is sufficient to precipitate all the lead in the form of iodide. The mass of the resulting iodide was 2.305 g. The volume of the initial nitrate solution was equal.

We have already encountered the equation of the reaction under consideration:

This equation shows that one mole of lead(II) nitrate is needed to produce one mole of iodide. Let us determine the molar amount of lead (II) iodide formed in the reaction. Insofar as

A chemical equation can be called a visualization of a chemical reaction using the signs of mathematics and chemical formulas. Such an action is a reflection of some kind of reaction, during which new substances appear.

Chemical tasks: types

A chemical equation is a sequence of chemical reactions. They are based on the law of conservation of mass of any substances. There are only two types of reactions:

• Compounds - these include (there is a replacement of atoms of complex elements with atoms of simple reagents), exchange (substitution of components of two complex substances), neutralization (the reaction of acids with bases, the formation of salt and water).
• Decomposition - the formation of two or more complex or simple substances from one complex, but their composition is simpler.

Chemical reactions can also be divided into types: exothermic (occur with the release of heat) and endothermic (absorption of heat).

This question worries many students. Here are some simple tips to help you learn how to solve chemical equations:

• Desire to understand and master. You can't deviate from your goal.
• Theoretical knowledge. Without them, it is impossible to compose even an elementary formula of the compound.
• The correctness of writing a chemical problem - even the slightest mistake in the condition will nullify all your efforts in solving it.

It is desirable that the process of solving chemical equations is exciting for you. Then the chemical equations (how to solve them and what points you need to remember, we will analyze in this article) will no longer be problematic for you.

Problems that are solved using the equations of chemical reactions

These tasks include:

• Finding the mass of a component given the mass of another reagent.
• Tasks for the combination "mass-mole".
• Calculations for the combination "volume-mole".
• Examples using the term "excess".
• Calculations using reagents, one of which is not devoid of impurities.
• Tasks for the decay of the result of the reaction and for production losses.
• Problems for finding a formula.
• Tasks where reagents are provided as solutions.
• Tasks containing mixtures.

Each of these types of tasks includes several subtypes, which are usually discussed in detail in the first school chemistry lessons.

Chemical Equations: How to Solve

There is an algorithm that helps to cope with almost any task from this difficult science. To understand how to solve chemical equations correctly, you need to follow a certain pattern:

• When writing the reaction equation, do not forget to set the coefficients.
• Determine how to find unknown data.
• The correctness of the application in the selected formula of proportions or the use of the concept of "amount of substance".
• Pay attention to units of measurement.

At the end, it is important to check the task. In the process of solving, you could make an elementary mistake that affected the result of the decision.

Basic rules for compiling chemical equations

If you follow the correct sequence, then the question of what chemical equations are, how to solve them, will not bother you:

• Formulas of substances that react (reagents) are written on the left side of the equation.
• The formulas of the substances that are formed as a result of the reaction are already written on the right side of the equation.

The formulation of the reaction equation is based on the law of conservation of mass of substances. Therefore, both sides of the equation must be equal, that is, with the same number of atoms. This can be achieved if the coefficients are correctly placed in front of the formulas of substances.

Arrangement of coefficients in a chemical equation

The algorithm for placing the coefficients is as follows:

• Count on the left and right side of the equation the atoms of each element.
• Determination of the changing number of atoms in an element. You also need to find N.O.K.
• Obtaining coefficients is achieved by dividing N.O.K. for indexes. Be sure to put these numbers in front of the formulas.
• The next step is to recalculate the number of atoms. Sometimes it becomes necessary to repeat an action.

The equalization of the parts of a chemical reaction occurs with the help of coefficients. Calculation of indexes is made through valency.

For the successful compilation and solution of chemical equations, it is necessary to take into account the physical properties of matter, such as volume, density, mass. You also need to know the state of the reacting system (concentration, temperature, pressure), understand the units of measurement of these quantities.

To understand the question of what chemical equations are, how to solve them, it is necessary to use the basic laws and concepts of this science. In order to successfully calculate such problems, it is also necessary to remember or master the skills of mathematical operations, to be able to perform actions with numbers. We hope that with our tips it will be easier for you to cope with chemical equations.

The main subject of comprehension in chemistry is the reactions between different chemical elements and substances. Great awareness of the validity of the interaction of substances and processes in chemical reactions makes it possible to manage them and apply them for their own purposes. A chemical equation is a method of expressing a chemical reaction, in which the formulas of the initial substances and products are written, indicators showing the number of molecules of any substance. Chemical reactions are divided into reactions of connection, substitution, decomposition and exchange. Also among them it is allowed to distinguish redox, ionic, reversible and irreversible, exogenous, etc.

Instruction

1. Determine which substances interact with each other in your reaction. Write them down on the left side of the equation. For example, consider the chemical reaction between aluminum and sulfuric acid. Arrange the reagents on the left: Al + H2SO4 Next, put an "equal" sign, as in a mathematical equation. In chemistry, you can find an arrow pointing to the right, or two oppositely directed arrows, a “sign of reversibility.” As a result of the interaction of a metal with an acid, a salt and hydrogen are formed. Write the reaction products after the equal sign, on the right. Al + H2SO4 \u003d Al2 (SO4) 3 + H2 The reaction scheme is obtained.

2. In order to write a chemical equation, you need to find the exponents. On the left side of the previously obtained scheme, sulfuric acid contains hydrogen, sulfur and oxygen atoms in a ratio of 2:1:4, on the right side there are 3 sulfur atoms and 12 oxygen atoms in the composition of the salt and 2 hydrogen atoms in the H2 gas molecule. On the left side, the ratio of these 3 elements is 2:3:12.

