Story
Title page of Tyrocinium Chymicum.
At first there was no concept of chemical equations, the basic chemical laws were not yet known, but already in ancient times, in the alchemical period of the development of chemistry, they began to designate chemical elements with symbols.
With the further development of chemistry, ideas about the symbolism of chemical elements changed, knowledge about their compounds expanded. With the discovery of many chemical phenomena, it became necessary to move from their verbal description to a more convenient mathematical notation using chemical formulas. Jean Beguin was the first to propose the use of chemical equations in 1615 in the first chemistry textbook Tyrocinium Chymicum ("Principles of Chemistry").
Late XVIII - early XIX centuries - the formation of the laws of stoichiometry. At the origins of these studies was the German scientist I. V. Richter. In his student years, he was greatly impressed by the words of his teacher, the philosopher I. Kant, that in certain areas of the natural sciences there is as much true science as there is mathematics in it. Richter devoted his dissertation to the use of mathematics in chemistry. Not being essentially a chemist, Richter introduced the first quantitative equations of chemical reactions, began to use the term stoichiometry.
Compilation rules
On the left side of the equation, write down the formulas (formula) of the substances that have entered into the reaction, connecting them with a plus sign. On the right side of the equation, write down the formulas (formula) of the formed substances, also connected by a plus sign. An arrow is placed between the parts of the equation. Then they find odds- the numbers in front of the formulas of substances so that the number of atoms of the same elements in the left and right parts of the equation is equal.
To compile the equations of chemical reactions, in addition to knowing the formulas of the reactants and reaction products, it is necessary to choose the right coefficients. This can be done using simple rules:
1. Before the formula of a simple substance, you can write a fractional coefficient, which shows the amount of the substance of the reacting and formed substances.
2. If there is a salt formula in the reaction scheme, then first equalize the number of ions that form the salt.
3. If the substances involved in the reaction contain hydrogen and oxygen, then the hydrogen atoms are equalized in the penultimate turn, and the oxygen atoms - in the last.
4. If there are several salt formulas in the reaction scheme, then it is necessary to start equalizing with the ions that are part of the salt containing a larger number of them.
Symbols in chemical equations
To designate various types reactions, the following symbols are used:
Arrangement of coefficients in equations
The law of conservation of mass states that the amount of substance of each element before the reaction is equal to the amount of substance of each element after the reaction. Thus, the left and right sides of a chemical equation must have the same number of atoms of one or another element. The chemical equation must be electrically neutral, that is, the sum of the charges on the left and right sides of the equation must add up to zero. One way to equalize the number of atoms in a chemical equation is to select coefficients by trial and error. For more complex cases, a system of linear algebraic equations should be used. As a rule, chemical equations are written with the smallest integer coefficients. If there is no coefficient before the chemical formula, it is assumed that it is equal to one. Checking the material balance, that is, the number of atoms on the left and right sides, can be as follows: a coefficient 1 is placed before the most complex chemical formula. Next, the coefficients are placed in front of the formulas in such a way that the number of atoms of each of the elements on the left and right sides of the equation is equal to . If one of the coefficients is fractional, then all coefficients should be multiplied by the number in the denominator of the fractional coefficient. If the coefficient is 1 before the formula, then it is omitted. An example, the arrangement of coefficients in the chemical reaction of methane combustion:
1CH 4 + O 2 CO 2 + H 2 O
The number of carbon atoms on the left and right sides is the same. The next element to balance is hydrogen. There are 4 hydrogen atoms on the left, 2 on the right, to equalize the number of hydrogen atoms, put a factor of 2 in front of water, as a result:
1CH 4 + O 2 CO 2 + 2H 2 O
Checking the correct placement of the coefficients in any chemical equation is carried out by counting the number of oxygen atoms, if the number of oxygen atoms is the same on the left and right sides, then the coefficients are placed correctly.
1CH 4 + 2O 2 CO 2 + 2H 2 O
Before the CH 4 and CO 2 molecules, the coefficient 1 is omitted.
Redox reactions
Redox reactions (ORRs) are counter-parallel chemical reactions that occur with a change in the oxidation states of the atoms that make up the reactants, realized by the redistribution of electrons between the oxidizing atom and the reducing atom.
In the process of a redox reaction, the reducing agent gives up electrons, that is, it is oxidized; The oxidizing agent gains electrons, that is, it is reduced. Moreover, any redox reaction is a unity of two opposite transformations - oxidation and reduction, occurring simultaneously and without separation of one from the other.
Oxidation is the process of donating electrons, with an increase in the degree of oxidation. When a substance is oxidized, as a result of the return of electrons, its oxidation state increases. The atoms of the oxidized substance are called electron donors, and the atoms of the oxidizing agent are called electron acceptors. The oxidizing agent, accepting electrons, acquires reducing properties, turning into a conjugated reducing agent.
Recovery is the process of attaching electrons to an atom of a substance, while its oxidation state decreases. During reduction, atoms or ions gain electrons. In this case, the oxidation state of the element decreases. The reducing agent, donating electrons, acquires oxidizing properties, turning into a conjugated oxidizing agent.
When compiling an equation for a redox reaction, it is necessary to determine the reducing agent, oxidizing agent, and the number of given and received electrons. As a rule, the coefficients are selected using either the electron balance method or the electron-ion balance method (sometimes the latter is called the half-reaction method).
Selection of coefficients by the method of electronic balance.
