Stoichiometry coefficient. Stoichiometry is the basis for chemical calculations

When compiling an equation for a redox reaction, it is necessary to determine the reducing agent, oxidizing agent, and the number of given and received electrons. There are mainly two methods for compiling equations of redox reactions:
1) electronic balance– based on the determination of the total number of electrons moving from the reducing agent to the oxidizing agent;
2) ion-electronic balance- provides for the separate compilation of equations for the process of oxidation and reduction with their subsequent summation into a common ionic equation-half-reaction method. In this method, it is necessary to find not only the coefficients for the reducing agent and oxidizing agent, but also for the molecules of the medium. Depending on the nature of the medium, the number of electrons accepted by the oxidizing agent or lost by the reducing agent may vary.
1) Electronic balance - a method for finding the coefficients in the equations of redox reactions, which considers the exchange of electrons between atoms of elements that change their oxidation state. The number of electrons donated by the reducing agent is equal to the number of electrons received by the oxidizing agent.

The equation is compiled in several stages:

1. Write down the reaction scheme.

KMnO 4 + HCl → KCl + MnCl 2 + Cl 2 + H 2 O

2. Put down the oxidation states above the signs of the elements that change.

KMn +7 O 4 + HCl -1 → KCl + Mn +2 Cl 2 + Cl 2 0 + H 2 O

3. Allocate elements that change the degree of oxidation and determine the number of electrons acquired by the oxidizing agent and given away by the reducing agent.

Mn +7 + 5ē = Mn +2

2Cl -1 - 2ē \u003d Cl 2 0

4. Equalize the number of acquired and donated electrons, thereby establishing the coefficients for compounds in which there are elements that change the oxidation state.

Mn +7 + 5ē = Mn +2 2

2Cl -1 - 2ē \u003d Cl 2 0 5

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2Mn +7 + 10Cl -1 = 2Mn +2 + 5Cl 2 0

5. Coefficients are selected for all other participants in the reaction. In this case, 10 HCl molecules participate in the reduction process, and 6 in the ion exchange process (binding of potassium and manganese ions).

2KMn +7 O 4 + 16HCl -1 = 2KCl + 2Mn +2 Cl 2 + 5Cl 2 0 + 8H 2 O

2) Method of ion-electron balance.

1. Write down the reaction scheme.

K 2 SO 3 + KMnO 4 + H 2 SO 4 → K 2 SO 4 + MnSO 4 + H 2 O

2. Write down schemes of half-reactions, using actually present particles (molecules and ions) in solution. At the same time, we sum up the material balance, i.e. the number of atoms of the elements participating in the half-reaction on the left side must be equal to their number on the right. Oxidized and reduced forms oxidizer and reductant often differ in oxygen content (compare Cr 2 O 7 2− and Cr 3+). Therefore, when compiling half-reaction equations using the electron-ion balance method, they include H + /H 2 O pairs (for acidic environment) and OH - / H 2 O (for alkaline environment). If during the transition from one form to another, the original form (usually − oxidized) loses its oxide ions (shown below in square brackets), the latter, since they do not exist in free form, must be in acidic medium are combined with hydrogen cations, and in alkaline medium - with water molecules, which leads to the formation water molecules(in an acidic environment) and hydroxide ions(in an alkaline environment):

acid environment+ 2H + = H 2 O example: Cr 2 O 7 2− + 14H + = 2Cr 3+ + 7H 2 O
alkaline environment+ H 2 O \u003d 2 OH - example: MnO 4 - + 2H 2 O \u003d MnO 2 + 4OH -

lack of oxygen in the original form (more often in the restored form) compared to the final form is compensated by adding water molecules(in acidic environment) or hydroxide ions(in alkaline environment):

acid environment H 2 O = + 2H + example: SO 3 2- + H 2 O = SO 4 2- + 2H +
alkaline environment 2 OH - \u003d + H 2 O example: SO 3 2- + 2OH - \u003d SO 4 2- + H 2 O

MnO 4 - + 8H + → Mn 2+ + 4H 2 O reduction

SO 3 2- + H 2 O → SO 4 2- + 2H + oxidation

3. We sum up the electronic balance, following the need for the equality of the total charge in the right and left parts of the half-reaction equations.