3. In order to equalize the number of sulfur and oxygen atoms in the composition of aluminum (III) sulfate, put the indicator 3 on the left side of the equation in front of the acid. Now there are six hydrogen atoms on the left side. In order to equalize the number of hydrogen elements, put the indicator 3 in front of it on the right side. Now the ratio of atoms in both parts is 2:1:6.

4. It remains to equalize the number of aluminum. Because the salt contains two metal atoms, put a 2 in front of aluminum on the left side of the diagram. As a result, you will get the reaction equation for this scheme. 2Al + 3H2SO4 \u003d Al2 (SO4) 3 + 3H2

A reaction is the transformation of one chemical into another. And the formula for writing them with the help of special symbols is the equation of this reaction. There are different types of chemical interactions, but the rule for writing their formulas is identical.

You will need

• periodic system of chemical elements D.I. Mendeleev

Instruction

1. The initial substances that react are written on the left side of the equation. They are called reagents. The recording is made with the help of special symbols that denote any substance. A plus sign is placed between reagent substances.

2. On the right side of the equation, the formula of the resulting one or more substances is written, which are called reaction products. Instead of an equal sign, an arrow is placed between the left and right sides of the equation, which indicates the direction of the reaction.

3. Later, writing the formulas of the reactants and reaction products, you need to arrange the indicators of the reaction equation. This is done so that, according to the law of conservation of mass of matter, the number of atoms of the same element in the left and right parts of the equation remains identical.

4. In order to correctly arrange the indicators, you need to make out any of the substances that enter into the reaction. To do this, one of the elements is taken and the number of its atoms on the left and right is compared. If it is different, then it is necessary to find a multiple of the numbers denoting the number of atoms of a given substance in the left and right parts. After that, this number is divided by the number of atoms of the substance in the corresponding part of the equation, and an indicator is obtained for any of its parts.

5. Since the indicator is placed in front of the formula and applies to each substance included in it, the next step will be to compare the data obtained with the number of another substance that is part of the formula. This is carried out in the same way as with the first element and taking into account the existing indicator for each formula.

6. Later, after all the elements of the formula have been parsed, a final check of the correspondence of the left and right parts is carried out. Then the reaction equation can be considered complete.

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Note!
In the equations of chemical reactions, it is impossible to swap the left and right sides. Otherwise, a scheme of a completely different process will turn out.

Helpful advice
The number of atoms of both individual reagent substances and substances that make up the reaction products is determined using the periodic system of chemical elements of D.I. Mendeleev

How unsurprising nature is for a person: in winter it wraps the earth in a snowy duvet, in spring it reveals everything that is alive, like popcorn flakes, in summer it rages with a riot of colors, in autumn it sets plants on fire with red fire ... And only if you think about it and look closely, you can see what are standing Behind all these habitual changes are difficult physical processes and CHEMICAL REACTIONS. And in order to study all living things, you need to be able to solve chemical equations. The main requirement for equalizing chemical equations is the knowledge of the law of conservation of the number of matter: 1) the number of matter before the reaction is equal to the number of matter after the reaction; 2) the total number of substances before the reaction is equal to the total number of substances after the reaction.

Instruction

1. In order to equalize the chemical "example" you need to follow a few steps. Write down the equation reactions in general. For this, unknown indicators in front of the formulas of substances are denoted by the letters of the Latin alphabet (x, y, z, t, etc.). Let it be required to equalize the reaction of the combination of hydrogen and oxygen, as a result of which water will be obtained. Before the molecules of hydrogen, oxygen and water, put the Latin letters (x, y, z) - indicators.

2. For any element, on the basis of physical equilibrium, compose mathematical equations and obtain a system of equations. In this example, for hydrogen on the left, take 2x, because it has the index “2”, on the right - 2z, tea also has the index “2”, it turns out 2x=2z, otsel, x=z. For oxygen, take 2y on the left, because there is an index “2”, on the right - z, there is no index for tea, which means it is equal to one, which is usually not written. It turns out, 2y=z, and z=0.5y.

Note!
If a greater number of chemical elements are involved in the equation, then the task does not become more complicated, but increases in volume, which should not be frightened.

Helpful advice
It is also possible to equalize reactions with the help of probability theory, using the valencies of chemical elements.

Tip 4: How to compose a redox reaction

Redox reactions are reactions with a change in oxidation states. It often happens that the initial substances are given and it is necessary to write the products of their interaction. Occasionally, the same substance can give different final products in different environments.

Instruction

1. Depending not only on the reaction medium, but also on the degree of oxidation, the substance behaves differently. A substance in its highest oxidation state is invariably an oxidizing agent, and in its lowest oxidation state it is a reducing agent. In order to make an acidic environment, sulfuric acid (H2SO4) is traditionally used, less often nitric acid (HNO3) and hydrochloric acid (HCl). If necessary, create an alkaline environment, use sodium hydroxide (NaOH) and potassium hydroxide (KOH). Let's take a look at some examples of substances.

2. MnO4(-1) ion. In an acidic environment, it turns into Mn (+2), a colorless solution. If the medium is neutral, then MnO2 is formed, a brown precipitate forms. In an alkaline medium, we obtain MnO4 (+2), a green solution.

3. Hydrogen peroxide (H2O2). If it is an oxidizing agent, i.e. accepts electrons, then in neutral and alkaline media it turns according to the scheme: H2O2 + 2e = 2OH (-1). In an acidic environment, we get: H2O2 + 2H(+1) + 2e = 2H2O. Provided that hydrogen peroxide is a reducing agent, i.e. donates electrons; in an acidic medium, O2 is formed; in an alkaline medium, O2 + H2O. If H2O2 enters an environment with a strong oxidizing agent, it will itself be a reducing agent.