AT simple equations coefficients are selected element by element in accordance with the formula of the final product. In more complex equations of redox reactions, the coefficients are selected using the electronic balance method:
1. Write down the reaction scheme (the formula of the reactants and products), and then find the elements that increase and decrease their oxidation states, and write them out separately;
2. Compose equations for half-reactions of reduction and oxidation, observing the laws of conservation of the number of atoms and charge in each half-reaction;
3. Additional factors are selected to equalize the half-reactions so that the charge conservation law is fulfilled for the reaction as a whole, for which the number of elements accepted in the reduction half-reactions is made equal to the number given elements in the oxidation half-reaction;
4. Put down (according to the factors found) stoichiometric coefficients in the reaction scheme (coefficient 1 is omitted);
5. Equalize the number of atoms of those elements that do not change their oxidation state during the course of the reaction (if there are two such elements, then it is enough to equalize the number of atoms of one of them, and check the second one). Get the chemical reaction equations;
6. They check for an element that has not changed its oxidation state (most often it is oxygen).
Arrangement of coefficients in ionic equations
Ionic equations are chemical equations in which electrolytes are written as dissociated ions. Ionic equations are used to write substitution reactions and exchange reactions in aqueous solutions. Example, exchange reaction, interaction of calcium chloride and silver nitrate with the formation of a precipitate of silver chloride:
CaCl 2 (l) + 2AgNO 3 (l) Ca(NO 3) 2 (l) + 2AgCl (tv)
full ionic equation:
Ca 2+ + 2Cl - + 2Ag + + 2NO 3 - Ca 2+ + 2NO 3 - + 2AgCl (solid)
reduced ionic equation:
2Cl − (l) + 2Ag + (l) 2AgCl(s)
ionic equation:
Ag + + Cl − AgCl(s)
Ions Ca 2+ and NO 3 - remain in solution, therefore they are not participants in the chemical reaction. In neutralization reactions, the ionic reaction equation is as follows:
H + + OH - H 2 O
There are several neutralization reactions that produce another low dissociating substance besides water. An example is the reaction of barium hydroxide with phosphoric acid, as water-insoluble barium phosphate is formed.
Literature
- Levitsky M. The language of chemists // Chemistry and life. - 2000. -№1. - P.50-52.
- Kudryavtsev A.A. Writing Chemical Equations - 4th Edition, Revised. and additional, 1968 - 359s.
- Berg L.G. Gromakov S.D. Zoroatskaya I.V. Averko-Antonovich I.N. Methods for selecting coefficients in chemical equations - Kazan: Publishing House of Kazan University, 1959.- 148 p.
- Leenson I.A. Even or odd - M .: Chemistry, 1987. - 176s.
- Chemistry, 8th grade textbook. ARC Publishing. 2003.
- Chemistry, 8th grade textbook. Drofa Publishing. 2009.
- Chemistry, 8th grade textbook. Publishing house "Mektep" Almaty. 2012.
- Chemistry, 9th grade textbook. Publishing house "Enlightenment" 2008.
see also
Links
- // Encyclopedic Dictionary of Brockhaus and Efron: In 86 volumes (82 volumes and 4 additional). - St. Petersburg. , 1890-1907.
Definition
chemical equation is a conditional record of a chemical reaction using chemical formulas and coefficients.
In order to correctly place the coefficients in a chemical equation, one should understand the difference between coefficients and indices.
Definition
Coefficient- shows the number of molecules and is represented by a large number in front of the molecular formula of the substance. Index- shows the number of atoms of an element in one molecule of a substance, is depicted to the right below the symbol of the element.
To calculate the total number of atoms, you need to multiply the number of molecules by the number of atoms of the element in one molecule. For example, the record of three molecules of sulfuric acid (the gross formula) is shown on the right, and a variant of the structural record is shown below. So, one molecule of sulfuric acid consists of three of three elements and in total contains (2 + 1 + 4) \u003d 7 atoms: 2 hydrogen atoms, one sulfur atom and four oxygen atoms. There will be three times as many atoms in three molecules, that is, 3*2=6 hydrogen atoms, 3*1=3 sulfur atoms and 3*4=12 oxygen atoms. This is clearly seen from the structural formula below.
To understand the logic of equalizing chemical reactions, try to practice at home with self-made models of atoms and molecules: prepare balls of different colors (gray, red and black) from plasticine. Try to carry out the combustion reaction of methane, the scheme of which is shown below.
When modeling, it will be obvious that the number of atoms (homemade plasticine balls) of each element (color) does not change during the reaction. That is, the number of carbon atoms before and after the transformation remains unchanged and is equal to one (one black ball). Two oxygen molecules on the left side of the equation consist of 4 atoms, on the right side of the equation, two oxygen atoms are contained in carbon dioxide ($CO_2$) and two atoms are in two water molecules, that is, there are also 4 oxygen atoms on the right.
Law of acting masses
When compiling reaction equations, it is necessary to use the law of conservation of the mass of substances (the law of mass action or LMA), discovered by M.V. Lomonosov and A. Lavoisier.
Law of acting masses: the mass of the substances that entered into the reaction is equal to the mass of the substances resulting from it.
Since substances are composed of atoms, when compiling chemical equations, we will use the rule: the number of atoms of each chemical element starting substances should be equal to the number of atoms in the reaction products. In a chemical reaction, the number of interacting atoms remains unchanged, only their rearrangement occurs with the destruction of the starting substances
Algorithm for compiling reaction equations.
Consider the algorithm for compiling chemical equations using the example of the interaction of simple substances: metals and non-metals with each other. Let phosphorus and oxygen interact (combustion reaction).
1. Write down the starting substances (reagents) side by side, put a "+" sign between them (here we will take into account the fact that oxygen is a diatomic molecule), and after them an arrow - as an equal sign.
$P+O_2 \rightarrow$
2. We write down the formula of the reaction product after the arrow:
$P+O_2\rightarrow P_2O_5$
3. It can be seen from the diagram that oxygen is 2 atoms on the left, 5 on the right, and in accordance with the law of conservation of mass of substances, the number of atoms of a given chemical element should be the same. To equalize their number, we find the least common multiple. For 2 and 5, this will be the number 10. Divide the least common multiple by the number of atoms in the formulas. 10:2=5, 10:5=2, these will be the coefficients that are placed respectively in front of oxygen $O_2$ and phosphorus oxide (V) $P_2O_5$.
