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
–––––––––––––––––––––––––––––––––––––––––––––––––––
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 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.
One of the most important chemical concepts on which stoichiometric calculations are based is chemical amount of a substance. The amount of some substance X is denoted by n(X). The unit for measuring the amount of a substance is mole.
A mole is the amount of a substance that contains 6.02 10 23 molecules, atoms, ions or other structural units that make up the substance.
The mass of one mole of some substance X is called molar mass M(X) of this substance. Knowing the mass m(X) of some substance X and its molar mass, we can calculate the amount of this substance using the formula:
The number 6.02 10 23 is called Avogadro's number(Na); its dimension mol –1.
By multiplying the Avogadro number N a by the amount of substance n(X), we can calculate the number of structural units, for example, molecules N(X) of some substance X:
N(X) = N a · n(X) .
By analogy with the concept of molar mass, the concept of molar volume was introduced: molar volume V m (X) of some substance X is the volume of one mole of this substance. Knowing the volume of a substance V(X) and its molar volume, we can calculate the chemical amount of a substance:
In chemistry, one often has to deal with the molar volume of gases. According to Avogadro's law, equal volumes of any gases taken at the same temperature and equal pressure contain the same number of molecules. Under equal conditions, 1 mole of any gas occupies the same volume. Under normal conditions (n.s.) - temperature 0 ° C and pressure 1 atmosphere (101325 Pa) - this volume is 22.4 liters. Thus, at n.o. V m (gas) = 22.4 l / mol. It should be emphasized that the molar volume value of 22.4 l/mol is applied only for gases.
Knowing the molar masses of substances and the Avogadro number allows you to express the mass of a molecule of any substance in grams. Below is an example of calculating the mass of a hydrogen molecule.
1 mol of hydrogen gas contains 6.02 10 23 H 2 molecules and has a mass of 2 g (because M (H 2) = 2 g / mol). Hence,
6.02·10 23 H 2 molecules have a mass of 2 g;
1 H 2 molecule has a mass x g; x \u003d 3.32 10 -24 g.
The concept of "mole" is widely used to carry out calculations according to the equations of chemical reactions, since the stoichiometric coefficients in the reaction equation show in what molar ratios substances react with each other and are formed as a result of the reaction.
For example, the reaction equation 4 NH 3 + 3 O 2 → 2 N 2 + 6 H 2 O contains the following information: 4 mol of ammonia react without excess and deficiency with 3 mol of oxygen, and 2 mol of nitrogen and 6 mol of water are formed.
Example 4.1 Calculate the mass of the precipitate formed during the interaction of solutions containing 70.2 g of calcium dihydrogen phosphate and 68 g of calcium hydroxide. What substance will be left in excess? What is its mass?
3 Ca(H 2 PO 4) 2 + 12 KOH ® Ca 3 (PO 4) 2 ¯ + 4 K 3 PO 4 + 12 H 2 O
It can be seen from the reaction equation that 3 mol Ca(H 2 PO 4) 2 reacts with 12 mol KOH. Let us calculate the amounts of reacting substances, which are given according to the condition of the problem:
n (Ca (H 2 PO 4) 2) \u003d m (Ca (H 2 PO 4) 2) / M (Ca (H 2 PO 4) 2) \u003d 70.2 g: 234 g / mol \u003d 0.3 mol ;
n(KOH) = m(KOH) / M(KOH) = 68 g: 56 g/mol = 1.215 mol.
3 mol Ca(H 2 PO 4) 2 requires 12 mol KOH
0.3 mol Ca (H 2 PO 4) 2 requires x mol KOH
x \u003d 1.2 mol - so much KOH will be required in order for the reaction to proceed without excess and deficiency. And according to the condition of the problem, there are 1.215 mol of KOH. Therefore, KOH is in excess; the amount of KOH remaining after the reaction:
n(KOH) \u003d 1.215 mol - 1.2 mol \u003d 0.015 mol;
its mass is m(KOH) = n(KOH) × M(KOH) = 0.015 mol × 56 g/mol = 0.84 g.
The calculation of the resulting reaction product (precipitate Ca 3 (PO 4) 2) should be carried out according to the substance that is in short supply (in this case, Ca (H 2 PO 4) 2), since this substance will react completely. It can be seen from the reaction equation that the number of moles of the resulting Ca 3 (PO 4) 2 is 3 times less than the number of moles of the reacted Ca (H 2 PO 4) 2:
n (Ca 3 (PO 4) 2) = 0.3 mol: 3 = 0.1 mol.
Therefore, m (Ca 3 (PO 4) 2) \u003d n (Ca 3 (PO 4) 2) × M (Ca 3 (PO 4) 2) \u003d 0.1 mol × 310 g / mol \u003d 31 g.
Task number 5
a) Calculate the chemical quantities of the reactants given in Table 5 (volumes of gaseous substances are given under normal conditions);
b) arrange the coefficients in a given reaction scheme and, using the reaction equation, determine which of the substances is in excess and which is in short supply;
c) find the chemical amount of the reaction product indicated in table 5;
d) calculate the mass or volume (see Table 5) of this reaction product.