In the above example, on the right side of the reduction half-reaction equation, the total charge of the ions is +7, on the left - +2, which means that five electrons must be added on the right side:

MnO 4 - + 8H + + 5ē → Mn 2+ + 4H 2 O

In the oxidation half-reaction equation, the total charge on the right side is -2, on the left side 0, which means that two electrons must be subtracted on the right side:

SO 3 2- + H 2 O - 2ē → SO 4 2- + 2H +

Thus, in both equations, the ion-electron balance is implemented and it is possible to put equal signs instead of arrows in them:

MnO 4 - + 8H + + 5ē \u003d Mn 2+ + 4H 2 O

SO 3 2- + H 2 O - 2ē \u003d SO 4 2- + 2H +

4. Following the rule about the need for equality of the number of electrons received by the oxidizing agent and given away by the reducing agent, we find the least common multiple for the number of electrons in both equations (2∙5 = 10).

5. We multiply by the coefficients (2.5) and sum both equations by adding the left and right parts of both equations.

MnO 4 - + 8H + + 5ē \u003d Mn 2+ + 4H 2 O 2

SO 3 2- + H 2 O - 2ē \u003d SO 4 2- + 2H + 5

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2MnO 4 - + 16H + + 5SO 3 2- + 5H 2 O = 2Mn 2+ + 8H 2 O + 5SO 4 2- + 10H +

2MnO 4 - + 6H + + 5SO 3 2- = 2Mn 2+ + 3H 2 O + 5SO 4 2-

or in molecular form:

5K 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 = 6K 2 SO 4 + 2MnSO 4 + 3H 2 O

This method considers the transition of electrons from one atom or ion to another, taking into account the nature of the medium (acidic, alkaline or neutral) in which the reaction takes place. In an acidic medium, in the half-reaction equations, to equalize the number of hydrogen and oxygen atoms, hydrogen ions H + and water molecules should be used, in the basic one, hydroxide ions OH - and water molecules. Accordingly, in the products obtained, on the right side of the electron-ionic equation, there will be hydrogen ions (and not hydroxide ions) and water molecules (acidic medium) or hydroxide ions and water molecules (alkaline medium). So, for example, the equation for the reduction half-reaction of a permanganate ion in an acidic medium cannot be compiled with the presence of hydroxide ions on the right side:

MnO 4 - + 4H 2 O + 5ē \u003d Mn 2+ + 8OH -.

Correctly: MnO 4 - + 8H + + 5ē \u003d Mn 2+ + 4H 2 O

That is, when writing electron-ionic equations, one must proceed from the composition of the ions actually present in the solution. In addition, as in the preparation of abbreviated ionic equations, substances that are poorly dissociating, poorly soluble or liberated in the form of a gas should be written in molecular form.

Drawing up the equations of redox reactions using the half-reaction method leads to the same result as the electron balance method.

Let's compare both methods. The advantage of the half-reaction method in comparison with the electron balance method is that that it uses not hypothetical ions, but real ones.

When using the half-reaction method, it is not necessary to know the oxidation state of the atoms. Writing separate ionic half-reaction equations is necessary to understand the chemical processes in a galvanic cell and during electrolysis. With this method, the role of the environment as an active participant in the entire process is visible. Finally, when using the half-reaction method, it is not necessary to know all the resulting substances, they appear in the reaction equation when deriving it. Therefore, the method of half-reactions should be preferred and used in the preparation of equations for all redox reactions occurring in aqueous solutions

In this method, the oxidation states of atoms in the initial and final substances are compared, guided by the rule: the number of electrons donated by the reducing agent must be equal to the number of electrons attached to the oxidizing agent. To draw up an equation, you need to know the formulas of the reactants and reaction products. The latter are determined either empirically or on the basis of known properties of the elements.