$P+5O_2\rightarrow 2P_2O_5$
oxygen on the left and on the right became 10 (5 2=10, 2 5=10)
4. The coefficient refers to the entire formula and is placed in front of it. After it was placed on the right, there were 2 2 = 4 atoms of phosphorus. And on the left 1 (coefficient 1 is not set). So we put a coefficient 4 in front of phosphorus.
$4P + 5O_2\rightarrow 2P_2O_5$
This is the final record of the chemical equation.
It reads: four pe plus five o-two equals two pe-two o-five.
Let's analyze the algorithm for putting down the coefficients on another example:
$KNO_3 = KNO_2 + O_2$
When potassium nitrate decomposes, potassium nitrite and oxygen are formed.
There is one potassium atom on the left side of the equation, and one on the right side. The number of nitrogen atoms on the left and right is the same and equal to one. But the number of oxygen atoms is different: on the left - 3, on the right - 4. In such cases, you can resort to doubling, that is, put the coefficient \u003d 2 in front of potassium nitrate.
Methodology for solving problems in chemistry
When solving problems, you need to be guided by a few simple rules:
- Carefully read the condition of the problem;
- Write down what is given;
- Convert, if necessary, units of physical quantities to SI units (some non-systemic units are allowed, for example, liters);
- Write down, if necessary, the reaction equation and arrange the coefficients;
- Solve the problem using the concept of the amount of substance, and not the method of drawing up proportions;
- Write down the answer.
In order to successfully prepare in chemistry, you should carefully consider the solutions to the problems given in the text, as well as independently solve a sufficient number of them. It is in the process of solving problems that the main theoretical provisions of the chemistry course will be fixed. It is necessary to solve problems throughout the entire time of studying chemistry and preparing for the exam.
You can use the tasks on this page, or you can download a good collection of tasks and exercises with the solution of typical and complicated tasks (M. I. Lebedeva, I. A. Ankudimova): download.
Mole, molar mass
Molar mass is the ratio of the mass of a substance to the amount of a substance, i.e.
М(х) = m(x)/ν(x), (1)
where M(x) is the molar mass of substance X, m(x) is the mass of substance X, ν(x) is the amount of substance X. The SI unit for molar mass is kg/mol, but g/mol is commonly used. The unit of mass is g, kg. The SI unit for the amount of a substance is the mole.
Any chemistry problem solved through the amount of matter. Remember the basic formula:
ν(x) = m(x)/ М(х) = V(x)/V m = N/N A , (2)
where V(x) is the volume of substance Х(l), Vm is the molar volume of gas (l/mol), N is the number of particles, N A is the Avogadro constant.
1. Determine the mass sodium iodide NaI amount of substance 0.6 mol.
Given: ν(NaI)= 0.6 mol.
To find: m(NaI) =?
Decision. The molar mass of sodium iodide is:
M(NaI) = M(Na) + M(I) = 23 + 127 = 150 g/mol
Determine the mass of NaI:
m(NaI) = ν(NaI) M(NaI) = 0.6 150 = 90 g.
2. Determine the amount of substance atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g.
Given: m(Na 2 B 4 O 7) \u003d 40.4 g.
To find: ν(B)=?
Decision. The molar mass of sodium tetraborate is 202 g/mol. Determine the amount of substance Na 2 B 4 O 7:
ν (Na 2 B 4 O 7) \u003d m (Na 2 B 4 O 7) / M (Na 2 B 4 O 7) \u003d 40.4 / 202 \u003d 0.2 mol.
Recall that 1 mol of sodium tetraborate molecule contains 2 mol of sodium atoms, 4 mol of boron atoms and 7 mol of oxygen atoms (see the formula of sodium tetraborate). Then the amount of atomic boron substance is: ν (B) \u003d 4 ν (Na 2 B 4 O 7) \u003d 4 0.2 \u003d 0.8 mol.
Calculations for chemical formulas. Mass share.
The mass fraction of a substance is the ratio of the mass of a given substance in the system to the mass of the entire system, i.e. ω(X) =m(X)/m, where ω(X) is the mass fraction of substance X, m(X) is the mass of substance X, m is the mass of the entire system. Mass fraction is a dimensionless quantity. It is expressed as a fraction of a unit or as a percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. ω(O)=0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. ω(Cl)=0.607.
3. Determine the mass fraction water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.
Decision: The molar mass of BaCl 2 2H 2 O is:
M (BaCl 2 2H 2 O) \u003d 137+ 2 35.5 + 2 18 \u003d 244 g / mol
From the formula BaCl 2 2H 2 O it follows that 1 mol of barium chloride dihydrate contains 2 mol of H 2 O. From this we can determine the mass of water contained in BaCl 2 2H 2 O:
m(H 2 O) \u003d 2 18 \u003d 36 g.
We find the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.
ω (H 2 O) \u003d m (H 2 O) / m (BaCl 2 2H 2 O) \u003d 36/244 \u003d 0.1475 \u003d 14.75%.
4. From a rock sample weighing 25 g containing the mineral argentite Ag 2 S, silver weighing 5.4 g was isolated. Determine the mass fraction argentite in the sample.
Given: m(Ag)=5.4 g; m = 25 g.
To find: ω(Ag 2 S) =?
Decision: we determine the amount of silver substance in argentite: ν (Ag) \u003d m (Ag) / M (Ag) \u003d 5.4 / 108 \u003d 0.05 mol.