Table 5 - Conditions of task No. 5
| option number | Reactive Substances | Reaction scheme | Calculate |
| m(Fe)=11.2 g; V (Cl 2) \u003d 5.376 l | Fe + Cl 2 ® FeCl 3 | m(FeCl 3) | |
| m(Al)=5.4 g; m(H 2 SO 4) \u003d 39.2 g | Al + H 2 SO 4 ® Al 2 (SO 4) 3 + H 2 | V(H2) | |
| V(CO)=20 l; m(O 2) \u003d 20 g | CO+O2 ® CO2 | V(CO2) | |
| m(AgNO 3)=3.4 g; m(Na 2 S)=1.56 g | AgNO 3 +Na 2 S®Ag 2 S+NaNO 3 | m(Ag 2 S) | |
| m(Na 2 CO 3)=53 g; m(HCl)=29.2 g | Na 2 CO 3 +HCl®NaCl+CO 2 +H 2 O | V(CO2) | |
| m (Al 2 (SO 4) 3) \u003d 34.2 g; m (BaCl 2) \u003d 52 g | Al 2 (SO 4) 3 + BaCl 2 ®AlCl 3 + BaSO 4 | m(BaSO4) | |
| m(KI)=3.32 g; V(Cl 2) \u003d 448 ml | KI+Cl 2 ® KCl+I 2 | m(I2) | |
| m(CaCl 2)=22.2 g; m(AgNO 3) \u003d 59.5 g | CaCl 2 + AgNO 3 ®AgCl + Ca (NO 3) 2 | m(AgCl) | |
| m(H 2 )=0.48 g; V (O 2) \u003d 2.8 l | H 2 + O 2 ® H 2 O | m(H 2 O) | |
| m (Ba (OH) 2) \u003d 3.42 g; V(HCl)=784ml | Ba(OH) 2 +HCl ® BaCl 2 +H 2 O | m(BaCl2) |
Table 5 continued
| option number | Reactive Substances | Reaction scheme | Calculate |
| m(H 3 PO 4)=9.8 g; m(NaOH)=12.2 g | H 3 PO 4 + NaOH ® Na 3 PO 4 + H 2 O | m(Na3PO4) | |
| m(H 2 SO 4)=9.8 g; m(KOH)=11.76 g | H 2 SO 4 +KOH ® K 2 SO 4 +H 2 O | m(K 2 SO 4) | |
| V(Cl 2)=2.24 l; m(KOH)=10.64 g | Cl 2 +KOH ® KClO + KCl + H 2 O | m(KClO) | |
| m ((NH 4) 2 SO 4) \u003d 66 g; m (KOH) \u003d 50 g | (NH 4) 2 SO 4 +KOH®K 2 SO 4 +NH 3 +H 2 O | V(NH3) | |
| m(NH 3)=6.8 g; V (O 2) \u003d 7.84 l | NH 3 + O 2 ® N 2 + H 2 O | V(N2) | |
| V(H 2 S)=11.2 l; m(O 2) \u003d 8.32 g | H 2 S+O 2 ® S+H 2 O | m(S) | |
| m(MnO 2)=8.7 g; m(HCl)=14.2 g | MnO 2 +HCl ® MnCl 2 +Cl 2 +H 2 O | V(Cl2) | |
| m(Al)=5.4 g; V (Cl 2) \u003d 6.048 l | Al+Cl 2 ® AlCl 3 | m(AlCl 3) | |
| m(Al)=10.8 g; m(HCl)=36.5 g | Al+HCl ® AlCl 3 +H 2 | V(H2) | |
| m(P)=15.5 g; V (O 2) \u003d 14.1 l | P+O 2 ® P 2 O 5 | m(P 2 O 5) | |
| m (AgNO 3) \u003d 8.5 g; m (K 2 CO 3) \u003d 4.14 g | AgNO 3 + K 2 CO 3 ®Ag 2 CO 3 + KNO 3 | m(Ag 2 CO 3) | |
| m(K 2 CO 3)=69 g; m(HNO 3) \u003d 50.4 g | K 2 CO 3 + HNO 3 ®KNO 3 + CO 2 + H 2 O | V(CO2) | |
| m(AlCl 3)=2.67 g; m(AgNO 3) \u003d 8.5 g | AlCl 3 + AgNO 3 ®AgCl + Al (NO 3) 3 | m(AgCl) | |
| m(KBr)=2.38 g; V(Cl 2) \u003d 448 ml | KBr+Cl 2 ® KCl+Br 2 | m(Br2) | |
| m(CaBr 2)=40 g; m(AgNO 3) \u003d 59.5 g | CaBr 2 + AgNO 3 ®AgBr + Ca (NO 3) 2 | m(AgBr) | |
| m(H 2 )=1.44 g; V (O 2) \u003d 8.4 l | H 2 + O 2 ® H 2 O | m(H 2 O) | |
| m (Ba (OH) 2) \u003d 6.84 g; V (HI) \u003d 1.568 l | Ba(OH) 2 +HI ® BaI 2 +H 2 O | m(BaI 2) | |
| m(H 3 PO 4)=9.8 g; m(KOH)=17.08 g | H 3 PO 4 +KOH ® K 3 PO 4 +H 2 O | m(K 3 PO 4) | |
| m(H 2 SO 4)=49 g; m(NaOH)=45 g | H 2 SO 4 + NaOH ® Na 2 SO 4 + H 2 O | m(Na 2 SO 4) | |
| V(Cl 2)=2.24 l; m(KOH)=8.4 g | Cl 2 +KOH ® KClO 3 +KCl + H 2 O | m(KClO 3) | |
| m(NH 4 Cl)=43 g; m (Ca (OH) 2) \u003d 37 g | NH 4 Cl + Ca (OH) 2 ®CaCl 2 + NH 3 + H 2 O | V(NH3) | |
| V(NH 3) \u003d 8.96 l; m(O 2) \u003d 14.4 g | NH 3 + O 2 ® NO + H 2 O | V(NO) | |
| V(H 2 S)=17.92 l; m(O 2) \u003d 40 g | H 2 S + O 2 ® SO 2 + H 2 O | V(SO2) | |
| m(MnO 2)=8.7 g; m(HBr)=30.8 g | MnO 2 +HBr ® MnBr 2 +Br 2 +H 2 O | m(MnBr 2) | |
| m(Ca)=10 g; m(H 2 O)=8.1 g | Ca + H 2 O ® Ca (OH) 2 + H 2 | V(H2) |
SOLUTION CONCENTRATION
As part of the general chemistry course, students learn 2 ways to express the concentration of solutions - mass fraction and molar concentration.