4. The Cr2O7 ion is an oxidizing agent; in an acidic environment, it turns into 2Cr(+3), which are green in color. From the Cr(+3) ion in the presence of hydroxide ions, i.e. in an alkaline medium, yellow CrO4(-2) is formed.

5. Let's give an example of the composition of the reaction. KI + KMnO4 + H2SO4 - In this reaction, Mn is in its highest oxidation state, that is, it is an oxidizing agent, accepting electrons. The environment is acidic, sulfuric acid (H2SO4) shows us this. The reducing agent here is I (-1), it donates electrons, while increasing its oxidation state. We write down the reaction products: KI + KMnO4 + H2SO4 - MnSO4 + I2 + K2SO4 + H2O. We arrange the indicators using the electronic equilibrium method or the half-reaction method, we get: 10KI + 2KMnO4 + 8H2SO4 = 2MnSO4 + 5I2 + 6K2SO4 + 8H2O.

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Note!
Don't forget to add indicators to your reactions!

Chemical reactions are the interaction of substances, accompanied by a change in their composition. In other words, the substances entering into the reaction do not correspond to the substances resulting from the reaction. A person encounters similar interactions hourly, every minute. Tea processes occurring in his body (respiration, protein synthesis, digestion, etc.) are also chemical reactions.

Instruction

1. Any chemical reaction must be written correctly. One of the main requirements is that the number of atoms of the entire element of substances on the left side of the reaction (they are called “initial substances”) corresponds to the number of atoms of the same element in the substances on the right side (they are called “reaction products”). In other words, the record of the reaction must be equalized.

2. Let's look at a specific example. What happens when a gas burner is lit in the kitchen? Natural gas reacts with oxygen in the air. This oxidation reaction is so exothermic, that is, accompanied by the release of heat, that a flame appears. With the support of which you either cook food or heat up already cooked food.

3. For simplicity, assume that natural gas consists of only one of its components - methane, which has the formula CH4. Because how to compose and equalize this reaction?

4. When carbon-containing fuels are burned, that is, when carbon is oxidized by oxygen, carbon dioxide is formed. You know his formula: CO2. What is formed when hydrogen contained in methane is oxidized with oxygen? Definitely water in the form of steam. Even the most distant person from chemistry knows its formula by heart: H2O.

5. It turns out that write down the initial substances on the left side of the reaction: CH4 + O2. On the right side, respectively, there will be reaction products: CO2 + H2O.

6. Advance recording of this chemical reaction will be further: CH4 + O2 = CO2 + H2O.

7. Equalize the above reaction, that is, achieve the basic rule: the number of atoms of the entire element in the left and right parts of the chemical reaction must be identical.

8. You can see that the number of carbon atoms is the same, but the number of oxygen and hydrogen atoms is different. There are 4 hydrogen atoms on the left side, and only 2 on the right side. Therefore, put the indicator 2 in front of the water formula. Get: CH4 + O2 \u003d CO2 + 2H2O.

9. The carbon and hydrogen atoms are equalized, now it remains to do the same with oxygen. There are 2 oxygen atoms on the left side, and 4 on the right. Putting the indicator 2 in front of the oxygen molecule, you will get the final record of the methane oxidation reaction: CH4 + 2O2 = CO2 + 2H2O.

A reaction equation is a conditional record of a chemical process in which some substances are converted into others with a change in properties. To record chemical reactions, formulas of substances and skills about the chemical properties of compounds are used.

Instruction

1. Write the formulas correctly according to their names. Let's say aluminum oxide Al? O?, index 3 from aluminum (corresponding to its oxidation state in this compound) put near oxygen, and index 2 (oxidation state of oxygen) near aluminum. If the oxidation state is +1 or -1, then the index is not set. For example, you need to write down the formula for ammonium nitrate. Nitrate is the acid residue of nitric acid (-NO?, s.o. -1), ammonium (-NH?, s.o. +1). So the formula for ammonium nitrate is NH? NO?. Occasionally, the oxidation state is indicated in the name of the compound. Sulfur oxide (VI) - SO?, silicon oxide (II) SiO. Some primitive substances (gases) are written with index 2: Cl?, J?, F?, O?, H? etc.

2. You need to know which substances are reacting. Visible signs of reaction: gas evolution, color metamorphosis and precipitation. Quite often the reactions pass without visible changes. Example 1: neutralization reaction H?SO? + 2 NaOH? Na?SO? + 2 H?O Sodium hydroxide reacts with sulfuric acid to form a soluble salt of sodium sulfate and water. The sodium ion is split off and combined with the acid residue, replacing the hydrogen. The reaction proceeds without external signs. Example 2: iodoform test С?H?OH + 4 J? + 6 NaOH?CHJ?? + 5 NaJ + HCOONa + 5 H?O The reaction proceeds in several stages. The final result is the precipitation of yellow iodoform crystals (good reaction to alcohols). Example 3: Zn + K?SO? ? The reaction is unthinkable, because in a series of metal stresses, zinc is later than potassium and cannot displace it from compounds.

3. The law of conservation of mass states that the mass of the reactants is equal to the mass of the formed substances. A competent record of a chemical reaction is half the furore. You need to set up indicators. Start equalizing with those compounds in the formulas of which there are large indices. K?Cr?O? + 14 HCl? 2CrCl? + 2 KCl + 3 Cl?? + 7 H?O its formula contains the largest index (7). Such accuracy in recording reactions is needed to calculate mass, volume, concentration, released energy, and other quantities. Be careful. Remember especially common formulas of acids and bases, as well as acid residues.