The ion-electron balance method is more versatile than the electron balance method and has an undeniable advantage in the selection of coefficients in many redox reactions, in particular, with the participation of organic compounds, in which even the procedure for determining oxidation states is very complicated.

Consider, for example, the process of ethylene oxidation that occurs when it is passed through an aqueous solution of potassium permanganate. As a result, ethylene is oxidized to ethylene glycol HO-CH 2 -CH 2 -OH, and permanganate is reduced to manganese (IV) oxide, in addition, as will be obvious from the final balance equation, potassium hydroxide is also formed on the right:

KMnO 4 + C 2 H 4 + H 2 O → C 2 H 6 O 2 + MnO 2 + KOH

Reduction and oxidation half-reaction equation:

MnO 4 - + 2H 2 O + 3e \u003d MnO 2 + 4OH - 2 recovery

C 2 H 4 + 2OH - - 2e \u003d C 2 H 6 O 2 3 oxidation

We summarize both equations, subtract the hydroxide ions present on the left and right sides.

We get the final equation:

2KMnO 4 + 3C 2 H 4 + 4H 2 O → 3C 2 H 6 O 2 + 2MnO 2 + 2KOH

When using the ion-electron balance method to determine the coefficients in reactions involving organic compounds, it is convenient to consider the oxidation states of hydrogen atoms equal to +1, oxygen -2, and calculate carbon using the balance of positive and negative charges in the molecule (ion). So, in an ethylene molecule, the total charge is zero:

4 ∙ (+1) + 2 ∙ X \u003d 0,

means the degree of oxidation of two carbon atoms - (-4), and one (X) - (-2).

Similarly, in the ethylene glycol molecule C 2 H 6 O 2 we find the oxidation state of carbon (X):

2 ∙ X + 2 ∙ (-2) + 6 ∙ (+1) = 0, X = -1

In some molecules of organic compounds, such a calculation leads to a fractional value of the oxidation state of carbon, for example, for an acetone molecule (C 3 H 6 O), it is -4/3. The electronic equation estimates the total charge of carbon atoms. In an acetone molecule, it is -4.

Similar information.

