Stoichiometric coefficient. Selection of stoichiometric coefficients ovr. Calculation scheme according to the equations of chemical reactions

The excess air coefficient with this method of organizing the combustion process should correspond to rich mixtures close to stoichiometric. In this case, it will be very difficult to organize efficient combustion of lean mixtures due to the insufficiently high speed of flame front propagation with a high probability of attenuation of ignition sources, significant cyclic non-uniformity of combustion and, ultimately, misfires. Thus, this direction can be called extremely slow combustion of rich gas-air mixtures.[ ...]

The excess air coefficient (a) significantly affects the combustion process and the composition of the combustion products. It is obvious that at a 1.0) it practically does not affect the component composition of flue gases and only leads to a decrease in the concentration of components due to dilution with air not used in the combustion process.[ ...]

Based on the stoichiometric coefficients of the reaction for obtaining dialkylchlorothiophosphate and the optimal solution for criterion 2, we impose the restriction X3 = -0.26 (1.087 mol/mol).[ ...]

This gives the value of the stoichiometric coefficient for the intake of polyphosphate 1/us,p = g P/g COD(HAc).[ ...]

In table. 24.5 shows the stoichiometric yield factors determined in experiments carried out in pure culture batch reactors. These values ​​are in fairly good agreement, despite various conditions microbiological growth.[ ...]

From expression (3.36) we find the stoichiometric coefficient "sat.r = 0.05 g P / g COD (HAc).[ ...]

[ ...]

From example 3.2, one can find the stoichiometric coefficients of the removal equation acetic acid: 1 mol of HAs (60 g of HAs) requires 0.9 mol 02 and 0.9 32 = 29 g 02.[ ...]

In these formulas, the first starting material is included in all stoichiometric equations and its stoichiometric coefficient in them is V/, = -1. For this substance, the degrees of transformation lu in each stoichiometric equation are given (all of them - K). In equations (3.14) and (3.15) it is assumed that the i-th component - the product for which the selectivity and yield are determined, is formed only in the 1st stoichiometric equation (then E / \u003d x (). The amounts of components in these formulas are measured in moles (designation LO, as is traditionally accepted in the chemical sciences.[ ...]

When compiling redox equations, stoichiometric coefficients are found for the oxidation of the element before and after the reaction. The oxidation of an element in compounds is determined by the number of electrons spent by the atom on the formation of polar and ionic bonds, and the sign of oxidation is determined by the direction of displacement of the binding electron pairs. For example, the oxidation of the sodium ion in the NaCl compound is +1, and that of chlorine is -I.[ ...]

It is more convenient to represent the stoichiometry of a microbiological reaction with a stoichiometric balance equation, rather than in the form of yield factor tables. Such a description of the composition of the components of a microbiological cell required the use of an empirical formula. The formula of the substance of the cell C5H702N was experimentally established, which is often used in the preparation of stoichiometric equations.[ ...]

In table. Figure 3.6 shows typical values ​​for kinetic and other constants, as well as stoichiometric coefficients, for an aerobic urban wastewater treatment process. It should be noted that there is a certain correlation between individual constants, so it is necessary to use a set of constants from one source, and not to select individual constants from different sources. In table. 3.7 shows similar correlations.[ ...]

The method is standardized by known amounts of iodine, converted to ozone, based on a stoichiometric coefficient equal to one (1 mole of ozone releases 1 mole of iodine). This coefficient is supported by the results of a number of studies, on the basis of which the stoichiometric reactions of ozone with olefins were established. With a different coefficient, these results would be difficult to explain. However, in the work it was found that the indicated coefficient is 1.5. This is consistent with the data, according to which a stoichiometric coefficient equal to one is obtained at pH 9, and much more iodine is released in an acidic environment than in a neutral and alkaline one.[ ...]

The tests were carried out at full load and constant frequency crankshaft rotation 1,500 min1. The excess air coefficient varied in the range of 0.8 [ ...]

Material processes in living nature, cycles of biogenic elements are associated with energy flows by stoichiometric coefficients that vary at the most various organisms only within the same order. At the same time, due to the high efficiency of catalysis, the energy costs for the synthesis of new substances in organisms are much less than in the technical analogues of these processes.[ ...]

Measurements of engine characteristics and emissions of harmful emissions for all combustion chambers were carried out in a wide range of changes in the excess air coefficient from a stoichiometric value to an extremely lean mixture. On fig. 56 and 57 show the main results depending on a, obtained at a speed of 2000 min and a wide open throttle. The value of the ignition advance angle was chosen from the condition of obtaining the maximum torque.[ ...]