Tip 7: How to Determine Redox Equations

A chemical reaction is a process of reincarnation of substances that occurs with a change in their composition. Those substances that enter into the reaction are called initial, and those that are formed as a result of this process are called products. It happens that in the course of a chemical reaction, the elements that make up the initial substances change their oxidation state. That is, they can accept other people's electrons and give their own. In both cases, their charge changes. Such reactions are called redox reactions.

Instruction

1. Write down the exact equation for the chemical reaction you are considering. Look at what elements are included in the composition of the initial substances, and what are the oxidation states of these elements. Later, compare these figures with the oxidation states of the same elements on the right side of the reaction.

2. If the oxidation state has changed, this reaction is redox. If the oxidation states of all the elements remained the same, then no.

3. Here, for example, is the widely known good quality reaction for the detection of the sulfate ion SO4 ^2-. Its essence is that barium sulfate, which has the formula BaSO4, is virtually insoluble in water. When formed, it immediately precipitates in the form of a dense, heavy white precipitate. Write down some equation for a similar reaction, say, BaCl2 + Na2SO4 = BaSO4 + 2NaCl.

4. It turns out that from the reaction you see that in addition to the precipitate of barium sulfate, sodium chloride was formed. Is this reaction a redox reaction? No, it is not, because not a single element that is part of the initial substances has changed its oxidation state. Both on the left and on the right side of the chemical equation, barium has an oxidation state of +2, chlorine -1, sodium +1, sulfur +6, oxygen -2.

5. And here is the reaction Zn + 2HCl = ZnCl2 + H2. Is it redox? Elements of initial substances: zinc (Zn), hydrogen (H) and chlorine (Cl). See what their oxidation states are? For zinc, it is equal to 0 as in any simple substance, for hydrogen it is +1, for chlorine it is -1. And what are the oxidation states of these same elements in the right side of the reaction? In chlorine, it remained unshakable, that is, equal to -1. But for zinc it became equal to +2, and for hydrogen - 0 (from the fact that hydrogen was released in the form of a simple substance - gas). Therefore, this reaction is a redox reaction.

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The canonical equation of an ellipse is compiled from those considerations that the sum of the distances from any point of the ellipse to 2 of its foci is invariably continuous. By fixing this value and moving the point along the ellipse, it is possible to determine the equation of the ellipse.

You will need

• Sheet of paper, ballpoint pen.

Instruction

1. Specify two fixed points F1 and F2 on the plane. Let the distance between the points be equal to some fixed value F1F2= 2s.

2. Draw a straight line on a piece of paper, which is the coordinate line of the abscissa axis, and draw the points F2 and F1. These points are the foci of the ellipse. The distance from the entire focus point to the origin must be the same value, c.

3. Draw the y-axis, thus forming a Cartesian coordinate system, and write the basic equation that defines the ellipse: F1M + F2M = 2a. The M point represents the current point of the ellipse.

4. Determine the value of the segments F1M and F2M using the Pythagorean theorem. Keep in mind that point M has current coordinates (x, y) relative to the origin, and regarding, say, point F1, point M has coordinates (x + c, y), that is, the “x” coordinate acquires a shift. Thus, in the expression of the Pythagorean theorem, one of the terms must be equal to the square of the value (x + c), or the value (x-c).

5. Substitute the expressions for the moduli of the vectors F1M and F2M into the basic ratio of the ellipse and square both sides of the equation, moving one of the square roots to the right side of the equation in advance and opening the brackets. After reducing the identical terms, divide the resulting ratio by 4a and again raise to the second power.

6. Give similar terms and collect terms with the same factor of the square of the "x" variable. Take out the square of the "X" variable.

7. Take the square of some quantity (say b) to be the difference between the squares of a and c, and divide the resulting expression by the square of this new quantity. Thus, you have obtained the canonical equation of an ellipse, on the left side of which is the sum of the squares of the coordinates divided by the magnitudes of the axes, and on the left side is one.

Helpful advice
In order to check the performance of the task, you can use the law of conservation of mass.

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 all got wet, as they say, “to the skin”, then is this not a transformation: after all, the clothes were dry, but became wet.

If, for example, you take an iron nail, process it with a file, and then assemble iron filings (Fe) , then this is also not a transformation: there was a nail - it became powder. But if after that to assemble the device and hold obtaining oxygen (O 2): heat up potassium permanganate(KMpo 4) and collect oxygen in a test tube, and then place these iron filings red-hot "to red" in it, then they will flare up with a bright flame and after combustion 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 state of clothing (dry, wet) change, these are not transformations. The fact is that the nail itself, as it was a substance (iron), remained so, despite its different form, and our clothes soaked up the water from the rain, and then it evaporated into the atmosphere. The water itself has not changed. So what are transformations in terms of chemistry?

From the point of view of chemistry, transformations are such phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It does not matter what form it took after being filed, but after being collected from it iron filings placed in an atmosphere of oxygen - it turned into iron oxide(Fe 2 O 3 ) . So, has something really changed? Yes, it has. There was a nail substance, 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 a person uninitiated in chemistry, questions immediately arise. What is the "molecular equation", what is Fe? Why are there numbers "4", "3", "2"? What are the small numbers "2" and "3" in the formula Fe 2 O 3? This means that the time has come to sort things out in order.

Signs of chemical elements.

Despite the fact that they begin to study chemistry 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 "Mendeleev's Table".