From the formula Ag 2 S it follows that the amount of argentite substance is half the amount of silver substance. Determine the amount of argentite substance:
ν (Ag 2 S) \u003d 0.5 ν (Ag) \u003d 0.5 0.05 \u003d 0.025 mol
We calculate the mass of argentite:
m (Ag 2 S) \u003d ν (Ag 2 S) M (Ag 2 S) \u003d 0.025 248 \u003d 6.2 g.
Now we determine the mass fraction of argentite in a rock sample, weighing 25 g.
ω (Ag 2 S) \u003d m (Ag 2 S) / m \u003d 6.2 / 25 \u003d 0.248 \u003d 24.8%.
Derivation of compound formulas
5. Determine the simplest compound formula potassium with manganese and oxygen, if the mass fractions of elements in this substance are 24.7, 34.8 and 40.5%, respectively.
Given: ω(K)=24.7%; ω(Mn)=34.8%; ω(O)=40.5%.
To find: compound formula.
Decision: for calculations, we select the mass of the compound, equal to 100 g, i.e. m=100 g. Masses of potassium, manganese and oxygen will be:
m (K) = m ω (K); m (K) \u003d 100 0.247 \u003d 24.7 g;
m (Mn) = m ω(Mn); m (Mn) = 100 0.348 = 34.8 g;
m (O) = m ω(O); m (O) \u003d 100 0.405 \u003d 40.5 g.
We determine the amount of substances of atomic potassium, manganese and oxygen:
ν (K) \u003d m (K) / M (K) \u003d 24.7 / 39 \u003d 0.63 mol
ν (Mn) \u003d m (Mn) / M (Mn) \u003d 34.8 / 55 \u003d 0.63 mol
ν (O) \u003d m (O) / M (O) \u003d 40.5 / 16 \u003d 2.5 mol
We find the ratio of the amounts of substances:
ν(K) : ν(Mn) : ν(O) = 0.63: 0.63: 2.5.
Dividing the right side of the equation by a smaller number (0.63) we get:
ν(K) : ν(Mn) : ν(O) = 1: 1: 4.
Therefore, the simplest formula of the KMnO 4 compound.
6. During the combustion of 1.3 g of the substance, 4.4 g of carbon monoxide (IV) and 0.9 g of water were formed. Find the molecular formula substance if its hydrogen density is 39.
Given: m(in-va) \u003d 1.3 g; m(CO 2)=4.4 g; m(H 2 O)=0.9 g; D H2 \u003d 39.
To find: the formula of the substance.
Decision: Assume that the substance you are looking for contains carbon, hydrogen and oxygen, because during its combustion, CO 2 and H 2 O were formed. Then it is necessary to find the amounts of substances CO 2 and H 2 O in order to determine the amounts of substances of atomic carbon, hydrogen and oxygen.
ν (CO 2) \u003d m (CO 2) / M (CO 2) \u003d 4.4 / 44 \u003d 0.1 mol;
ν (H 2 O) \u003d m (H 2 O) / M (H 2 O) \u003d 0.9 / 18 \u003d 0.05 mol.
We determine the amount of substances of atomic carbon and hydrogen:
ν(C)= ν(CO 2); v(C)=0.1 mol;
ν(H)= 2 ν(H 2 O); ν (H) \u003d 2 0.05 \u003d 0.1 mol.
Therefore, the masses of carbon and hydrogen will be equal:
m(C) = ν(C) M(C) = 0.1 12 = 1.2 g;
m (H) \u003d ν (H) M (H) \u003d 0.1 1 \u003d 0.1 g.
We determine the qualitative composition of the substance:
m (in-va) \u003d m (C) + m (H) \u003d 1.2 + 0.1 \u003d 1.3 g.
Consequently, the substance consists only of carbon and hydrogen (see the condition of the problem). Let us now determine its molecular weight, based on the given in the condition tasks density of a substance with respect to hydrogen.
M (in-va) \u003d 2 D H2 \u003d 2 39 \u003d 78 g / mol.
ν(C) : ν(H) = 0.1: 0.1
Dividing the right side of the equation by the number 0.1, we get:
ν(C) : ν(H) = 1: 1
Let's take the number of carbon (or hydrogen) atoms as "x", then, multiplying "x" by the atomic masses of carbon and hydrogen and equating this sum molecular weight substances, we solve the equation:
12x + x \u003d 78. Hence x \u003d 6. Therefore, the formula of the substance C 6 H 6 is benzene.
Molar volume of gases. Laws of ideal gases. Volume fraction.
The molar volume of a gas is equal to the ratio of the volume of gas to the amount of substance of this gas, i.e.
Vm = V(X)/ ν(x),
where V m is the molar volume of gas - a constant value for any gas under given conditions; V(X) is the volume of gas X; ν(x) - the amount of gas substance X. The molar volume of gases under normal conditions (normal pressure p n \u003d 101 325 Pa ≈ 101.3 kPa and temperature Tn \u003d 273.15 K ≈ 273 K) is V m \u003d 22.4 l /mol.
In calculations involving gases, it is often necessary to switch from these conditions to normal conditions or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:
──── = ─── (3)
Where p is pressure; V is the volume; T is the temperature in the Kelvin scale; the index "n" indicates normal conditions.
The composition of gas mixtures is often expressed using a volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.
where φ(X) is the volume fraction of the X component; V(X) is the volume of the X component; V is the volume of the system. The volume fraction is a dimensionless quantity, it is expressed in fractions of a unit or as a percentage.
7. What volume takes at a temperature of 20 ° C and a pressure of 250 kPa ammonia weighing 51 g?
Given: m(NH 3)=51 g; p=250 kPa; t=20°C.
To find: V(NH 3) \u003d?
Decision: determine the amount of ammonia substance:
ν (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 51/17 \u003d 3 mol.
The volume of ammonia under normal conditions is:
V (NH 3) \u003d V m ν (NH 3) \u003d 22.4 3 \u003d 67.2 l.