Mass fraction of the dissolved substance X is calculated as the ratio of the mass of this substance to the mass of the solution:
,
where ω(X) is the mass fraction of the dissolved substance X;
m(X) is the mass of the dissolved substance X;
m solution - the mass of the solution.
The mass fraction of a substance calculated according to the above formula is a dimensionless quantity expressed in fractions of a unit (0< ω(X) < 1).
The mass fraction can be expressed not only in fractions of a unit, but also as a percentage. In this case, the calculation formula looks like:
Mass fraction, expressed as a percentage, is often called percentage concentration . Obviously, the percentage concentration of the solute is 0%< ω(X) < 100%.
Percent concentration shows how many mass parts of a solute are contained in 100 mass parts of a solution. If you choose grams as the unit of mass, then this definition can also be written as follows: percentage concentration shows how many grams of a solute are contained in 100 grams of solution.
It is clear that, for example, a 30% solution corresponds to a mass fraction of a dissolved substance equal to 0.3.
Another way of expressing the content of a solute in a solution is the molar concentration (molarity).
The molar concentration of a substance, or the molarity of a solution, shows how many moles of a solute are contained in 1 liter (1 dm 3) of a solution
where C(X) is the molar concentration of solute X (mol/l);
n(X) is the chemical amount of dissolved substance X (mol);
V solution - the volume of the solution (l).
Example 5.1 Calculate the molar concentration of H 3 PO 4 in the solution, if it is known that the mass fraction of H 3 PO 4 is 60%, and the density of the solution is 1.43 g / ml.
By definition of percentage concentration
100 g of solution contains 60 g of phosphoric acid.
n (H 3 PO 4) \u003d m (H 3 PO 4) : M (H 3 PO 4) \u003d 60 g: 98 g / mol \u003d 0.612 mol;
V solution \u003d m solution: ρ solution \u003d 100 g: 1.43 g / cm 3 \u003d 69.93 cm 3 \u003d 0.0699 l;
C (H 3 PO 4) \u003d n (H 3 PO 4): V solution \u003d 0.612 mol: 0.0699 l \u003d 8.755 mol / l.
Example 5.2 There is a 0.5 M solution of H 2 SO 4 . What is the mass fraction of sulfuric acid in this solution? Take the density of the solution equal to 1 g/ml.
By definition of molar concentration
1 liter of solution contains 0.5 mol H 2 SO 4
(The entry "0.5 M solution" means that C (H 2 SO 4) \u003d 0.5 mol / l).
m solution = V solution × ρ solution = 1000 ml × 1 g/ml = 1000 g;
m (H 2 SO 4) \u003d n (H 2 SO 4) × M (H 2 SO 4) \u003d 0.5 mol × 98 g / mol \u003d 49 g;
ω (H 2 SO 4) \u003d m (H 2 SO 4) : m solution \u003d 49 g: 1000 g \u003d 0.049 (4.9%).
Example 5.3 What volumes of water and a 96% solution of H 2 SO 4 with a density of 1.84 g / ml should be taken to prepare 2 liters of a 60% solution of H 2 SO 4 with a density of 1.5 g / ml.
When solving problems for the preparation of a dilute solution from a concentrated one, it should be taken into account that the initial solution (concentrated), water and the resulting solution (diluted) have different densities. In this case, it should be borne in mind that V of the original solution + V of water ≠ V of the resulting solution,
because in the course of mixing a concentrated solution and water, a change (increase or decrease) in the volume of the entire system occurs.
The solution of such problems must begin with finding out the parameters of a dilute solution (i.e., the solution that needs to be prepared): its mass, the mass of the dissolved substance, if necessary, and the amount of the dissolved substance.
M 60% solution = V 60% solution ∙ ρ 60% solution = 2000 ml × 1.5 g/ml = 3000 g
m (H 2 SO 4) in 60% solution \u003d m 60% solution w (H 2 SO 4) in 60% solution \u003d 3000 g 0.6 \u003d 1800 g.