In this table, in the appropriate order, the elements are located. 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 hesitation, identifying them with objects: an iron bolt, aluminum wire, oxygen in the atmosphere, a golden ring, etc. etc. But in fact, all these substances (bolt, wire, ring) consist of their respective elements. The whole paradox is that the element cannot be touched, 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, however, as in other sciences, for calculations, compiling equations, and solving problems. Each element differs from the other in that it is characterized by its own 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 #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 number 3. Its atom consists of 3 protons and 3 electrons. Darmstadtium - element number 110. Its atom consists of 110 protons and 110 electrons.

Each element is denoted by a certain symbol, Latin letters, and has a certain reading in translation 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". Etc. All these designations can be found in any chemistry textbook for the 8th grade. For us now, the main thing is to understand that when compiling 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 iron and sulfur substances interact, then the equation will take the following 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). And you should pay attention
special attention to the fact that all metals are indicated by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals - either by 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 be of great importance in the formulation of equations. It is not at all difficult to guess that complex substances are substances formed from atoms of different types, for example,

one). Oxides:
aluminium oxide Al 2 O 3,

sodium oxide Na 2 O
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 potassium alkali 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 Na 2 SO 4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl 2

5). organic matter:
sodium acetate CH 3 COOHa,
methane CH 4,
acetylene C 2 H 2,
glucose C 6 H 12 O 6

Finally, after we have clarified 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 are almost the very essence of this science. For example, you can give such 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 number of atoms, the 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 record of a chemical reaction using chemical formulas and mathematical signs. A chemical equation inherently reflects a particular 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 that 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:

ВаСl 2 + H 2 SO 4 = BaSO 4 + 2НCl (3)

First of all, it is necessary to understand that the large number “2” in front of the HCl substance is called the coefficient, and the small numbers “2”, “4” under the formulas ВаСl 2, H 2 SO 4 , BaSO 4 are called indices. Both the coefficients and indices in chemical equations play the role of factors, not terms. In order to correctly write a chemical equation, it is necessary arrange the 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). Whence it follows that on the right side of the equation the number of hydrogen and chlorine atoms is half that 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 the number of atoms of the elements involved in this reaction, both on the left and on the right, we get the following balance:

In both parts of the equation, the number of atoms of the elements participating in the reaction are equal, therefore it is correct.

Chemical equation and chemical reactions

As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are such phenomena in the process of which the transformation of one substance into another occurs. Among their diversity, two main types can be distinguished:

one). Connection reactions
2). decomposition reactions.

The vast majority of chemical reactions belong to addition reactions, since changes in its composition can rarely occur with a single substance if it is not subjected to external influences (dissolution, heating, light). Nothing characterizes a chemical phenomenon, or reaction, as much as the changes that occur when two or more substances interact. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, color change, sedimentation, release of gaseous products, noise.

For clarity, we present several equations that reflect the processes of compound reactions, during which we obtain sodium chloride(NaCl), zinc chloride(ZnCl 2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)

Cl 2 + 2Nа = 2NaCl (4)

CuCl 2 + Zn \u003d ZnCl 2 + Cu (5)

AgNO 3 + KCl \u003d AgCl + 2KNO 3 (6)

3HCl + Al(OH) 3 \u003d AlCl 3 + 3H 2 O (7)

Among the reactions of the compound, the following should be especially noted : substitution (5), exchange (6), and as a special case of the exchange reaction, the reaction neutralization (7).

Substitution reactions include those in which atoms of a simple substance replace the 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 ZnCl 2 salt, and copper is released from the solution in the metallic state.

Exchange reactions are those reactions in which two complex substances exchange their constituents. In the case of reaction (6), the soluble salts of AgNO 3 and KCl, when both solutions are drained, form an insoluble precipitate of the AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are attached to NO 3 anions, and silver cations Ag + - to Cl - anions.

A special, particular case of exchange reactions is the neutralization reaction. Neutralization reactions are reactions in which acids react with bases to form salt and water. In example (7), hydrochloric acid HCl reacts with base Al(OH) 3 to form AlCl 3 salt and water. In this case, aluminum cations Al 3+ from the base are exchanged with Cl anions - from the acid. As a result, it happens hydrochloric acid neutralization.

Decomposition reactions include those in which two or more new simple or complex substances, but of a simpler composition, are formed from one complex one. As reactions, one can cite those in the process of which 1) decompose. potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate is formed (K 2 MnO 4), manganese oxide(MnO 2) and oxygen (O 2); 3). calcium carbonate or marble; in the process are formed carbonicgas(CO 2) and calcium oxide(Cao)

2KNO 3 \u003d 2KNO 2 + O 2 (8)
2KMnO 4 \u003d K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 \u003d CaO + CO 2 (10)

In reaction (8), one complex and one simple substance is 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 undergo decomposition:

one). 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 \u003d SO 3 + H 2 O (13)

4). Salts: calcium carbonate CaCO 3 \u003d CaO + CO 2 (14)

5). organic matter: alcoholic fermentation of glucose

C 6 H 12 O 6 \u003d 2C 2 H 5 OH + 2CO 2 (15)

According to another classification, all chemical reactions can be divided into two types: reactions that take place with the release of heat, they are called exothermic, and reactions that go 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. interactions with oxygen methane combustion:

CH 4 + 2O 2 \u003d 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 during the reaction (+Q) or absorbed (-Q):

CaCO 3 \u003d 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 \u003d 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 \u003d 2Mg +2 O -2

These types of reactions are redox . They will be considered separately. To formulate equations for reactions of this type, it is necessary to use half-reaction method and apply electronic balance equation.

