Stoichiometry includes finding chemical formulas, drawing up equations for chemical reactions, and calculations used in preparative chemistry and chemical analysis.
At the same time, many inorganic compounds, for various reasons, can 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 (belonging to sodium tungstate oxide bronzes), 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 violet colors. And even sodium chloride can have a non-stoichiometric composition, acquiring a blue color when the metal is in excess. Deviations from the laws of stoichiometry are observed for condensed phases and are associated with the formation of solid solutions (for crystalline substances), dissolution of an excess reaction component in a liquid, or thermal dissociation of the resulting compound (in the liquid phase, in the melt).
If the starting substances enter chemical reaction in strictly defined proportions, 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 that describe them are called stoichiometric equations. By 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 are shown by coefficients that are called stoichiometric (they are also the coefficients of chemical equations, they are also the coefficients of chemical reaction equations). If substances react in a 1:1 ratio, then their stoichiometric amounts 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 und Dritter Theil. Breßlau und Hirschberg, 1792–93), who summarized the results of his determinations of the masses of acids and bases during the formation of salts.
Stoichiometry is based on the laws of conservation of mass, equivalents, the law of Avogadro, Gay-Lussac, the law of constancy of composition, the law of multiple ratios. 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 chemical reaction equations 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. Let's consider the combustion reaction of the 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 a 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
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We know what a chemical equation is, and we've 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 is simply concerned with studying or calculating the relationships 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 in chemistry the word reagents is often used. 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 guess 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 get into 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. In it, two iron atoms are bonded to three oxygen atoms. Plus aluminum... aluminum. The result is Al2O3 plus iron. Let me remind you that when we do stoichiometry, the first thing we have to do is balance the equations. A large number of stoichiometry problems will be given using an already balanced equation. But I find it a useful practice to balance the equations themselves. Let's try to balance it out. 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 here. Great, now we have three oxygens in this part. And three oxygens in this 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're ready to tackle stoichiometry. Let's get started. There is more than one type of stoichiometric problem, but they all follow this pattern: If I am given x grams of this, how many grams of aluminum must I add for the reaction to occur? Or if I give you 10 grams of these molecules and 30 grams of these molecules, which of them will be used up first? It's all stoichiometry. We will deal with exactly these two tasks in this video tutorial. Let's say we were given 85 grams of iron trioxide. Let's write this 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 can immediately see the mole ratio. For every mole of this, so for every mole of that... for every atom of iron trioxide we use, we need two atoms of aluminum. So we need to figure out 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 just look at the odds, we just look at the numbers. One molecule of iron trioxide combines with two molecules of aluminum to create 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 iron and three oxygen. Let me write out the atomic masses of iron and oxygen. Iron is here, 55.85. And I think it’s quite enough to round up to 56. Let’s imagine that we are dealing with a type of iron, more precisely, with an isotope of iron that 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, is 16. Iron was 56. This mass will be... will be 2 times 56 plus 3 times 16. We can do this in our minds. 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 equals 160. Is that correct? That's 48 plus 112, right, 160. So one molecule of iron trioxide will 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... iron, iron dioxide, yeah... 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 this? 85 grams of iron trioxide... 85 grams of iron trioxide is equal to the fraction 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 that has been shown in green and blue was 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 another. For every mole of one substance we need two moles of the other. 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 get 1.06 and called it a day. Aluminum. In the UK this word is pronounced slightly differently. Actually, I like British pronunciation. Aluminum has an atomic weight of 26.98. Let's imagine that the aluminum we are dealing with has a mass of 27 atomic mass units. So. Aluminum alone 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 is it? Let's do the math. 1.06 times 27 equals 28.62. We need 28.62 grams of aluminum... aluminum to fully utilize our 85 grams of iron trioxide. If we had more than 28.62 grams of aluminum, it would remain after the reaction has occurred. Let's assume that we mix everything as needed and the reaction proceeds to completion. 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 reactant. Since we have an excess of this, this is what will limit this process. If we have less than 28.63 grams of aluminum, then aluminum will be the limiting reactant 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 look at a problem entirely devoted to limiting reagents. Subtitles by the Amara.org community
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 propagation of the flame front with a high probability of attenuation of ignition sources, significant cyclic unevenness 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 component composition of combustion products. Obviously, at a 1.0) it has practically no effect on the component composition of the 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 the production of dialkyl chlorothiophosphate and the optimal solution for criterion 2, we impose the restriction X3 = -0.26 (1.087 mol/mol).[...]
| 24.5 |
This gives the value of the stoichiometric coefficient for polyphosphate consumption 1/us,p = g P/g COD(NAs).[...]