Using formula (3), we bring the volume of ammonia to these conditions [temperature T \u003d (273 + 20) K \u003d 293 K]:
p n TV n (NH 3) 101.3 293 67.2
V (NH 3) \u003d ──────── \u003d ────────── \u003d 29.2 l.
8. Determine volume, which will take under normal conditions a gas mixture containing hydrogen, weighing 1.4 g and nitrogen, weighing 5.6 g.
Given: m(N 2)=5.6 g; m(H2)=1.4; well.
To find: V(mixture)=?
Decision: find the amount of substance hydrogen and nitrogen:
ν (N 2) \u003d m (N 2) / M (N 2) \u003d 5.6 / 28 \u003d 0.2 mol
ν (H 2) \u003d m (H 2) / M (H 2) \u003d 1.4 / 2 \u003d 0.7 mol
Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of gases, i.e.
V (mixtures) \u003d V (N 2) + V (H 2) \u003d V m ν (N 2) + V m ν (H 2) \u003d 22.4 0.2 + 22.4 0.7 \u003d 20.16 l.
Calculations by chemical equations
Calculations according to chemical equations (stoichiometric calculations) are based on the law of conservation of the mass of substances. However, in real chemical processes, due to an incomplete reaction and various losses of substances, the mass of the resulting products is often less than that which should be formed in accordance with the law of conservation of the mass of substances. The yield of the reaction product (or the mass fraction of the yield) is the ratio of the mass of the actually obtained product, expressed as a percentage, to its mass, which should be formed in accordance with the theoretical calculation, i.e.
η = /m(X) (4)
Where η is the product yield, %; m p (X) - the mass of the product X obtained in the real process; m(X) is the calculated mass of substance X.
In those tasks where the product yield is not specified, it is assumed that it is quantitative (theoretical), i.e. η=100%.
9. What mass of phosphorus should be burned to receive phosphorus oxide (V) weighing 7.1 g?
Given: m(P 2 O 5) \u003d 7.1 g.
To find: m(P) =?
Decision: we write the equation for the combustion reaction of phosphorus and arrange the stoichiometric coefficients.
4P+ 5O 2 = 2P 2 O 5
We determine the amount of substance P 2 O 5 obtained in the reaction.
ν (P 2 O 5) \u003d m (P 2 O 5) / M (P 2 O 5) \u003d 7.1 / 142 \u003d 0.05 mol.
It follows from the reaction equation that ν (P 2 O 5) \u003d 2 ν (P), therefore, the amount of phosphorus substance required in the reaction is:
ν (P 2 O 5) \u003d 2 ν (P) \u003d 2 0.05 \u003d 0.1 mol.
From here we find the mass of phosphorus:
m(Р) = ν(Р) М(Р) = 0.1 31 = 3.1 g.
10. Magnesium weighing 6 g and zinc weighing 6.5 g were dissolved in an excess of hydrochloric acid. What volume hydrogen, measured under normal conditions, stand out wherein?
Given: m(Mg)=6 g; m(Zn)=6.5 g; well.
To find: V(H 2) =?
Decision: we write down the reaction equations for the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients.
Zn + 2 HCl \u003d ZnCl 2 + H 2
Mg + 2 HCl \u003d MgCl 2 + H 2
We determine the amount of magnesium and zinc substances that reacted with hydrochloric acid.
ν(Mg) \u003d m (Mg) / M (Mg) \u003d 6/24 \u003d 0.25 mol
ν (Zn) \u003d m (Zn) / M (Zn) \u003d 6.5 / 65 \u003d 0.1 mol.
It follows from the reaction equations that the amount of the substance of the metal and hydrogen are equal, i.e. ν (Mg) \u003d ν (H 2); ν (Zn) \u003d ν (H 2), we determine the amount of hydrogen resulting from two reactions:
ν (Н 2) \u003d ν (Mg) + ν (Zn) \u003d 0.25 + 0.1 \u003d 0.35 mol.
We calculate the volume of hydrogen released as a result of the reaction:
V (H 2) \u003d V m ν (H 2) \u003d 22.4 0.35 \u003d 7.84 l.
11. When passing hydrogen sulfide with a volume of 2.8 liters (normal conditions) through an excess of copper (II) sulfate solution, a precipitate was formed weighing 11.4 g. Determine the exit reaction product.
Given: V(H 2 S)=2.8 l; m(precipitate)= 11.4 g; well.
To find: η =?
Decision: we write the reaction equation for the interaction of hydrogen sulfide and copper (II) sulfate.
H 2 S + CuSO 4 \u003d CuS ↓ + H 2 SO 4
Determine the amount of hydrogen sulfide substance involved in the reaction.
ν (H 2 S) \u003d V (H 2 S) / V m \u003d 2.8 / 22.4 \u003d 0.125 mol.
It follows from the reaction equation that ν (H 2 S) \u003d ν (СuS) \u003d 0.125 mol. So you can find the theoretical mass of CuS.
m(CuS) \u003d ν (CuS) M (CuS) \u003d 0.125 96 \u003d 12 g.
Now we determine the product yield using formula (4):
η = /m(X)= 11.4 100/ 12 = 95%.
12. What weight ammonium chloride is formed by the interaction of hydrogen chloride weighing 7.3 g with ammonia weighing 5.1 g? What gas will be left in excess? Determine the mass of the excess.
Given: m(HCl)=7.3 g; m(NH 3) \u003d 5.1 g.
To find: m(NH 4 Cl) =? m(excess) =?
Decision: write the reaction equation.
HCl + NH 3 \u003d NH 4 Cl
This task is for "excess" and "deficiency". We calculate the amount of hydrogen chloride and ammonia and determine which gas is in excess.