The mass of pure sulfuric acid in the prepared solution should be equal to the mass of sulfuric acid in that portion of the 96% solution that must be taken to prepare the dilute solution. Thus,
m (H 2 SO 4) in 60% solution \u003d m (H 2 SO 4) in 96% solution \u003d 1800 g.
m 96% solution = m (H 2 SO 4) in 96% solution: w (H 2 SO 4) in 96% solution = 1800 g: 0.96 = 1875 g.
m (H 2 O) \u003d m 40% solution - m 96% solution \u003d 3000 g - 1875 g \u003d 1125 g.
V 96% solution \u003d m 96% solution: ρ 96% solution \u003d 1875 g: 1.84 g / ml \u003d 1019 ml » 1.02 l.
V water \u003d m water: ρ water \u003d 1125g: 1 g / ml \u003d 1125 ml \u003d 1.125 l.
Example 5.4 Mixed 100 ml of a 0.1 M solution of CuCl 2 and 150 ml of a 0.2 M solution of Cu(NO 3) 2 Calculate the molar concentration of Cu 2+, Cl - and NO 3 - ions in the resulting solution.
When solving a similar problem of mixing dilute solutions, it is important to understand that dilute solutions have approximately the same density, approximately equal to the density of water. When they are mixed, the total volume of the system practically does not change: V 1 of a dilute solution + V 2 of a dilute solution + ... "V of the resulting solution.
In the first solution:
n (CuCl 2) \u003d C (CuCl 2) V solution of CuCl 2 \u003d 0.1 mol / l × 0.1 l \u003d 0.01 mol;
CuCl 2 - strong electrolyte: CuCl 2 ® Cu 2+ + 2Cl -;
Therefore, n (Cu 2+) \u003d n (CuCl 2) \u003d 0.01 mol; n(Cl -) \u003d 2 × 0.01 \u003d 0.02 mol.
In the second solution:
n (Cu (NO 3) 2) \u003d C (Cu (NO 3) 2) × V solution Cu (NO 3) 2 \u003d 0.2 mol / l × 0.15 l \u003d 0.03 mol;
Cu(NO 3) 2 - strong electrolyte: CuCl 2 ® Cu 2+ + 2NO 3 -;
Therefore, n (Cu 2+) \u003d n (Cu (NO 3) 2) \u003d 0.03 mol; n (NO 3 -) \u003d 2 × 0.03 \u003d 0.06 mol.
After mixing solutions:
n(Cu2+)tot. = 0.01 mol + 0.03 mol = 0.04 mol;
V common. » Vsolution CuCl 2 + Vsolution Cu(NO 3) 2 \u003d 0.1 l + 0.15 l \u003d 0.25 l;
C(Cu 2+) = n(Cu 2+) : Vtot. \u003d 0.04 mol: 0.25 l \u003d 0.16 mol / l;
C(Cl -) = n(Cl -) : Vtot. \u003d 0.02 mol: 0.25 l \u003d 0.08 mol / l;
C (NO 3 -) \u003d n (NO 3 -): V total. \u003d 0.06 mol: 0.25 l \u003d 0.24 mol / l.
Example 5.5 684 mg of aluminum sulfate and 1 ml of a 9.8% sulfuric acid solution with a density of 1.1 g/ml were added to the flask. The resulting mixture was dissolved in water; The volume of the solution was made up to 500 ml with water. Calculate the molar concentrations of H + , Al 3+ SO 4 2– ions in the resulting solution.
Calculate the amount of dissolved substances:
n (Al 2 (SO 4) 3) \u003d m (Al 2 (SO 4) 3) : M (Al 2 (SO 4) 3) \u003d 0.684 g: 342 g mol \u003d 0.002 mol;
Al 2 (SO 4) 3 - strong electrolyte: Al 2 (SO 4) 3 ® 2Al 3+ + 3SO 4 2–;
Therefore, n(Al 3+)=2×0.002 mol=0.004 mol; n (SO 4 2–) \u003d 3 × 0.002 mol \u003d 0.006 mol.
m solution of H 2 SO 4 \u003d V solution of H 2 SO 4 × ρ solution of H 2 SO 4 \u003d 1 ml × 1.1 g / ml \u003d 1.1 g;
m (H 2 SO 4) \u003d m solution of H 2 SO 4 × w (H 2 SO 4) \u003d 1.1 g 0.098 \u003d 0.1078 g.
n (H 2 SO 4) \u003d m (H 2 SO 4) : M (H 2 SO 4) \u003d 0.1078 g: 98 g / mol \u003d 0.0011 mol;
H 2 SO 4 is a strong electrolyte: H 2 SO 4 ® 2H + + SO 4 2–.
Therefore, n (SO 4 2–) \u003d n (H 2 SO 4) \u003d 0.0011 mol; n(H +) \u003d 2 × 0.0011 \u003d 0.0022 mol.
According to the condition of the problem, the volume of the resulting solution is 500 ml (0.5 l).
n(SO 4 2–)tot. \u003d 0.006 mol + 0.0011 mol \u003d 0.0071 mol.
C (Al 3+) \u003d n (Al 3+): V solution \u003d 0.004 mol: 0.5 l \u003d 0.008 mol / l;
C (H +) \u003d n (H +) : V solution \u003d 0.0022 mol: 0.5 l \u003d 0.0044 mol / l;
C (SO 4 2–) \u003d n (SO 4 2–) total. : V solution \u003d 0.0071 mol: 0.5 l \u003d 0.0142 mol / l.