After bringing the various types of chemical reactions, you can proceed to the principle of compiling chemical equations, in other words, the selection of coefficients in their left and right parts.

Mechanisms for compiling chemical equations.

Whatever type this or that chemical reaction belongs to, its record (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 adjustment, 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 parts of the equation. What principles should be followed in such cases? Is there any system in the selection of coefficients? There is, and not one. These systems include:

one). Selection of coefficients according to given formulas.

2). Compilation according to the valencies of the reactants.

3). Compilation according to the oxidation states of the reactants.

In the first case, it is assumed that we know the formulas of the reactants 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 the equality between the atoms of the elements before and after the reaction is established, the equal sign (=) is not put in the equation, but is replaced by an arrow (→). Now let's get down to the actual balancing. 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). It is not necessary to equalize it by the number of nitrogen atoms, but by oxygen it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were three atoms. Let's make the following diagram:

before reaction after reaction
O 2 O 3

Let's define the smallest multiple between the given numbers of atoms, it will be "6".

O 2 O 3
\ 6 /

Divide this number on the left side of the oxygen equation by "2". We get the number "3", 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", just put it in the equation to be solved:

N 2 + 3O 2 → 2N 2 O 3

The number of oxygen atoms in both the left and right parts of the equation became equal, respectively, 6 atoms:

But the number of nitrogen atoms in both sides of the equation will not match:

On the left side there are two atoms, on the right side there are four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation, putting the coefficient "2":

Thus, the equality for nitrogen is observed and, in general, the equation will take the form:

2N 2 + 3O 2 → 2N 2 O 3

Now in the equation, instead of an arrow, you can put an equal sign:

2N 2 + 3O 2 \u003d 2N 2 O 3 (20)

Let's take 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). It is not necessary to equalize it by the number of phosphorus atoms, but for chlorine it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were five atoms. Let's make the following diagram:

before reaction after reaction
Cl 2 Cl 5

Let's define 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 equation for chlorine by "2". We get the number "5", put it in the equation to be solved:

Р + 5Cl 2 → РCl 5

We also divide the number "10" for the right side of the equation by "5". We get the number "2", just put it in the equation to be solved:

Р + 5Cl 2 → 2РCl 5

The number of chlorine atoms in both the left and right parts of the equation became equal, respectively, 10 atoms:

But the number of phosphorus atoms in both sides of the equation will not match:

Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation, putting the coefficient "2":

Thus, the equality for phosphorus is observed and, in general, the equation will take the form:

2Р + 5Cl 2 = 2РCl 5 (21)

When writing equations by valency must be given definition of valency and set values ​​for the most famous elements. Valence is one of the previously used concepts, currently not used in a number of school programs. But with its help it is easier to explain the principles of compiling equations of chemical reactions. By valency is meant the number of chemical bonds that an atom can form with another, or other atoms . Valence has no 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 apply them in the preparation of chemical equations? The numerical values ​​of the valencies of the elements coincide with their group number of the Periodic system of chemical elements of D. I. Mendeleev (Table 1).

For other elements valency values may have other values, but never greater than the number of the group in which they are located. Moreover, for even numbers of groups (IV and VI), the valencies 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 the general principle of compiling chemical equations for given valences for certain elements. Most often, this method is acceptable in the case of compiling equations for chemical reactions of combining simple substances, for example, when interacting with oxygen ( oxidation reactions). Suppose you want to display the oxidation reaction aluminum. But recall that metals are denoted by single atoms (Al), and non-metals that are in a gaseous state - with indices "2" - (O 2). First, we write the general scheme of the reaction:

Al + O 2 → AlO

At this stage, it is not yet known what the correct spelling should be for alumina. And it is precisely at this stage that knowledge of the valencies of the elements will come to our aid. For aluminum and oxygen, we put them above the proposed formula for this oxide:

IIIII
Al O

After that, "cross"-on-"cross" these symbols of the elements will put the corresponding indices below:

IIIII
Al 2 O 3

Composition of a chemical compound Al 2 O 3 determined. The further scheme of the reaction equation will take the form:

Al + O 2 → Al 2 O 3

It remains only to equalize the left and right parts of it. We proceed in the same way as in the case of formulating equation (19). We equalize the number of oxygen atoms, resorting to finding the smallest multiple:

before reaction after reaction

O 2 O 3
\ 6 /

Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in the equation to be solved. We also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:

Al + 3O 2 → 2Al 2 O 3

In order to achieve equality for aluminum, it is necessary to adjust its amount on the left side of the equation by setting the coefficient "4":

4Al + 3O 2 → 2Al 2 O 3

Thus, the equality for aluminum and oxygen is observed and, in general, the equation will take the final form:

4Al + 3O 2 \u003d 2Al 2 O 3 (22)

Using the valency method, it is possible to predict which substance is formed in the course of a chemical reaction, what its formula will look like. Suppose nitrogen and hydrogen with the corresponding valences III and I entered into the reaction of the compound. Let's write the general reaction scheme:

N 2 + H 2 → NH

For nitrogen and hydrogen, we put down the valencies over the proposed formula of this compound:

As before, "cross"-on-"cross" for these element symbols, we put the corresponding indices below:

III I
N H 3

The further scheme of the reaction equation will take the form:

N 2 + H 2 → NH 3

Equalizing in the already known way, through the smallest multiple for hydrogen, equal to "6", we obtain the desired coefficients, and the equation as a whole:

N 2 + 3H 2 \u003d 2NH 3 (23)

When compiling equations for oxidation states reacting substances, it must be recalled that the degree of oxidation of an element is the number of electrons received or given away in the process of a chemical reaction. The oxidation state in compounds basically, numerically coincides with the values ​​of the element's valences. 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 valencies 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 commonly used in equations are shown in Table 3.