ν(HCl) \u003d m (HCl) / M (HCl) \u003d 7.3 / 36.5 \u003d 0.2 mol;
ν (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 5.1 / 17 \u003d 0.3 mol.
Ammonia is in excess, so the calculation is based on the deficiency, i.e. by hydrogen chloride. It follows from the reaction equation that ν (HCl) \u003d ν (NH 4 Cl) \u003d 0.2 mol. Determine the mass of ammonium chloride.
m (NH 4 Cl) \u003d ν (NH 4 Cl) M (NH 4 Cl) \u003d 0.2 53.5 \u003d 10.7 g.
We determined that ammonia is in excess (according to the amount of substance, the excess is 0.1 mol). Calculate the mass of excess ammonia.
m (NH 3) \u003d ν (NH 3) M (NH 3) \u003d 0.1 17 \u003d 1.7 g.
13. Technical calcium carbide weighing 20 g was treated with excess water, obtaining acetylene, passing through which through an excess of bromine water formed 1,1,2,2-tetrabromoethane weighing 86.5 g. Determine mass fraction SaS 2 in technical carbide.
Given: m = 20 g; m(C 2 H 2 Br 4) \u003d 86.5 g.
To find: ω (CaC 2) =?
Decision: we write down the equations of interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients.
CaC 2 +2 H 2 O \u003d Ca (OH) 2 + C 2 H 2
C 2 H 2 +2 Br 2 \u003d C 2 H 2 Br 4
Find the amount of substance tetrabromoethane.
ν (C 2 H 2 Br 4) \u003d m (C 2 H 2 Br 4) / M (C 2 H 2 Br 4) \u003d 86.5 / 346 \u003d 0.25 mol.
It follows from the reaction equations that ν (C 2 H 2 Br 4) \u003d ν (C 2 H 2) \u003d ν (CaC 2) \u003d 0.25 mol. From here we can find the mass of pure calcium carbide (without impurities).
m (CaC 2) \u003d ν (CaC 2) M (CaC 2) \u003d 0.25 64 \u003d 16 g.
We determine the mass fraction of CaC 2 in technical carbide.
ω (CaC 2) \u003d m (CaC 2) / m \u003d 16/20 \u003d 0.8 \u003d 80%.
Solutions. Mass fraction of the solution component
14. Sulfur weighing 1.8 g was dissolved in benzene with a volume of 170 ml. The density of benzene is 0.88 g / ml. Determine mass fraction sulfur in solution.
Given: V(C 6 H 6) =170 ml; m(S) = 1.8 g; ρ(C 6 C 6)=0.88 g/ml.
To find: ω(S) =?
Decision: to find the mass fraction of sulfur in the solution, it is necessary to calculate the mass of the solution. Determine the mass of benzene.
m (C 6 C 6) \u003d ρ (C 6 C 6) V (C 6 H 6) \u003d 0.88 170 \u003d 149.6 g.
Find the total mass of the solution.
m (solution) \u003d m (C 6 C 6) + m (S) \u003d 149.6 + 1.8 \u003d 151.4 g.
Calculate the mass fraction of sulfur.
ω(S) =m(S)/m=1.8 /151.4 = 0.0119 = 1.19%.
15. Iron sulfate FeSO 4 7H 2 O weighing 3.5 g was dissolved in water weighing 40 g. Determine mass fraction of iron sulfate (II) in the resulting solution.
Given: m(H 2 O)=40 g; m (FeSO 4 7H 2 O) \u003d 3.5 g.
To find: ω(FeSO 4) =?
Decision: find the mass of FeSO 4 contained in FeSO 4 7H 2 O. To do this, calculate the amount of substance FeSO 4 7H 2 O.
ν (FeSO 4 7H 2 O) \u003d m (FeSO 4 7H 2 O) / M (FeSO 4 7H 2 O) \u003d 3.5 / 278 \u003d 0.0125 mol
From the formula of ferrous sulfate it follows that ν (FeSO 4) \u003d ν (FeSO 4 7H 2 O) \u003d 0.0125 mol. Calculate the mass of FeSO 4:
m (FeSO 4) \u003d ν (FeSO 4) M (FeSO 4) \u003d 0.0125 152 \u003d 1.91 g.
Given that the mass of the solution consists of the mass of ferrous sulfate (3.5 g) and the mass of water (40 g), we calculate the mass fraction of ferrous sulfate in the solution.
ω (FeSO 4) \u003d m (FeSO 4) / m \u003d 1.91 / 43.5 \u003d 0.044 \u003d 4.4%.
Tasks for independent solution
- 50 g of methyl iodide in hexane were treated with sodium metal, and 1.12 liters of gas, measured under normal conditions, were released. Determine the mass fraction of methyl iodide in the solution. Answer: 28,4%.
- Some alcohol was oxidized to form a monobasic carboxylic acid. When burning 13.2 g of this acid, carbon dioxide, for the complete neutralization of which it took 192 ml of a solution of KOH with a mass fraction of 28%. The density of the KOH solution is 1.25 g/ml. Determine the formula for alcohol. Answer: butanol.
- The gas obtained by the interaction of 9.52 g of copper with 50 ml of an 81% solution of nitric acid, with a density of 1.45 g / ml, was passed through 150 ml of a 20% NaOH solution with a density of 1.22 g / ml. Determine the mass fractions of dissolved substances. Answer: 12.5% NaOH; 6.48% NaNO 3 ; 5.26% NaNO 2 .
- Determine the volume of gases released during the explosion of 10 g of nitroglycerin. Answer: 7.15 l.
- A sample of organic matter weighing 4.3 g was burned in oxygen. The reaction products are carbon monoxide (IV) with a volume of 6.72 liters (normal conditions) and water with a mass of 6.3 g. The vapor density of the starting substance for hydrogen is 43. Determine the formula of the substance. Answer: C 6 H 14 .