Example 5.6 What mass of ferrous sulfate (FeSO 4 7H 2 O) and what volume of water must be taken to prepare 3 liters of a 10% solution of iron (II) sulfate. Take the density of the solution equal to 1.1 g/ml.
The mass of the solution to be prepared is:
m solution = V solution ∙ ρ solution = 3000 ml ∙ 1.1 g/ml = 3300 g.
The mass of pure iron (II) sulfate in this solution is:
m (FeSO 4) \u003d m solution × w (FeSO 4) \u003d 3300 g × 0.1 \u003d 330 g.
The same mass of anhydrous FeSO 4 must be contained in the amount of crystalline hydrate that must be taken to prepare the solution. From a comparison of the molar masses M (FeSO 4 7H 2 O) \u003d 278 g / mol and M (FeSO 4) \u003d 152 g / mol,
we get the proportion:
278 g FeSO 4 7H 2 O contains 152 g FeSO 4;
x g of FeSO 4 7H 2 O contains 330 g of FeSO 4;
x \u003d (278 330) : 152 \u003d 603.6 g.
m water \u003d m solution - m ferrous sulfate \u003d 3300 g - 603.6 g \u003d 2696.4 g.
Because the density of water is 1 g / ml, then the volume of water that must be taken to prepare the solution is: V water \u003d m water: ρ water \u003d 2696.4 g: 1 g / ml \u003d 2696.4 ml.
Example 5.7 What mass of Glauber's salt (Na 2 SO 4 10H 2 O) must be dissolved in 500 ml of 10% sodium sulfate solution (solution density 1.1 g / ml) to obtain a 15% Na 2 SO 4 solution?
Let x grams of Glauber's salt Na 2 SO 4 10H 2 O be required. Then the mass of the resulting solution is:
m 15% solution = m original (10%) solution + m Glauber's salt = 550 + x (g);
m initial (10%) solution = V 10% solution × ρ 10% solution = 500 ml × 1.1 g/ml = 550 g;
m (Na 2 SO 4) in the original (10%) solution \u003d m 10% solution a w (Na 2 SO 4) \u003d 550 g 0.1 \u003d 55 g.
Express through x the mass of pure Na 2 SO 4 contained in x grams of Na 2 SO 4 10H 2 O.
M (Na 2 SO 4 10H 2 O) \u003d 322 g / mol; M (Na 2 SO 4) \u003d 142 g / mol; hence:
322 g of Na 2 SO 4 10H 2 O contains 142 g of anhydrous Na 2 SO 4;
x g of Na 2 SO 4 10H 2 O contains m g of anhydrous Na 2 SO 4.
m(Na 2 SO 4) \u003d 142 x: 322 \u003d 0.441 x x.
The total mass of sodium sulfate in the resulting solution will be equal to:
m (Na 2 SO 4) in 15% solution = 55 + 0.441 × x (g).
In the resulting solution: 
= 0,15
, whence x = 94.5 g.
Task number 6
Table 6 - Conditions of task No. 6
| option number | Condition text |
| 5 g of Na 2 SO 4 × 10H 2 O were dissolved in water, and the volume of the resulting solution was brought to 500 ml with water. Calculate the mass fraction of Na 2 SO 4 in this solution (ρ = 1 g/ml) and the molar concentrations of Na + and SO 4 2– ions. | |
| Mixed solutions: 100 ml of 0.05M Cr 2 (SO 4) 3 and 100 ml of 0.02M Na 2 SO 4 . Calculate the molar concentrations of Cr 3+ , Na + and SO 4 2– ions in the resulting solution. | |
| What volumes of water and a 98% solution (density 1.84 g/ml) of sulfuric acid should be taken to prepare 2 liters of a 30% solution with a density of 1.2 g/ml? | |
| 50 g of Na 2 CO 3 × 10H 2 O were dissolved in 400 ml of water. What are the molar concentrations of Na + and CO 3 2– ions and the mass fraction of Na 2 CO 3 in the resulting solution (ρ = 1.1 g / ml)? | |
| Mixed solutions: 150 ml of 0.05 M Al 2 (SO 4) 3 and 100 ml of 0.01 M NiSO 4 . Calculate the molar concentrations of Al 3+ , Ni 2+ , SO 4 2- ions in the resulting solution. | |
| What volumes of water and a 60% solution (density 1.4 g/ml) of nitric acid will be required to prepare 500 ml of a 4 M solution (density 1.1 g/ml)? | |
| What mass of copper sulfate (CuSO 4 × 5H 2 O) is needed to prepare 500 ml of a 5% solution of copper sulfate with a density of 1.05 g / ml? | |
| 1 ml of a 36% solution (ρ = 1.2 g/ml) of HCl and 10 ml of a 0.5 M solution of ZnCl 2 were added to the flask. The volume of the resulting solution was brought to 50 ml with water. What are the molar concentrations of H + , Zn 2+ , Cl - ions in the resulting solution? | |
| What is the mass fraction of Cr 2 (SO 4) 3 in a solution (ρ » 1 g / ml), if it is known that the molar concentration of sulfate ions in this solution is 0.06 mol / l? | |
| What volumes of water and 10 M solution (ρ=1.45 g/ml) of sodium hydroxide will be required to prepare 2 liters of 10% NaOH solution (ρ= 1.1 g/ml)? | |
| How many grams of ferrous sulfate FeSO 4 × 7H 2 O can be obtained by evaporating water from 10 liters of a 10% iron (II) sulfate solution (solution density 1.2 g / ml)? | |