The biological process of phosphorus removal is complex, so, of course, our approach is greatly simplified. In table. 8.1 presents a set of stoichiometric coefficients describing the processes occurring with the participation of FAO. The table looks complicated, but simplifications have already been made in it.[ ...]

In one of recent works it is assumed that 1 mol of NO2 gives 0.72 g-ion of NO7. According to the data provided international organization standardization, the stoichiometric coefficient depends on the composition of the Griess-type reagents. Six variants of this reagent are proposed, differing in the composition of its components, and it is indicated that the absorption efficiency for all types of absorption solutions is 90%, and the stoichiometric coefficient, taking into account the absorption efficiency, varies from 0.8 to 1. Reducing the amount of NEDA and replacing sulfanilic acid with sulfanilamide (white streptocide) gives a greater value of this coefficient. The authors of the work explain this by the loss of HN02 due to the formation of NO during side reactions.[ ...]

When designing structures biochemical treatment Wastewater and the analysis of their work, the following calculation parameters are usually used: the rate of biological oxidation, stoichiometric coefficients for electron acceptors, growth rate and physical properties activated sludge biomass. The study of chemical changes in connection with biological transformations occurring in a bioreactor makes it possible to obtain a fairly complete picture of the structure's operation. For anaerobic systems, which include anaerobic filters, such information is needed to ensure the optimal pH value of the environment, which is the main factor in the normal operation of treatment facilities. In some aerobic systems, such as those in which nitrification occurs, control of the pH of the medium is also necessary to ensure optimal microbial growth rates. For closed treatment plants, which came into practice in the late 60s, which use pure oxygen (oxy-tank), the study of chemical interactions became necessary not only for pH control, but also for the engineering calculation of gas pipeline equipment.[ ...]

The catalytic conversion rate constant k in the general case at a given temperature is a function of the rate constants of the direct, reverse, and side reactions, as well as the diffusion coefficients of the initial reagents and their interaction products. The rate of a heterogeneous catalytic process is determined, as noted above, by the relative rates of its individual stages and is limited by the slowest of them. As a result, the order of the catalytic reaction almost never coincides with the molecularity of the reaction corresponding to the stoichiometric ratio in the equation for this reaction, and the expressions for calculating the rate constant of the catalytic conversion are specific for specific stages and conditions for its implementation.[ ...]

To control the neutralization reaction, one must know how much acid or base to add to the solution to obtain the desired pH value. To solve this problem, the method of empirical evaluation of stoichiometric coefficients can be used, which is carried out using titration.[ ...]

The equilibrium composition of the combustion products in the chamber is determined by the law of mass action. According to this law, the rate of chemical reactions is directly proportional to the concentration of the initial reagents, each of which is taken to a degree equal to the stoichiometric coefficient with which the substance enters the chemical reaction equation. Based on the composition of the fuels, it can be assumed that the products of combustion, for example, liquid rocket fuels in the chamber will consist of CO2, H20, CO, NO, OH, N2, H2, N. H, O, for solid rocket fuel - from A1203, N2, H2, HC1, CO, CO2, H20 at T= 1100...2200 K.[ ...]

To substantiate the possibility of using two-stage combustion of natural gas, experimental studies of the distribution of local temperatures, concentrations of nitrogen oxides and combustible substances along the length of the flame depending on the coefficient of excess air supplied through the burner were carried out. The experiments were carried out with the combustion of natural gas in the furnace of a PTVM-50 boiler equipped with a VTI vortex burner with peripheral gas jet discharge into a swirling transverse air flow. It has been established that at ag O.wb the process of fuel burn-up ends at a distance 1f/X>out = 4.2, and at ag = 1.10 - at a distance bf10out = 3.6. This indicates the prolongation of the combustion process under conditions significantly different from stoichiometric ones.[ ...]

A simplified matrix of process parameters with activated sludge without nitrification is presented in Table. 4.2. It is assumed here that three main factors contribute to the conversion process: biological growth, degradation, and hydrolysis. The reaction rates are indicated in the right column, and the coefficients presented in the table are stoichiometric. Using the table data, one can write the mass balance equation, for example, for the easily decomposable organic matter Be in a perfectly stirred reactor. The expressions responsible for the transport need no explanation. We find two expressions describing the transformations of a substance by multiplying the stoichiometric coefficients from (in this case) "component" columns by the corresponding reaction rates from the right column of Table. 4.2.[ ...]