In order to learn how to equalize chemical equations, you first need to highlight the main points and use the correct algorithm.
Key points
Building the logic of the process is easy. To do this, we distinguish the following stages:
- Determining the type of reagents (all reagents are organic, all reagents are inorganic, organic and inorganic reagents in one reaction)
- Determining the type of chemical reaction (reaction with a change in the oxidation states of the components or not)
- Isolation of a check atom or group of atoms
Examples
- All components are inorganic, without changing the oxidation state, the test atom will be oxygen - O (it was not affected by any interactions:
NaOH + HCl = NaCl + H2O
Let's count the number of atoms of each element of the right and left parts and make sure that the coefficients are not required here (by default, the absence of a coefficient is a coefficient equal to 1)
NaOH + H2SO4 = Na 2 SO4 + H2O
In this case, on the right side of the equation we see 2 sodium atoms, so on the left side of the equation we need to substitute a factor of 2 in front of the compound containing sodium:
2 NaOH + H2SO4 = Na 2 SO4 + H2O
We check for oxygen - O: on the left side 2O from NaOH and 4 from the sulfate ion SO4, and on the right side 4 from SO4 and 1 in water. Add 2 before water:
2 NaOH + H2SO4 = Na 2 SO4+ 2 H2O
- All components are organic, without changing the oxidation state:
HOOC-COOH + CH3OH = CH3OOC-COOCH3 + H2O (reaction possible under certain conditions)
In this case, we see that on the right side there are 2 groups of CH3 atoms, and on the left side there is only one. Add a factor of 2 to the left side before CH3OH, check for oxygen and add 2 before water
HOOC-COOH + 2CH3OH = CH3OOC-COOCH3 + 2H2O
- Organic and inorganic components without changing oxidation states:
CH3NH2 + H2SO4 = (CH3NH2)2∙SO4
In this reaction, a check atom is optional. On the left side there is a molecule of methylamine CH3NH2, and on the right side 2. So we need a coefficient of 2 in front of methylamine.
2CH3NH2 + H2SO4 = (CH3NH2)2∙SO4
- Organic component, inorganic, change in oxidation state.
CuO + C2H5OH = Cu + CH3COOH + H2O
In this case, it is necessary to draw up an electronic balance, and the formulas organic matter better to convert to gross. The test atom will be oxygen - its quantity shows that the coefficients are not required, the electronic balance confirms
CuO + C2H6O = Cu + C2H4O2
2C +2 - 2e = 2C0
C3H8 + O2 = CO2 + H2O
Here O cannot be a test, since it changes its oxidation state itself. Checking for N.
O2 0 + 2 * 2 e \u003d 2O-2 (we are talking about oxygen from CO2)
3C (-8/3) - 20e \u003d 3C +4 (conditional fractional oxidation states are used in organic redox reactions)
It can be seen from the electronic balance that 5 times more oxygen is required to oxidize carbon. We put 5 in front of O2, also from the electronic balance m should put 3 in front of C from CO2, check for H, and put 4 in front of water
C3H8 + 5O2 = 3CO2 + 4H2O
- Inorganic compounds, change in oxidation states.
Na2SO3 + KMnO4 + H2SO4 = Na2SO4 + K2SO4 + H2O + MnO2
The test will be hydrogens in water and acid residues SO4 2- from sulfuric acid.
S + 4 (from SO3 2-) - 2e \u003d S + 6 (from Na2SO4)
Mn+7 + 3e = Mn+4
Thus, you need to put 3 before Na2SO3 and Na2SO4, 2 before KMnO4 and MNO2.
3Na2SO3 + 2KMnO4 + H2SO4 = 3Na2SO4 + K2SO4 + H2O + 2MnO2
Quite often, schoolchildren and students have to make up the so-called. ionic reaction equations. In particular, problem 31, proposed at the Unified State Examination in Chemistry, is devoted to this topic. In this article, we will discuss in detail the algorithm for writing short and complete ionic equations, we will analyze many examples different levels difficulties.
Why ionic equations are needed
Let me remind you that when many substances are dissolved in water (and not only in water!) A process of dissociation occurs - substances break up into ions. For example, HCl molecules in an aqueous medium dissociate into hydrogen cations (H + , more precisely, H 3 O +) and chlorine anions (Cl -). Sodium bromide (NaBr) is in an aqueous solution not in the form of molecules, but in the form of hydrated Na + and Br - ions (by the way, ions are also present in solid sodium bromide).
When writing the "ordinary" (molecular) equations, we do not take into account that not molecules but ions enter into the reaction. Here, for example, is the equation for the reaction between hydrochloric acid and sodium hydroxide:
HCl + NaOH = NaCl + H 2 O. (1)
Of course, this diagram does not quite correctly describe the process. As we have already said, there are practically no HCl molecules in an aqueous solution, but there are H + and Cl - ions. The same is true for NaOH. It would be better to write the following:
H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O. (2)
That's what it is complete ionic equation. Instead of "virtual" molecules, we see particles that are actually present in the solution (cations and anions). We will not dwell on the question why we have written H 2 O in molecular form. This will be explained a little later. As you can see, there is nothing complicated: we have replaced the molecules with ions, which are formed during their dissociation.
However, even the complete ionic equation is not perfect. Indeed, take a closer look: both in the left and in the right parts of equation (2) there are identical particles - Na + cations and Cl - anions. These ions do not change during the reaction. Why then are they needed at all? Let's remove them and get short ionic equation:
H + + OH - = H 2 O. (3)
As you can see, it all comes down to the interaction of H + and OH - ions with the formation of water (neutralization reaction).
All complete and short ionic equations are written down. If we solved problem 31 at the exam in chemistry, we would get the maximum mark for it - 2 points.