| Mixed solutions: 100 ml of 0.1 M Cr 2 (SO 4) 3 and 50 ml of 0.2 M CuSO 4 . Calculate the molar concentrations of Cr 3+ , Cu 2+ , SO 4 2- ions in the resulting solution. |
Table 6 continued
| option number | Condition text |
| What volumes of water and a 40% solution of phosphoric acid with a density of 1.35 g / ml will be required to prepare 1 m 3 of a 5% solution of H 3 PO 4, the density of which is 1.05 g / ml? | |
| 16.1 g of Na 2 SO 4 × 10H 2 O were dissolved in water and the volume of the resulting solution was brought to 250 ml with water. Calculate the mass fraction and molar concentration of Na 2 SO 4 in the resulting solution (assume that the density of the solution is 1 g/ml). | |
| Mixed solutions: 150 ml of 0.05 M Fe 2 (SO 4) 3 and 100 ml of 0.1 M MgSO 4 . Calculate the molar concentrations of Fe 3+ , Mg 2+ , SO 4 2– ions in the resulting solution. | |
| What volumes of water and 36% hydrochloric acid (density 1.2 g/ml) are needed to prepare 500 ml of a 10% solution with a density of 1.05 g/ml? | |
| 20 g of Al 2 (SO 4) 3 × 18H 2 O were dissolved in 200 ml of water. What is the mass fraction of the solute in the resulting solution, the density of which is 1.1 g / ml? Calculate the molar concentrations of Al 3+ and SO 4 2– ions in this solution. | |
| Mixed solutions: 100 ml of 0.05 M Al 2 (SO 4) 3 and 150 ml of 0.01 M Fe 2 (SO 4) 3 . Calculate the molar concentrations of Fe 3+ , Al 3+ and SO 4 2– ions in the resulting solution. | |
| What volumes of water and 80% solution of acetic acid (density 1.07 g/ml) will be required to prepare 0.5 l of table vinegar, in which the mass fraction of acid is 7%? Take the density of table vinegar equal to 1 g/ml. | |
| What mass of ferrous sulfate (FeSO 4 × 7H 2 O) is needed to prepare 100 ml of a 3% solution of ferrous sulfate? The density of the solution is 1 g/ml. | |
| 2 ml of 36% HCl solution (density 1.2 g/cm 3 ) and 20 ml of 0.3 M CuCl 2 solution were added to the flask. The volume of the resulting solution was brought to 200 ml with water. Calculate the molar concentrations of H + , Cu 2+ and Cl - ions in the resulting solution. | |
| What is the percentage concentration of Al 2 (SO 4) 3 in a solution in which the molar concentration of sulfate ions is 0.6 mol / l. The density of the solution is 1.05 g/ml. | |
| What volumes of water and 10 M KOH solution (solution density 1.4 g/ml) will be required to prepare 500 ml of 10% KOH solution with a density of 1.1 g/ml? | |
| How many grams of copper sulfate CuSO 4 × 5H 2 O can be obtained by evaporating water from 15 liters of 8% copper sulfate solution, the density of which is 1.1 g / ml? | |
| Mixed solutions: 200 ml of 0.025 M Fe 2 (SO 4) 3 and 50 ml of 0.05 M FeCl 3 . Calculate the molar concentration of Fe 3+ , Cl - , SO 4 2- ions in the resulting solution. | |
| What volumes of water and a 70% solution of H 3 PO 4 (density 1.6 g/ml) will be required to prepare 0.25 m 3 of a 10% solution of H 3 PO 4 (density 1.1 g/ml)? | |
| 6 g of Al 2 (SO 4) 3 × 18H 2 O were dissolved in 100 ml of water. Calculate the mass fraction of Al 2 (SO 4) 3 and the molar concentrations of Al 3+ and SO 4 2– ions in the resulting solution, the density of which is 1 g /ml | |
| Mixed solutions: 50 ml of 0.1 M Cr 2 (SO 4) 3 and 200 ml of 0.02 M Cr(NO 3) 3 . Calculate the molar concentrations of Cr 3+ , NO 3 - , SO 4 2- ions in the resulting solution. | |
| What volumes of a 50% solution of perchloric acid (density 1.4 g/ml) and water are needed to prepare 1 liter of an 8% solution with a density of 1.05 g/ml? | |
| How many grams of Glauber's salt Na 2 SO 4 × 10H 2 O must be dissolved in 200 ml of water to obtain a 5% sodium sulfate solution? | |
| 1 ml of 80% solution of H 2 SO 4 (solution density 1.7 g/ml) and 5000 mg of Cr 2 (SO 4) 3 were added to the flask. The mixture was dissolved in water; the volume of the solution was brought to 250 ml. Calculate the molar concentrations of H + , Cr 3+ and SO 4 2– ions in the resulting solution. |
Table 6 continued
CHEMICAL EQUILIBRIUM
All chemical reactions can be divided into 2 groups: irreversible reactions, i.e. reactions proceeding until the complete consumption of at least one of the reacting substances, and reversible reactions in which none of the reacting substances is completely consumed. This is due to the fact that a reversible reaction can proceed both in the forward and reverse directions. A classic example of a reversible reaction is the synthesis of ammonia from nitrogen and hydrogen:
N 2 + 3 H 2 ⇆ 2 NH 3.