In table Table 24.5 shows the stoichiometric yield coefficients determined in experiments conducted in continuous batch reactors with pure culture. These values are in fairly good agreement despite the different microbiological growth conditions.[...]
From expression (3.36) we find the stoichiometric coefficient “sat.p = 0.05 g P/g COD(NAs).[...]
[ ...]
From Example 3.2 you 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.[...]
| 3.12 |
In these formulas, the first starting substance is included in all stoichiometric equations and its stoichiometric coefficient in them is V/, = -1. For this substance, the degrees of conversion of lu in each stoichiometric equation are given (there are K in total). In equations (3.14) and (3.15), it is assumed that the ith component, the product for which selectivity and yield are determined, is formed only in the 1st stoichiometric equation (then E/ = x(). The quantities of components in these formulas are measured in moles (designation LO, as is traditionally accepted in the chemical sciences. [...]
When drawing up redox equations, find the stoichiometric coefficients 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 bonding 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 using a stoichiometric balance equation rather than in the form of tables of yield coefficient values. Such a description of the composition of the components of a microbiological cell required the use of an empirical formula. The formula of the cell substance C5H702N was experimentally established, which is often used in the preparation of stoichiometric equations.[...]
In table 3.6 presents typical values of kinetic and other constants, as well as stoichiometric coefficients for the aerobic process of urban wastewater treatment. 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, rather than selecting 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 unity (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 stoichiometry of ozone reactions with olefins was established. With a different coefficient, these results would be difficult to explain. However, the work found that the specified coefficient is 1.5. This is consistent with data according to which the stoichiometric coefficient, equal to one, is obtained at pH 9, and in an acidic environment significantly more iodine is released than in a neutral and alkaline one.[...]
Tests were carried out at full load and a constant crankshaft speed of 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 with stoichiometric coefficients that vary within the most diverse organisms only within one order of magnitude. Moreover, due to the high efficiency of catalysis, the energy consumption for the synthesis of new substances in organisms is much less than in the technical analogues of these processes.[...]
Measurements of engine characteristics and harmful emissions for all combustion chambers were carried out over a wide range of changes in the excess air ratio from the stoichiometric value to an extremely lean mixture. In Fig. 56 and 57 show the main results depending on a, obtained at a rotation speed of 2,000 min and a fully open throttle valve. The value of the ignition timing angle was selected from the condition of obtaining maximum torque.[...]
The biological process of phosphorus removal is complex, so of course the approach we use is greatly simplified. In table Figure 8.1 presents a set of stoichiometric coefficients that describe the processes occurring with the participation of FAO. The table looks complicated, but simplifications have already been made in it.[...]
In one of the latest works, it was accepted that 1 mol of N02 gives 0.72 g of N07 ion. According to data provided International organization standardization, the stoichiometric coefficient depends on the composition of 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 higher 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 biochemical treatment facilities waste water and analysis of their operation, the following calculated parameters are usually used: the rate of biological oxidation, stoichiometric coefficients for electron acceptors, growth rate and physical properties activated sludge biomass. Studying chemical changes in conjunction with the biological transformations occurring in the bioreactor, it makes it possible to obtain a fairly complete picture of the operation of the structure. 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, pH control 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 has become necessary not only for pH regulation, but also for the engineering calculation of gas pipeline equipment. [...]
The rate constant of the catalytic transformation k in the general case is, at a given temperature, a function of the rate constants of the forward, reverse and side reactions, as well as the diffusion coefficients of the initial reagents and the products of their interaction. 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 of this reaction, and expressions for calculating the rate constant of the catalytic transformation are specific to the specific stages and conditions of its implementation. [...]
To control the neutralization reaction, you need to know how much acid or alkali should be added to the solution to obtain the required pH value. To solve this problem, a method of empirical assessment of stoichiometric coefficients can be used, which is carried out using titration.[...]
The equilibrium composition of 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 equation chemical reaction. Based on the composition of the fuels, we can assume that the combustion products of, for example, liquid rocket fuels in the chamber will consist of CO2, H20, CO, N0, OH, Li2, H2, N. H, O, for solid rocket fuel - from A1203, N2, H2, HC1, CO, C02, H20 at T = 1100...2200 K. [...]
To justify the possibility of using two-stage combustion natural gas were carried out experimental studies distribution of local temperatures, concentrations of nitrogen oxides and combustible substances along the length of the torch, depending on the excess air ratio supplied through the burner. The experiments were carried out by burning natural gas in the furnace of a PTVM-50 boiler, equipped with a VTI vortex burner with peripheral delivery of gas jets into a swirling transverse air flow. It has been established that at ag O.bb the fuel burnout process ends at a distance 1ph/X>Out = 4.2, and at ag=1.10 - at a distance bph10out = 3.6. This indicates an extended 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. Reaction rates are indicated in the right column, and the coefficients presented in the table are stoichiometric. Using the table data, you can write a mass balance equation, for example, for the easily decomposable organic matter Be in an ideal mixing reactor. The transport expressions are self-explanatory. 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 the table. 4.2.[...]