On fig. 50 shows the change in the content of Wx in the combustion products (g / kWh) depending on the composition of the mixture and the ignition timing. Because the formation of NOx largely depends on the gas temperature, with early ignition, the emission of NOx increases. The dependence of the formation of 1 Ux on the coefficient of excess air is more complex, because There are two opposing factors. The formation of 1NHOx depends on the oxygen concentration in the combustible mixture and the temperature. Leaning the mixture increases the oxygen concentration but reduces the maximum combustion temperature. This leads to the fact that the maximum content is achieved when working with mixtures slightly poorer than stoichiometric. At the same values ​​of the excess air coefficient, the effective efficiency has a maximum.[ ...]

On fig. Figure 7.2 shows the experimental dependences of the methanol concentration on the NO3-N concentration at the outlet of the complete displacement biofilter. The lines connecting the experimental points characterize the distribution of the substance along the filter at different Smc/Sn ratios. The slope of the curves corresponds to the value of the stoichiometric coefficient: 3.1 kg CH3OH/kg NO -N.

The relation connecting the concentrations of the reacting substances with the equilibrium constant is a mathematical expression of the law of mass action, which can be formulated as follows: for a given reversible reaction in a state of chemical equilibrium, the ratio of the product of the equilibrium concentrations of the reaction products to the product of the equilibrium concentrations of the starting substances at a given temperature is a constant value, and the concentration of each substance must be raised to the power of its stoichiometric coefficient.[ ...]

In the Soviet Union, the method of Polezhaev and Girina is used to determine NO¡¡ in the atmosphere. This method uses an 8% solution of KJ to capture nitrogen dioxide. The determination of nitrite ions in the resulting solution is carried out using the Griess-Ilosvay reagent. Potassium iodide solution is a much more effective NO2 absorber than alkali solution. With its volume (only 6 ml) and air flow rate (0.25 l / min), no more than 2% NO2 slips through the absorption device with a porous glass plate. The selected samples are well preserved (about a month). The stoichiometric coefficient for the absorption of NOa by the KJ solution is 0.75, taking into account the breakthrough. According to our data, NO does not interfere with this method at a ratio of NO: NOa concentrations of 3: 1.[ ...]

The disadvantages of this method, widely introduced into the practice of high-temperature waste processing, is the need to use expensive alkaline reagents (NaOH and Na2CO3). In this way, it is possible to meet the needs of many industries that need to neutralize small amounts of liquid waste with a wide range components chemical composition and any content of organochlorine compounds. However, the combustion of chlorine-containing solvents should be approached with caution, since under certain conditions (1 > 1200 ° C, excess air coefficient > 1.5), exhaust gases may contain phosgene - highly toxic carbon chlorine, or carbonic acid chloride (COC12). The life-threatening concentration of this substance is 450 mg per 1 m3 of air.[ ...]

The processes of leaching or chemical weathering of sparingly soluble minerals or their associations are characterized by the formation of new solid phases; equilibria between them and dissolved components are analyzed using thermodynamic state diagrams. Fundamental difficulties here usually arise in connection with the need to describe the kinetics of processes, without which their consideration is often not justified. The corresponding kinetic models require the reflection of chemical interactions in an explicit form - through the partial concentrations of the reactants cx, taking into account the stoichiometric coefficients V. of specific reactions.

Which studies the quantitative relationships between the substances that entered into the reaction and formed during it (from other Greek "stechion" - "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 the laws of Avogadro (1811), Gay-Lussac (1802), the law of constancy of composition (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 not calculable. 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 are stoichiometric if they have molecular structure.

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.

Stoichiometry includes finding chemical formulas, drawing up equations of chemical reactions, calculations used in preparative chemistry and chemical analysis.

At the same time, many inorganic compounds by virtue of different reasons may have a variable composition (berthollides). Substances for which deviations from the laws of stoichiometry are observed are called non-stoichiometric. Thus, titanium(II) oxide has a variable composition, in which there can be from 0.65 to 1.25 oxygen atoms per titanium atom. Sodium tungsten bronze (related to oxide bronzes, sodium tungstate ), as sodium is removed from it, changes its color from golden yellow (NaWO 3) to dark blue-green (NaO 3WO 3), passing through intermediate red and purple colors. And even sodium chloride can have a non-stoichiometric composition, acquiring a blue color with an excess of metal. Deviations from the laws of stoichiometry are observed for condensed phases and are associated with the formation of solid solutions (for crystalline substances), with the dissolution of an excess of a reaction component in a liquid, or with thermal dissociation of the resulting compound (in a liquid phase, in a melt).