In the case of compound reactions, the principle of compiling equations in terms of oxidation states is the same as in compiling in terms of valencies. For example, let's give the reaction equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write the proposed equation:

Cl 2 + O 2 → ClO

We put the oxidation states of the corresponding atoms over the proposed ClO compound:

As in the previous cases, we establish that the desired 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

Equalizing for oxygen, finding the smallest multiple between two and seven, equal to "14", we finally establish the equality:

2Cl 2 + 7O 2 \u003d 2Cl 2 O 7 (24)

A slightly different method must be used with oxidation states when compiling 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 do you know what happens in a reaction?

Indeed, how do you know: what reaction products can arise in the course of a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?

Ba (NO 3) 2 + K 2 SO 4 →?

Maybe VAC 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, during this reaction, compounds are formed: BaSO 4 and KNO 3. And how is this known? And how to write formulas of substances? Let's start with what is most often overlooked: the very concept of "exchange reaction". This means that in these reactions, the substances change with each other in constituent parts. Since the exchange reactions are mostly carried out between bases, acids or salts, the parts with which they will change 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, the exchange reaction can be given in the following notation:

Kt1An1 + Kt2An1 = Kt1An2 + Kt2An1 (25)

Where Kt1 and Kt2 are the metal cations (1) and (2), and An1 and An2 are the anions (1) and (2) corresponding to them. In this case, it must be taken into account that in compounds before the reaction and after the reaction, cations are always installed in the first place, and anions - in the second. Therefore, if it reacts potassium chloride and silver nitrate, both in solution

KCl + AgNO 3 →

then in the process of it substances KNO 3 and AgCl are formed and the corresponding equation will take the form:

KCl + AgNO 3 \u003d KNO 3 + AgCl (26)

In neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):

HCl + KOH \u003d KCl + H 2 O (27)

The oxidation states of metal cations and the charges of anions of acid residues are indicated in the table of the solubility of substances (acids, salts and bases in water). Metal cations are shown horizontally, and anions of acid residues are shown vertically.

Based on this, when compiling the exchange reaction equation, it is first necessary to establish the oxidation states of the particles receiving in this chemical process on its left side. For example, you need to write an equation for the interaction between calcium chloride and sodium carbonate. Let's draw up the initial scheme for this reaction:

CaCl + NaCO 3 →

Ca 2+ Cl - + Na + CO 3 2- →

Having performed the already known “cross”-to-“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 establish the preliminary formulas of the substances formed during the reaction:

CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl

We put down the corresponding charges over their cations and anions:

Ca 2+ CO 3 2- + Na + Cl -

Substance formulas are written correctly, in accordance with the charges of cations and anions. Let's make a complete equation by equating the left and right parts of it in terms of sodium and chlorine:

CaCl 2 + Na 2 CO 3 \u003d CaCO 3 + 2NaCl (28)

As another example, here is the equation for the neutralization reaction between barium hydroxide and phosphoric acid:

VaON + NPO 4 →

We put the corresponding charges over cations and anions:

Ba 2+ OH - + H + RO 4 3- →

Let's define the real formulas of the starting substances:

Va (OH) 2 + H 3 RO 4 →

Based on the principle of exchange of cations and anions (25), we establish the preliminary formulas of the substances formed during the reaction, taking into account that in the exchange reaction, one of the substances must necessarily be water:

Ba (OH) 2 + H 3 RO 4 → Ba 2+ RO 4 3- + H 2 O

Let's determine the correct record of the formula of the salt formed during the reaction:

Ba (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

Equate the left side of the equation for barium:

3VA (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

Since on the right side of the equation the residue of phosphoric acid is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:

3VA (OH) 2 + 2H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

It remains to match the number of hydrogen and oxygen atoms on the right side of the water. Since the total number of hydrogen atoms on the left is 12, on the right it must also correspond to twelve, therefore, before the formula of water, it is necessary put a coefficient"6" (since there are already 2 hydrogen atoms in the water molecule). For oxygen, equality is also observed: on the left 14 and on the right 14. So, the equation has the correct form of writing:

3Ва (ОН) 2 + 2Н 3 РО 4 → Ва 3 (РО 4) 2 + 6Н 2 O (29)

Possibility of chemical reactions

The world is made up 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, assert that a chemical reaction will correspond to it? There is a misconception that if the right arrange odds in the equation, then it will be feasible in practice. For example, if we take sulfuric acid solution and drop into it zinc, then we can observe the process of hydrogen evolution:

Zn + H 2 SO 4 \u003d ZnSO 4 + H 2 (30)

But if copper is lowered 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 \u003d CuSO 4 + SO 2 + 2H 2 O (31)

In reaction (23) between nitrogen and hydrogen gases, thermodynamic balance, those. how many molecules ammonia NH 3 is formed per unit time, the same number of them will decompose back into nitrogen and hydrogen. Shift in chemical equilibrium can be achieved by increasing the pressure and decreasing the temperature

N 2 + 3H 2 \u003d 2NH 3

If you take potassium hydroxide solution and pour on it 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 attributed to a substitution reaction:

NaCl + Br 2 ≠

What are the reasons for such discrepancies? The fact is that it is not enough just to correctly define compound formulas, you need to know the specifics of the interaction of metals with acids, skillfully use the table of solubility of substances, know the rules of substitution in the series of activity of metals and halogens. This article outlines only the most basic principles of how arrange the coefficients in the reaction equations, as write molecular equations, as determine the composition of a chemical compound.