When drawing up an equation for a redox reaction (ORR), it is necessary to determine the reducing agent, oxidizing agent, and the number of given and received electrons. OVR stoichiometric coefficients are selected using either the electron balance method or the electron-ion balance method (the latter is also called the half-reaction method). Let's look at a few examples. As an example of compiling OVR equations and selecting stoichiometric coefficients, we analyze the process of oxidation of iron (II) disulfide (pyrite) with concentrated nitric acid: First of all, we determine the possible reaction products. Nitric acid is a strong oxidizing agent, so the sulfide ion can be oxidized either to the maximum oxidation state S (H2S04) or to S (SO2), and Fe to Fe, while HN03 can be reduced to N0 or N02 (the set of specific products is determined concentrations of reagents, temperature, etc.). Let's choose the following possible option: H20 will be on the left or right side of the equation, we don't know yet. There are two main methods for selecting coefficients. Let us first apply the method of electron-ion balance. The essence of this method is in two very simple and very important statements. First, in this method, the transition of electrons from one particle to another is considered with the obligatory consideration of the nature of the medium (acidic, alkaline, or neutral). Secondly, when compiling the equation of the electron-ion balance, only those particles are written that actually exist during the course of a given OVR - only really existing cations or annones are written in the form of ions; Substances that are poorly disociated, insoluble or liberated in the form of a gas are written in molecular form. When compiling an equation for the processes of oxidation and reduction, to equalize the number of hydrogen and oxygen atoms, one introduces (depending on the medium) either water molecules and hydrogen ions (if the medium is acidic), or water molecules and hydroxide ions (if the medium is alkaline). Consider for our case the oxidation half-reaction. FeS2 molecules (a poorly soluble substance) are converted into Fe3+ ions (iron nitrate (II) completely dissociates into ions) and sulfate ions S042" (dissociation of H2SO4): Consider now the nitrate reduction half-reaction: To equalize oxygen, add 2 to the right side water molecules, and to the left - 4 H + ions: To equalize the charge to the left side (charge +3), add 3 electrons: Finally, we have: Reducing both parts by 16H + and 8H20, we get the final, reduced ionic equation of the redox reaction: By adding the corresponding number of NOJ nH+ ions to both sides of the equation, we find the molecular equation for the reaction: In addition, we took into account the influence of the environment and “automatically” determined that H20 is on the right side of the equation. There is no doubt that this method has a great chemical meaning. Empirical balance method. The essence of the method of finding the stoichiometric coefficients in the equations of the OVR is the obligatory determination of the oxidation states of the atoms of the elements involved in the OVR. Using this approach, we again equalize the reaction (11.1) (above we applied the method of half-reactions to this reaction). The reduction process is described simply: It is more difficult to draw up an oxidation scheme, since two elements are oxidized at once - Fe and S. You can assign iron an oxidation state of +2, sulfur - 1 and take into account that there are two S atoms per Fe atom: You can, however, do without determination of oxidation states and write down a scheme resembling scheme (11.2): The right side has a charge of +15, the left side has a charge of 0, so FeS2 must give up 15 electrons. We write down the overall balance: We need to “understand” the resulting balance equation a little more - it shows that 5 HN03 molecules are used to oxidize FeS2 and another 3 HNO molecules are needed to form Fe(N03)j: To equalize hydrogen and oxygen, to the right part you need to add 2 H2O molecules: The electron-ion balance method is more versatile than the electron balance method and has an undeniable advantage in the selection of coefficients in many OTS, in particular, with the participation of organic compounds, in which even the procedure for determining oxidation states is very complicated . - Consider, for example, the process of ethylene oxidation, which occurs when it is passed through an aqueous solution of potassium permanganate. As a result, ethylene is oxidized to ethylene glycol HO - CH2 - CH2 - OH, and permanganate is reduced to manganese oxide (TV), in addition, as will be obvious from the final balance equation, potassium hydroxide is also formed on the right: After making the necessary reductions of such terms, we write the equation in the final molecular form * Influence of the environment on the nature of the OVR flow. The examples (11.1) - (11.4) clearly illustrate the "technique" of using the electron-ion balance method in the case of OVR flow in an acidic or alkaline medium. The nature of the environment! influences the course of one or another OVR; in order to “feel” this influence, let us consider the behavior of one and the same oxidizing agent (KMnO4) in different environments. , recovering up to Mn+4(Mn0j), and the minimum - in the strength of the last one, in which the risen Shaiyaaapsya up to (mvnganat-nOn Mn042"). This is explained as follows. The acids of the dissociation line form hydroxide ions ffjO +, which strongly polarize 4 "MoOH ions Weaken the bonds of manganese with oxygen (thereby enhancing the action of the reducing agent) .. In a neutral medium, the polarizing effect of water molecules is significantly c-aafep. >"MnO ions; much less polarized. In a strongly alkaline medium, hydroxide ions “even strengthen the Mn-O bond, as a result of which the effectiveness of the reducing agent decreases and MnO^ accepts only one electron. An example of the behavior of potassium permanganate in a neutral medium is represented by the reaction (11.4). Let us also give one example of reactions involving KMnOA in acidic and alkaline media

For each substance in the reaction, there are the following quantities of the substance:

Initial amount of the i-th substance (amount of substance before the start of the reaction);

The final amount of the i-th substance (the amount of the substance at the end of the reaction);

The amount of reacted (for starting substances) or formed substance (for reaction products).

Since the amount of a substance cannot be negative, for the starting substances

Since >.

For reaction products >, therefore, .

Stoichiometric ratios - ratios between quantities, masses or volumes (for gases) of reacting substances or reaction products, calculated on the basis of the reaction equation. Calculations using reaction equations are based on the basic law of stoichiometry: the ratio of the amounts of reacting or formed substances (in moles) is equal to the ratio of the corresponding coefficients in the reaction equation (stoichiometric coefficients).