If the initial substances enter into chemical interaction in strictly defined ratios, and as a result of the reaction products are formed, the amount of which can be accurately calculated, then such reactions are called stoichiometric, and the chemical equations describing them are called stoichiometric equations. Knowing the relative molecular weights of various compounds, it is possible to calculate in what proportions these compounds will react. The molar ratios between the substances participating in the reaction show coefficients that are called stoichiometric (they are also the coefficients of chemical equations, they are also the coefficients of the equations of chemical reactions). If substances react in a ratio of 1:1, then their stoichiometric quantities are called equimolar.

The term "stoichiometry" was introduced by I. Richter in the book "The Beginnings of Stoichiometry, or the Art of Measurement". chemical elements» (J. B. Richter. Anfangsgründe der Stöchyometrie oder Meßkunst chymischer Elemente. Erster, Zweyter and Dritter Theil. Breßlau und Hirschberg, 1792–93), who summarized the results of his determinations of the masses of acids and bases in the formation of salts.

Stoichiometry is based on the laws of conservation mass, equivalents, the law Avogadro, Gay-Lussac, the law constancy composition, the law multiple relations. The discovery of the laws of stoichiometry, strictly speaking, marked the beginning of chemistry as an exact science. The rules of stoichiometry underlie all calculations related to the chemical equations of reactions and are used in analytical and preparative chemistry, chemical technology and metallurgy.

The laws of stoichiometry are used in calculations related to the formulas of substances and finding the theoretically possible yield of reaction products. Consider the combustion reaction of a thermite mixture:

Fe 2 O 3 + 2Al → Al 2 O 3 + 2Fe. (85.0 g F e 2 O 3 1) (1 m o l F e 2 O 3 160 g F e 2 O 3) (2 m o l A l 1 m o l F e 2 O 3) (27 g A l 1 m o l A l) = 28.7 g A l (\displaystyle \mathrm (\left((\frac (85.0\ g\ Fe_(2)O_(3))(1))\right)\left((\frac (1\ mol\ Fe_( 2)O_(3))(160\ g\ Fe_(2)O_(3)))\right)\left((\frac (2\ mol\ Al)(1\ mol\ Fe_(2)O_(3 )))\right)\left((\frac (27\ g\ Al)(1\ mol\ Al))\right)=28.7\ g\ Al) )

Thus, to carry out the reaction with 85.0 grams of iron (III) oxide, 28.7 grams of aluminum are needed.