Chemistry, as a science, is extremely diverse and multifaceted. This article reflects only a small part of the processes taking place in the real world. Types, thermochemical equations, electrolysis, organic synthesis processes and much, much more. But more on that in future articles.

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To characterize a certain chemical reaction, it is necessary to be able to make a record that will display the conditions for the course of a chemical reaction, show which substances have reacted and which have formed. For this, schemes of chemical reactions are used.

Scheme of a chemical reaction- a conditional record showing which substances enter into the reaction, which reaction products are formed, as well as the conditions for the reaction to take place. Consider, as an example, the reaction of the interaction of coal and oxygen. Scheme this reaction is written as follows:

C + O2 → CO2

coal reacts with oxygen to form carbon dioxide

Carbon and oxygen- in this reaction, the reagents, and the resulting carbon dioxide is the product of the reaction. Sign " → ” denotes the progress of the reaction. Often the conditions under which the reaction occurs are written above the arrow.

• Sign « t° → » means that the reaction proceeds when heated.
• Sign "R →" stands for pressure
• Sign «hv→»- that the reaction proceeds under the influence of light. Also above the arrow may indicate additional substances involved in the reaction.
• For example, "O2 →". If a gaseous substance is formed as a result of a chemical reaction, then in the reaction scheme, after the formula of this substance, the sign “ → ". If a precipitate forms during the course of the reaction, it is indicated by the sign " → ».
• For example, when chalk powder is heated (it contains a substance with the chemical formula CaCO3), two substances are formed: quicklime CaO and carbon dioxide. The reaction scheme is written as follows:

СaCO3 t° → CaO + CO2

So, natural gas, mainly consists of methane CH4, when it is heated to 1500 ° C, it turns into two other gases: hydrogen H2 and acetylene C2H2. The reaction scheme is written as follows:

CH4 t° → C2H2 + H2.

It is important not only to be able to draw up schemes of chemical reactions, but also to understand what they mean. Consider another reaction scheme:

H2O electric current → H2 + O2

This scheme means that under the influence of an electric current, water decomposes into two simple gaseous substances: hydrogen and oxygen. The scheme of a chemical reaction is a confirmation of the law of conservation of mass and shows that chemical elements do not disappear during a chemical reaction, but only rearrange into new chemical compounds.

Chemical reaction equations

According to the law of conservation of mass, the initial mass of the products is always equal to the mass of the obtained reagents. The number of atoms of elements before and after the reaction is always the same, the atoms only rearrange and form new substances. Let's go back to the reaction schemes written earlier:

СaCO3 t° → CaO + CO2

C + O2 CO2.

In these reaction schemes, the sign " → ” can be replaced with the “=” sign, since it is clear that the number of atoms before and after the reactions is the same. The entries will look like this:

СaCO3 = CaO + CO2

C + O2 = CO2.

It is these records that are called the equations of chemical reactions, that is, they are records of reaction schemes in which the number of atoms before and after the reaction is the same.

chemical reaction equation- conditional record of a chemical reaction by means of chemical formulas, which corresponds to the law of conservation of the mass of a substance

If we consider the other schemes of equations given earlier, we can see that on At first glance, the law of conservation of mass is not fulfilled in them:

CH4 t° → C2H2 + H2.

It can be seen that on the left side of the diagram, there is one carbon atom, and on the right side there are two. Hydrogen atoms are equally divided and there are four of them in the left and right parts. Let's turn this diagram into an equation. For this it is necessary equalize the number of carbon atoms. Equalize chemical reactions using coefficients that are written in front of the formulas of substances. Obviously, in order for the number of carbon atoms to become the same on the left and right, on the left side of the diagram, in front of the methane formula, it is necessary to put coefficient 2:

2CH4 t° → C2H2 + H2

It can be seen that the carbon atoms on the left and right are now equally divided, two each. But now the number of hydrogen atoms is not the same. On the left side of their equation 2∙4 = 8. There are 4 hydrogen atoms on the right side of the equation (two of them in the acetylene molecule, and two more in the hydrogen molecule). If you put a coefficient in front of acetylene, the equality of carbon atoms will be violated. We put a coefficient 3 in front of the hydrogen molecule:

2CH4 = C2H2 + 3H2

Now the number of carbon and hydrogen atoms on both sides of the equation is the same. The law of conservation of mass is fulfilled! Let's consider another example. reaction scheme Na + H2O → NaOH + H2 needs to be converted into an equation. In this scheme, the number of hydrogen atoms is different. There are two on the left and two on the right three atoms. Put a factor of 2 before NaOH.

Na + H2O → 2NaOH + H2

Then there will be four hydrogen atoms on the right side, therefore, coefficient 2 must be added before the water formula:

Na + 2H2O → 2NaOH + H2

Let's equalize the number of sodium atoms:

2Na + 2H2O = 2NaOH + H2

Now the number of all atoms before and after the reaction is the same. Thus, we can conclude: in order to turn a chemical reaction scheme into an equation of a chemical reaction, it is necessary to equalize the number of all atoms that make up the reactants and reaction products using coefficients. The coefficients are placed before the formulas of substances. Let's summarize about Chemical Reaction Equations

• A chemical reaction scheme is a conditional record showing which substances react, which reaction products are formed, as well as the conditions for the reaction to occur.
• Reaction schemes use symbols that indicate the features of their course.
• The equation of a chemical reaction is a conditional record of a chemical reaction by means of chemical formulas, which corresponds to the law of conservation of the mass of a substance
• The scheme of a chemical reaction is converted into an equation by placing the coefficients in front of the formulas of substances