At the start of the reaction, the concentrations of the initial substances in the system are maximum; at this moment, the rate of the forward reaction is also maximum. At the start of the reaction, there are still no reaction products in the system (in this example, ammonia), therefore, the rate of the reverse reaction is zero. As the initial substances interact with each other, their concentrations decrease, therefore, the rate of the direct reaction also decreases. The concentration of the reaction product gradually increases, therefore, the rate of the reverse reaction also increases. After some time, the rate of the forward reaction becomes equal to the rate of the reverse. This state of the system is called state of chemical equilibrium. The concentrations of substances in a system that is in a state of chemical equilibrium are called equilibrium concentrations. The quantitative characteristic of a system in a state of chemical equilibrium is equilibrium constant.
For any reversible reaction a A + b B+ ... ⇆ p P + q Q + …, the expression for the chemical equilibrium constant (K) is written as a fraction, in the numerator of which are the equilibrium concentrations of the reaction products, and in the denominator are the equilibrium concentrations of the starting substances, moreover, the concentration of each substance must be raised to a power equal to the stoichiometric coefficient in the reaction equation.
For example, for the reaction N 2 + 3 H 2 ⇆ 2 NH 3.
It should be borne in mind that the expression of the equilibrium constant includes the equilibrium concentrations of only gaseous substances or substances that are in a dissolved state . The concentration of a solid is assumed to be constant and is not written into the equilibrium constant expression.
CO 2 (gas) + C (solid) ⇆ 2CO (gas)
CH 3 COOH (solution) ⇆ CH 3 COO - (solution) + H + (solution)
Ba 3 (PO 4) 2 (solid) ⇆ 3 Ba 2+ (saturated solution) + 2 PO 4 3– (saturated solution) K \u003d C 3 (Ba 2+) C 2 (PO 4 3–)
There are two most important types of problems associated with calculating the parameters of an equilibrium system:
1) the initial concentrations of the starting substances are known; from the condition of the problem, one can find the concentrations of substances that have reacted (or formed) by the time equilibrium is reached; in the problem it is required to calculate the equilibrium concentrations of all substances and the numerical value of the equilibrium constant;
2) the initial concentrations of the initial substances and the equilibrium constant are known. The condition does not contain data on the concentrations of reacted or formed substances. It is required to calculate the equilibrium concentrations of all participants in the reaction.
To solve such problems, it is necessary to understand that the equilibrium concentration of any initial substances can be found by subtracting the concentration of the reacted substance from the initial concentration:
C equilibrium \u003d C initial - C of the reacted substance.
Equilibrium concentration reaction product is equal to the concentration of the product formed at the time of equilibrium:
C equilibrium \u003d C of the resulting product.
Thus, in order to calculate the parameters of an equilibrium system, it is very important to be able to determine how much of the initial substance had reacted by the time equilibrium was reached and how much of the reaction product was formed. To determine the amount (or concentration) of the reacted and formed substances, stoichiometric calculations are carried out according to the reaction equation.
Example 6.1 The initial concentrations of nitrogen and hydrogen in the equilibrium system N 2 + 3H 2 ⇆ 2 NH 3 are 3 mol/l and 4 mol/l, respectively. By the time the chemical equilibrium was reached, 70% of hydrogen from its initial amount remained in the system. Determine the equilibrium constant of this reaction.
It follows from the conditions of the problem that by the time equilibrium was reached, 30% of hydrogen had reacted (problem 1 type):
4 mol/l H 2 - 100%
x mol / l H 2 - 30%
x \u003d 1.2 mol / l \u003d C proreag. (H2)
As can be seen from the reaction equation, nitrogen should have reacted 3 times less than hydrogen, i.e. With proreact. (N 2) \u003d 1.2 mol / l: 3 \u003d 0.4 mol / l. Ammonia is formed 2 times more than nitrogen reacted:
From images. (NH 3) \u003d 2 × 0.4 mol / l \u003d 0.8 mol / l
The equilibrium concentrations of all participants in the reaction will be as follows:
Equal (H 2) \u003d C initial. (H 2) - C proreact. (H 2) \u003d 4 mol / l - 1.2 mol / l \u003d 2.8 mol / l;
Equal (N 2) \u003d C beg. (N 2) – C proreact. (N 2) \u003d 3 mol / l - 0.4 mol / l \u003d 2.6 mol / l;
Equal (NH 3) = C images. (NH 3) \u003d 0.8 mol / l.
Equilibrium constant = 
.
Example 6.2 Calculate the equilibrium concentrations of hydrogen, iodine and hydrogen iodine in the system H 2 + I 2 ⇆ 2 HI, if it is known that the initial concentrations of H 2 and I 2 are 5 mol/l and 3 mol/l, respectively, and the equilibrium constant is 1.
It should be noted that in the condition of this problem (task of type 2), the condition does not say anything about the concentrations of the reacted initial substances and the products formed. Therefore, when solving such problems, the concentration of some reacted substance is usually taken as x.