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stoichiometry

Chemistry 11 Stoichiometric Chemical Laws

Problems in chemistry. Mixtures of substances. Stoichiometric chains

Subtitles

We know what the chemical equation is and we have learned how to balance it. Now we are ready to study stoichiometry. This extremely fancy word often makes people think that stoichiometry is difficult. In reality, it simply deals with the study or calculation of the ratios between different molecules in a reaction. Here is the definition given by Wikipedia: Stoichiometry is the calculation of quantitative or measurable ratios of reactants and products. You will see that the word reagents is often used in chemistry. For most of our purposes, you can use the word reagents and reactants interchangeably. They are both reactants in the reaction. The term "reagents" is sometimes used for certain types of reactions where you want to add a reagent and see what happens. And check whether your assumption about the substance is correct or not. But for our purposes, reagent and reactant are the same concepts. There is a relationship between reactants and products in a balanced chemical equation. If we are given an unbalanced equation, then we know how to get a balanced one. Balanced chemical equation. Let's take a look at stoichiometry. So, to gain experience in balancing equations, I will always start with unbalanced equations. Let's say we have iron trioxide. I'll write it down. It has two iron atoms bonded to three oxygen atoms. Plus aluminum... aluminum. The result is Al2O3 plus iron. Let me remind you that when we deal with stoichiometry, first of all we must balance the equations. A large number of stoichiometry problems will be given using an already balanced equation. But I find it useful practice to balance the equations themselves. Let's try to balance it. We have two iron atoms here in this iron trioxide. How many iron atoms do we have on the right side of the equation? We only have one iron atom. Let's multiply it by 2 right here. Great, now we have three oxygens in this part. And three oxygens in that part of the equation. It looks good. Aluminum is on the left side of the equation. We only have one aluminum atom. On the right side of the equation, we have two aluminum atoms. We have to put 2 here. We have balanced this equation. Now we are ready to deal with stoichiometry. Let's get started. There is more than one type of stoichiometric problem, but they all follow these patterns: if I'm given x grams of this, how many grams of aluminum must be added for the reaction to occur? Or if I give you y grams of these molecules and z grams of these molecules, which one will be used up first? All this is stoichiometry. We will deal with these two tasks in this video tutorial. Suppose we were given 85 grams of iron trioxide. Let's write it down. 85 grams of iron trioxide. My question to you is how many grams of aluminum do we need? How many grams of aluminum do we need? It's simple. If you look at the equation, you will immediately see the molar ratio. For every mole of this, so for every mole of this... for every atom of iron trioxide used, we need two aluminum atoms. So we need to calculate how many moles of this molecule are in 85 grams. And then we need to have twice the moles of aluminum. Because for every mole of iron trioxide, we have two moles of aluminum. We're just looking at odds, we're just looking at numbers. One molecule of iron trioxide combines with two molecules of aluminum to make a reaction. Let's first calculate how many moles are in 85 grams. What is the atomic mass or mass number of this entire molecule? Let me do it below here. So we have two irons and three oxygens. Let me write down the atomic masses of iron and oxygen. The iron is right here, 55.85. And I think it's enough to round up to 56. Let's imagine that we are dealing with a type of iron, more specifically with an isotope of iron, which has 30 neutrons. It has an atomic mass number of 56. Iron has an atomic mass number of 56. Whereas oxygen, as we already know, it is 16. Iron was 56. That mass would be... would be 2 times 56 plus 3 times 16. We can do this mentally. But this is not a math lesson, so I'll calculate everything on a calculator. Let's see, 2 times 56... 2 times 56 plus 3 times 16 is 160. Is that right? That's 48 plus 112, that's right, 160. So one molecule of iron trioxide would have a mass equal to one hundred and sixty atomic mass units. One hundred and sixty atomic mass units. So one mole or... one mole or 6.02 times 10 to the 23rd power of iron oxide molecules would have a mass of... iron, iron dioxide, yes... would have a mass of 160 grams. In our reaction, we said that we start with 85 grams of iron oxide. How many moles is that? 85 grams of iron trioxide... 85 grams of iron trioxide is equal to 85/160 moles. This is equal to 85 divided by 160, which is 0.53. 0.53 mol. Everything we've worked with so far, which has been shown in green and blue, has been needed to determine how many moles are in 85 grams of iron trioxide. We determined that this is equal to 0.53 moles. Because a whole mole would be 160 grams. But we only have 85. We know from the balanced equation that for every mole of iron trioxide we need two moles of aluminum. If we have 0.53 moles of iron molecules, more precisely iron trioxide, then we will need twice the amount of aluminum. We need 1.06 moles of aluminum. I'll just take 0.53 times 2. Because the ratio is 1:2. For every molecule of one substance, we need two molecules of the other. For every mole of one substance, we need two moles of another. If we have 0.53 moles, you multiply that by 2 and you get 1.06 moles of aluminum. Great, so we just figured out how many grams a mole of aluminum contains and then multiplied it to 1.06 and that was it. Aluminum. In the UK, this word is pronounced a little differently. Actually, I like British pronunciation. Aluminum has an atomic weight of 26.98. Imagine that the aluminum we are dealing with has a mass of 27 atomic mass units. So. One aluminum has a mass of 27 atomic mass units. One mole of aluminum will be 27 grams. Or 6.02 times 10 to the 23rd power of aluminum atoms, which gives 27 grams. If we need 1.06 moles, how much will it be? 1.06 moles of aluminum equals 1.06 times 27 grams. How much it? Let's count. 1.06 multiplied by 27 equals 28.62. We need 28.62 grams of aluminium... aluminum to make the most of our 85 grams of iron trioxide. If we had more than 28.62 grams of aluminum, then they would be left after the reaction took place. Let's assume that we mix everything properly, and the reaction proceeds to the end. We'll talk more about this later. In a situation where we have more than 28.63 grams of aluminum, this molecule will be the limiting reagent. Since we have an excess of this, that's what will limit this process. If we have less than 28.63 grams of aluminum, then aluminum will be the limiting reagent, because we will not be able to use all 85 grams of our iron molecules, more precisely iron trioxide. In any case, I don't want to confuse you with these limiting reagents. In the next video tutorial, we will consider a problem entirely devoted to limiting reagents. Subtitles by the Amara.org community

When drawing up 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

––––––––––––––––––––––––

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(V acidic environment) or hydroxide ions(V 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 necessity of equality of the number of electrons accepted 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 -.

Right: 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, which occurs when it is passed through water solution 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, this method considers the transition of electrons from one particle to another, 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 recorded that actually exist during the course of a given OVR - only really existing cations or annones are recorded 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 the 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. Molecules of FeS2 (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 reduction half-reaction of the nitrate ion: 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: Adding the corresponding number of NOJ nH+ ions to both sides of the equation, we find the molecular reaction equation: Please note that to determine the number of given and received electrons, we never had to determine the oxidation state of the elements. 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 still need to “figure out” the resulting balance equation - 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 molecules of H20: 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 medium 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