In Fig. Figure 50 shows the change in the content of Shx in 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 Yux on the excess air coefficient is more complex, because there are two opposing factors. The formation of 1Ох depends on the oxygen concentration in the combustion mixture and 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 ones. At the same values of the excess air coefficient, the effective efficiency has a maximum.[...]
In Fig. Figure 7.2 shows the experimental dependence 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 relationship connecting the concentrations of reactants 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 at the state 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 Polezhaev and Girina method is used to determine NO¡¡ in the atmosphere. This method uses an 8% KJ solution to capture nitrogen dioxide. Determination of nitrite ions in the resulting solution is carried out using the Griess-Ilosvay reagent. Potassium iodide solution is a significantly more effective NO2 absorber than alkali solution. With its volume (only 6 ml) and air transmission rate (0.25 l/min), no more than 2% NO2 passes through the absorption device with a porous glass plate. The samples taken 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, this method does not interfere with NO at a concentration ratio of NO:NOa 3:1.[...]
The disadvantages of this method, which is widely used in the practice of high-temperature waste processing, is the need to use expensive alkaline reagents (NaOH and Na2CO3). Thus, it is possible to meet the needs of many industries that need to treat small quantities of liquid waste with a wide range of 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 ratio > 1.5) the exhaust gases may contain phosgene, a highly toxic carbon chloroxide, 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 phase 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 explicit form - through the partial concentrations of reacting substances cx, taking into account the stoichiometric coefficients V. of specific reactions.
Stoichiometry- quantitative relationships between reacting substances.
If reagents enter into a chemical interaction in strictly defined quantities, and as a result of the reaction substances are formed, the amount of which can be calculated, then such reactions are called stoichiometric.
Laws of stoichiometry:
The coefficients in chemical equations before the formulas of chemical compounds are called stoichiometric.
All calculations according to chemical equations are based on the use of stoichiometric coefficients and are associated with finding quantities of a substance (number of moles).
The amount of substance in the reaction equation (number of moles) = the coefficient in front of the corresponding molecule.
N A=6.02×10 23 mol -1.
η - ratio of the actual mass of the product m p to a theoretically possible m t, expressed in fractions of a unit or as a percentage.
If the yield of reaction products is not indicated in the condition, then in the calculations it is taken equal to 100% (quantitative yield).
Calculation scheme using chemical reaction equations:
- Write an equation for a chemical reaction.
- Above the chemical formulas of substances write known and unknown quantities with units of measurement.
- Under the chemical formulas of substances with known and unknowns, write down the corresponding values of these quantities found from the reaction equation.
- Compose and solve a proportion.
Example. Calculate the mass and amount of magnesium oxide formed during the complete combustion of 24 g of magnesium.
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Given: m(Mg) = 24 g Find: ν (MgO) m (MgO) |
Solution: 1. Let's create an equation for a chemical reaction: 2Mg + O 2 = 2MgO. 2. Under the formulas of substances we indicate the amount of substance (number of moles) that corresponds to the stoichiometric coefficients: 2Mg + O2 = 2MgO 2 mole 2 mole 3. Let's define molar mass magnesium: Relative atomic mass of magnesium Ar (Mg) = 24. Because the molar mass value is equal to the relative atomic or molecular mass, then M (Mg)= 24 g/mol. 4. Using the mass of the substance specified in the condition, we calculate the amount of the substance:
5. Above chemical formula magnesium oxide MgO, whose mass is unknown, we set xmole, above the magnesium formula Mg we write its molar mass: 1 mole xmole 2Mg + O2 = 2MgO 2 mole 2 mole
According to the rules for solving proportions:
Magnesium Oxide Amount ν (MgO)= 1 mol. 7. Calculate the molar mass of magnesium oxide: M (Mg)=24 g/mol, M(O)=16 g/mol. M(MgO)= 24 + 16 = 40 g/mol. We calculate the mass of magnesium oxide: m (MgO) = ν (MgO) × M (MgO) = 1 mol × 40 g/mol = 40 g. Answer: ν (MgO) = 1 mol; m (MgO) = 40 g. |
All quantitative relationships in the calculation chemical processes based on the stoichiometry of reactions. It is more convenient to express the amount of a substance in such calculations in moles, or derivative units (kmol, mmol, etc.). The mole is one of the SI base units. One mole of any substance corresponds to its quantity numerically equal to its molecular weight. Therefore, molecular weight in this case should be considered a dimensional value with units: g/mol, kg/kmol, kg/mol. For example, the molecular weight of nitrogen is 28 g/mol, 28 kg/kmol, but 0.028 kg/mol.