For the aluminothermic reaction described by the equation:

3Fe 3 O 4 + 8Al = 4Al 2 O 3 + 9Fe,

the amounts of reacted substances and reaction products are related as

For calculations, it is more convenient to use another formulation of this law: the ratio of the amount of a reacted or formed substance as a result of a reaction to its stoichiometric coefficient is a constant for a given reaction.

In general, for a reaction of the form

aA + bB = cC + dD,

where small letters denote coefficients and large letters denote chemicals, the amounts of reactants are related by:

Any two terms of this ratio, related by equality, form the proportion of a chemical reaction: for example,

If the mass of the formed or reacted substance of the reaction is known for the reaction, then its amount can be found by the formula

and then, using the proportion of the chemical reaction, can be found for the remaining substances of the reaction. A substance, by mass or quantity of which the masses, quantities or volumes of other participants in the reaction are found, is sometimes called a reference substance.

If the masses of several reagents are given, then the calculation of the masses of the remaining substances is carried out according to the substance that is in short supply, i.e., is completely consumed in the reaction. Amounts of substances that exactly match the reaction equation without excess or deficiency are called stoichiometric quantities.

Thus, in tasks related to stoichiometric calculations, the main action is to find the reference substance and calculate its amount that entered or formed as a result of the reaction.

Calculation of the amount of an individual solid

where is the amount of individual solid A;

Mass of individual solid A, g;

Molar mass of substance A, g/mol.

Calculation of the amount of natural mineral or mixture of solids

Let the natural mineral pyrite be given, the main component of which is FeS 2 . In addition to it, the composition of pyrite includes impurities. The content of the main component or impurities is indicated in mass percent, for example, .

If the content of the main component is known, then

If the content of impurities is known, then

where is the amount of individual substance FeS 2, mol;

Mass of the mineral pyrite, g.

Similarly, the amount of a component in a mixture of solids is calculated if its content in mass fractions is known.

Calculation of the amount of substance of a pure liquid

If the mass is known, then the calculation is similar to the calculation for an individual solid.

If the volume of the liquid is known, then

1. Find the mass of this volume of liquid:

m f = V f s f,

where m W is the mass of liquid g;

V W - volume of liquid, ml;

c w is the density of the liquid, g/ml.

2. Find the number of moles of liquid:

This technique is suitable for any aggregate state of matter.

Determine the amount of substance H 2 O in 200 ml of water.

Solution: if the temperature is not specified, then the density of water is assumed to be 1 g / ml, then:

Calculate the amount of a solute in a solution if its concentration is known

If the mass fraction of the solute, the density of the solution and its volume are known, then

m r-ra \u003d V r-ra s r-ra,

where m p-ra is the mass of the solution, g;

V p-ra - the volume of the solution, ml;

with r-ra - the density of the solution, g / ml.

where is the mass of the dissolved substance, g;

Mass fraction of the dissolved substance, expressed in%.

Determine the amount of nitric acid substance in 500 ml of a 10% acid solution with a density of 1.0543 g/ml.

Determine the mass of the solution

m r-ra \u003d V r-ra s r-ra \u003d 500 1.0543 \u003d 527.150 g

Determine the mass of pure HNO 3

Determine the number of moles of HNO 3

If the molar concentration of the solute and the substance and the volume of the solution are known, then

where is the volume of the solution, l;

Molar concentration of the i-th substance in solution, mol/l.

Calculation of the amount of an individual gaseous substance

If the mass of a gaseous substance is given, then it is calculated by formula (1).

If the volume measured under normal conditions is given, then according to formula (2), if the volume of a gaseous substance is measured under any other conditions, then according to formula (3), the formulas are given on pages 6-7.

Which studies the quantitative relationships between the substances that entered into the reaction and formed during it (from the other Greek "stekhion" - "elemental composition", "meitren" - "I measure").

Stoichiometry is the most important for material and energy calculations, without which it is impossible to organize any chemical production. Chemical stoichiometry allows you to calculate the amount of raw materials needed for a particular production, taking into account the desired performance and possible losses. No enterprise can be opened without preliminary calculations.