Let x mol/l H 2 have reacted by the time equilibrium is reached. Then, as follows from the reaction equation, x mol/l I 2 should react, and 2x mol/l HI should be formed. The equilibrium concentrations of all participants in the reaction will be as follows:
Equal (H 2) \u003d C beg. (H 2) - C proreact. (H 2) \u003d (5 - x) mol / l;
Equal (I 2) = C beg. (I 2) – C proreact. (I 2) \u003d (3 - x) mol / l;
Equal (HI) = C images. (HI) = 2x mol/l.
4x2 = 15 - 8x + x2
3x2 + 8x - 15 = 0
x 1 = -3.94 x 2 = 1.27
Only the positive root x = 1.27 has physical meaning.
Therefore, C equal. (H 2) \u003d (5 - x) mol / l \u003d 5 - 1.27 \u003d 3.73 mol / l;
Equal (I 2) \u003d (3 - x) mol / l \u003d 3 - 1.27 \u003d 1.73 mol / l;
Equal (HI) \u003d 2x mol / l \u003d 2 1.27 \u003d 2.54 mol / l.
Task number 7
Table 7 - Conditions of task No. 7
Table 7 continued
stoichiometry- quantitative ratios between reacting substances.
If the reactants enter into 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 in front of 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) = coefficient in front of the corresponding molecule.
N A=6.02×10 23 mol -1 .
η - the ratio of the actual mass of the product m p to the theoretically possible m t, expressed in fractions of a unit or as a percentage.
If the yield of reaction products is not specified in the condition, then in the calculations it is taken equal to 100% (quantitative yield).
Calculation scheme according to the equations of chemical reactions:
- 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 unknown, write down the corresponding values of these quantities found from the reaction equation.
- Compose and solve proportions.
Example. Calculate the mass and amount of magnesium oxide substance formed during the complete combustion of 24 g of magnesium.
|
Given: m(Mg) = 24 g Find: ν (MgO) m (MgO) |
Solution: 1. Let's make the equation of chemical reaction: 2Mg + O 2 \u003d 2MgO. 2. Under the formulas of substances, we indicate the amount of substance (number of moles), which corresponds to stoichiometric coefficients: 2Mg + O 2 \u003d 2MgO 2 mol 2 mol 3. Determine the molar mass of magnesium: Relative atomic mass of magnesium Ar(Mg) = 24. Because the value of the molar mass is equal to the relative atomic or molecular mass, then M(Mg)= 24 g/mol. 4. By the mass of the substance given in the condition, we calculate the amount of the substance:
5. Above the chemical formula of magnesium oxide MgO, whose mass is unknown, we set xmole, over magnesium formula mg write its molar mass: 1 mol xmole 2Mg + O 2 \u003d 2MgO 2 mol 2 mol
According to the rules for solving proportions:
The amount of magnesium oxide v(MgO)= 1 mol. 7. Calculate the molar mass of magnesium oxide: M (Mg)\u003d 24 g / mol, M (O)=16 g/mol. M(MgO)= 24 + 16 = 40 g/mol. Calculate the mass of magnesium oxide: m (MgO) \u003d ν (MgO) × M (MgO) \u003d 1 mol × 40 g / mol \u003d 40 g. Answer: ν (MgO) = 1 mol; m(MgO) = 40 g. |
When compiling the equations of redox reactions, the following two important rules must be observed:
Rule 1: In any ionic equation, charge conservation must be observed. This means that the sum of all charges on the left side of the equation ("left") must match the sum of all charges on the right side of the equation ("right"). This rule applies to any ionic equation, both for complete reactions and for half-reactions.
Charges from left to right
Rule 2: The number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction 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:
The following procedure can be used to derive the full redox equation from the equations of the two half-reactions:
1. The equations of each of the two half-reactions are balanced separately, and to fulfill the above rule 1, the corresponding number of electrons is added to the left or right side of each equation.
2. The equations of both half-reactions are balanced with respect to 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, 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 us balance the equations of the half-reactions given below and formulate an equation for the redox reaction of the oxidation of an aqueous solution of any ferrous salt into a ferric salt with the help of 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 its left side, or subtract the same number of electrons from the right side. After that 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 reduction half-reaction.
For equation (6), we can write
To balance this equation, you can add one electron to its right side. Then
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).[ ...]
| 24.5 |
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 different microbiological growth conditions.[ ...]
From expression (3.36) we find the stoichiometric coefficient "sat.r = 0.05 g P / g COD (HAc).[ ...]
[ ...]
From example 3.2, you can find the stoichiometric coefficients of the equation for the removal of acetic acid: 1 mol of HAs (60 g of HAs) requires 0.9 mol of 02 and 0.9 32 = 29 g of 02.[ ...]
| 3.12 |
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 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 a constant crankshaft speed of 1,500 min1. The excess air coefficient varied in the range of 0.8 [ ...]
Material processes in living nature, the cycles of biogenic elements are associated with energy flows by stoichiometric coefficients that vary in a wide variety of 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 the latest works, it is assumed that 1 mol of NO2 gives 0.72 g-ion of NO7. According to data provided by the International Organization for 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 biochemical wastewater treatment plants and analyzing their operation, the following design parameters are usually used: the rate of biological oxidation, stoichiometric coefficients for electron acceptors, growth rate and physical properties of 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 is at a given temperature 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, we can assume 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 Bae 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). Thus, it is possible to meet the needs of many industries that need to neutralize small amounts of liquid waste with a wide range of chemical composition components 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.