Mass and molar amounts of a substance are related by known relationships
N A = m A / M A; m A = N A M A,
where N A is the amount of component A, mol; m A is the mass of this component, kg;
M A - molecular weight of component A, kg/mol.
In continuous processes, the flow of substance A can be expressed by its mol-
quantity per unit of time
where W A is the molar flow of component A, mol/s; τ - time, s.
For a simple reaction that is practically irreversible, usually stoichiomet
the ric equation is written in the form
v A A + v B B = v R R + v S S.
However, it is more convenient to write the stoichiometric equation in the form of an algebraic
th, assuming that the stoichiometric coefficients of the reactants are negative, and those of the reaction products are positive:
Then for each simple reaction we can write the following equalities:
The index "0" refers to the initial quantity of the component.
These equalities give rise to the following equations of material balance for a component for a simple reaction:
Example 7.1. The hydrogenation reaction of phenol to cyclohexanol proceeds according to the equation
C 6 H 5 OH + ZH 2 = C 6 H 11 OH, or A + ZV = R.
Calculate the amount of product formed if the initial amount of component A was 235 kg, and the final amount was 18.8 kg
Solution: Let's write the reaction in the form
R - A - ZV = 0.
Molecular masses of the components: M A = 94 kg/kmol, M B = 2 kg/kmol and
M R = 100 kg/kmol. Then the molar amounts of phenol at the beginning and end of the reaction will be:
N A 0 = 235/94 = 2.5; N A 0 = 18.8/94 =0.2; n = (0.2 - 2.5)/(-1) = 2.3.
The amount of cyclohexanol formed will be equal to
N R = 0 +1∙2.3 = 2.3 kmol or m R = 100∙2.3 = 230 kg.
Determining stoichiometrically independent reactions in their system during material and thermal calculations of reaction apparatuses is necessary to exclude reactions that are the sum or difference of some of them. This assessment can most easily be carried out using the Gram criterion.
To avoid unnecessary calculations, it is necessary to evaluate whether the system is stoichiometrically dependent. For these purposes it is necessary:
Transpose the original matrix of the reaction system;
Multiply the original matrix by the transposed one;
Calculate the determinant of the resulting square matrix.
If this determinant is zero, then the reaction system is stoichiometrically dependent.
Example 7.2. We have a system of reactions:
FeO + H 2 = Fe + H 2 O;
Fe 2 O 3 + 3H 2 = 2Fe + 3H 2 O;
FeO + Fe 2 O 3 + 4H 2 = 3Fe + 4H 2 O.
This system is stoichiometrically dependent, since the third reaction is the sum of the other two. Let's create a matrix
When composing equations for redox reactions, the following two important rules must be observed:
Rule 1: In any ionic equation, conservation of charges must be observed. This means that the sum of all the charges on the left side of the equation (the "left") must be the same as the sum of all the charges on the right side of the equation (the "right"). This rule applies to any ionic equations, as for complete reactions, and for half-reactions.
Charges from left to right
Rule 2: The number of electrons lost in the oxidative half-reaction must be equal to the number of electrons gained in the reducing half-reaction. For example, in the first example given at the beginning of this section (the reaction between iron and hydrated cuprous ions), the number of electrons lost in the oxidative half-reaction is two:
Therefore, the number of electrons acquired in the reduction half-reaction must also be equal to two:
To construct the equation for a complete redox reaction from the equations for two half-reactions, the following procedure can be used:
1. The equations for each of the two half-reactions are balanced separately, with the appropriate number of electrons added to the left or right side of each equation to fulfill Rule 1 above.
2. The equations for both half-reactions are balanced against each other so that the number of electrons lost in one reaction becomes equal to the number of electrons gained in the other half-reaction, as required by Rule 2.
3. The equations for both half-reactions are summed to obtain the complete equation for the redox reaction. For example, by summing the equations of the two half-reactions above and removing from the left and right sides of the resulting equation
equal number of electrons, we find
Let's balance the equations of the half-reactions below and write the equation for the redox oxidation reaction aqueous solution any ferrous salt into a ferric salt using an acidic potassium solution.
Stage 1. First, we balance the equation of each of the two half-reactions separately. For equation (5) we have
To balance both sides of this equation, you need to add five electrons to the left side, or subtract the same number of electrons from the right side. After this we get
This allows us to write the following balanced equation:
Since electrons had to be added to the left side of the equation, it describes a reducing half-reaction.
For equation (6) we can write
To balance this equation, you can add one electron to the right side. Then