A bit of history

The very word "stoichiometry" is an invention of the German chemist Jeremy Benjamin Richter, proposed by him in his book, in which the idea of ​​​​the possibility of calculations using chemical equations was first described. Later, Richter's ideas received theoretical justification with the discovery of Avogadro's laws (1811), Gay-Lussac's (1802), the law of composition constancy (J.L. Proust, 1808), multiple ratios (J. Dalton, 1803), and the development of atomic and molecular theory. Now these laws, as well as the law of equivalents, formulated by Richter himself, are called the laws of stoichiometry.

The concept of "stoichiometry" is used in relation to both substances and chemical reactions.

Stoichiometric Equations

Stoichiometric reactions - reactions in which the starting substances interact in certain ratios, and the amount of products corresponds to theoretical calculations.

Stoichiometric equations are equations that describe stoichiometric reactions.

Stoichiometric equations) show the quantitative relationships between all participants in the reaction, expressed in moles.

Most inorganic reactions are stoichiometric. For example, three successive reactions to produce sulfuric acid from sulfur are stoichiometric.

S + O 2 → SO 2

SO 2 + ½O 2 → SO 3

SO 3 + H 2 O → H 2 SO 4

Calculations using these reaction equations can determine how much each substance needs to be taken in order to obtain a certain amount of sulfuric acid.

Most organic reactions are non-stoichiometric. For example, the reaction equation for cracking ethane looks like this:

C 2 H 6 → C 2 H 4 + H 2 .

However, in reality, during the reaction, different amounts of by-products will always be obtained - acetylene, methane and others, which cannot be calculated theoretically. Some inorganic reactions also defy calculations. For example, ammonium nitrate:

NH 4 NO 3 → N 2 O + 2H 2 O.

It goes in several directions, so it is impossible to determine how much starting material needs to be taken in order to obtain a certain amount of nitric oxide (I).

Stoichiometry is the theoretical basis of chemical production

All reactions that are used in or in production must be stoichiometric, that is, subject to accurate calculations. Will the plant or factory be profitable? Stoichiometry allows you to find out.

On the basis of stoichiometric equations, a theoretical balance is made. It is necessary to determine how much of the starting materials will be required to obtain the desired amount of the product of interest. Further, operational experiments are carried out, which will show the real consumption of the starting materials and the yield of products. The difference between theoretical calculations and practical data allows you to optimize production and evaluate the future economic efficiency of the enterprise. Stoichiometric calculations also make it possible to compile the heat balance of the process in order to select equipment, determine the masses of by-products formed that will need to be removed, and so on.

Stoichiometric substances

According to the law of composition constancy proposed by J.L. Proust, any chemical has a constant composition, regardless of the method of preparation. This means that, for example, in a molecule of sulfuric acid H 2 SO 4, regardless of the method by which it was obtained, there will always be one sulfur atom and four oxygen atoms per two hydrogen atoms. All substances that have a molecular structure are stoichiometric.

However, substances are widespread in nature, the composition of which may differ depending on the method of preparation or source of origin. The vast majority of them are crystalline substances. One could even say that for solids, stoichiometry is the exception rather than the rule.

For example, consider the composition of well-studied titanium carbide and oxide. In titanium oxide TiO x X=0.7-1.3, that is, from 0.7 to 1.3 oxygen atoms per titanium atom, in carbide TiC x X=0.6-1.0.

The nonstoichiometric nature of solids is explained by an interstitial defect at the nodes of the crystal lattice or, conversely, by the appearance of vacancies at the nodes. Such substances include oxides, silicides, borides, carbides, phosphides, nitrides and other inorganic substances, as well as high-molecular organic ones.

And although evidence for the existence of compounds with a variable composition was presented only at the beginning of the 20th century by I.S. Kurnakov, such substances are often called berthollides by the name of the scientist K.L. Berthollet, who suggested that the composition of any substance changes.