So, once again about the terminology:
- HCl + NaOH = NaCl + H 2 O - molecular equation ("usual" equation, schematically reflecting the essence of the reaction);
- H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O - complete ionic equation (real particles in solution are visible);
- H + + OH - = H 2 O - a short ionic equation (we removed all the "garbage" - particles that do not participate in the process).
Algorithm for writing ionic equations
- We compose the molecular equation of the reaction.
- All particles that dissociate in solution to a noticeable degree are written as ions; substances that are not prone to dissociation, we leave "in the form of molecules."
- We remove from the two parts of the equation the so-called. observer ions, i.e., particles that do not participate in the process.
- We check the coefficients and get the final answer - a short ionic equation.
Example 1. Write a complete and short ionic equation describing the interaction of aqueous solutions of barium chloride and sodium sulfate.
Decision. We will act in accordance with the proposed algorithm. Let's set up the molecular equation first. Barium chloride and sodium sulfate are two salts. Let's look at the section of the reference book "Properties of inorganic compounds". We see that salts can interact with each other if a precipitate forms during the reaction. Let's check:
Exercise 2. Complete the equations for the following reactions:
- KOH + H 2 SO 4 \u003d
- H 3 PO 4 + Na 2 O \u003d
- Ba(OH) 2 + CO 2 =
- NaOH + CuBr 2 =
- K 2 S + Hg (NO 3) 2 \u003d
- Zn + FeCl 2 =
Exercise 3. Write the molecular equations for the reactions (in aqueous solution) between: a) sodium carbonate and nitric acid, b) nickel (II) chloride and sodium hydroxide, c) orthophosphoric acid and calcium hydroxide, d) silver nitrate and potassium chloride, e) phosphorus oxide (V) and potassium hydroxide.
I sincerely hope that you had no problems completing these three tasks. If this is not the case, you need to return to the topic " Chemical properties main classes of inorganic compounds".
How to turn a molecular equation into a complete ionic equation
The most interesting begins. We must understand which substances should be written as ions and which should be left in "molecular form". You have to remember the following.
In the form of ions write:
- soluble salts (I emphasize that only salts are highly soluble in water);
- alkalis (let me remind you that water-soluble bases are called alkalis, but not NH 4 OH);
- strong acids (H 2 SO 4 , HNO 3 , HCl, HBr, HI, HClO 4 , HClO 3 , H 2 SeO 4 , ...).
As you can see, this list is not difficult to remember: it includes strong acids and bases and all soluble salts. By the way, to especially vigilant young chemists who may be outraged by the fact that strong electrolytes (insoluble salts) are not included in this list, I can tell you the following: NOT including insoluble salts in this list does not at all reject the fact that they are strong electrolytes.
All other substances must be present in the ionic equations in the form of molecules. Those demanding readers who are not satisfied with the vague term "all other substances", and who, following the example of the hero of a famous film, demand "announce full list I give the following information.
In the form of molecules, write:
- all insoluble salts;
- all weak bases (including insoluble hydroxides, NH 4 OH and similar substances);
- all weak acids (H 2 CO 3 , HNO 2 , H 2 S, H 2 SiO 3 , HCN, HClO, almost all organic acids ...);
- in general, all weak electrolytes (including water!!!);
- oxides (all types);
- all gaseous compounds (in particular H 2 , CO 2 , SO 2 , H 2 S, CO);
- simple substances (metals and non-metals);
- almost all organic compounds (with the exception of water-soluble salts of organic acids).
Phew, I don't think I forgot anything! Although it is easier, in my opinion, to remember list No. 1. Of the fundamentally important in list No. 2, I will once again note water.
Let's train!
Example 2. Make a complete ionic equation describing the interaction of copper (II) hydroxide and hydrochloric acid.
Decision. Let's start, of course, with the molecular equation. Copper (II) hydroxide is an insoluble base. All insoluble bases react with strong acids to form a salt and water:
Cu(OH) 2 + 2HCl = CuCl 2 + 2H 2 O.
And now we find out which substances to write in the form of ions, and which - in the form of molecules. The lists above will help us. Copper (II) hydroxide is an insoluble base (see solubility table), a weak electrolyte. Insoluble bases are written in molecular form. HCl- strong acid, in solution almost completely dissociates into ions. CuCl 2 is a soluble salt. We write in ionic form. Water - only in the form of molecules! We get the full ionic equation:
Cu (OH) 2 + 2H + + 2Cl - \u003d Cu 2+ + 2Cl - + 2H 2 O.
Example 3. Write a complete ionic equation for the reaction of carbon dioxide with an aqueous solution of NaOH.
Decision. Carbon dioxide is a typical acidic oxide, NaOH is an alkali. When acidic oxides interact with aqueous solutions of alkalis, salt and water are formed. We compose the molecular reaction equation (do not forget, by the way, about the coefficients):
CO 2 + 2NaOH \u003d Na 2 CO 3 + H 2 O.
CO 2 - oxide, gaseous compound; keep the molecular shape. NaOH - strong base (alkali); written in the form of ions. Na 2 CO 3 - soluble salt; write in the form of ions. Water is a weak electrolyte, practically does not dissociate; leave it in molecular form. We get the following:
CO 2 + 2Na + + 2OH - \u003d Na 2+ + CO 3 2- + H 2 O.
Example 4. Sodium sulfide in aqueous solution reacts with zinc chloride to form a precipitate. Write the complete ionic equation for this reaction.
Decision. Sodium sulfide and zinc chloride are salts. When these salts interact, zinc sulfide precipitates:
Na 2 S + ZnCl 2 \u003d ZnS ↓ + 2NaCl.
I will immediately write down the full ionic equation, and you will analyze it yourself:
2Na + + S 2- + Zn 2+ + 2Cl - = ZnS↓ + 2Na + + 2Cl - .
Here are some tasks for you to independent work and a little test.
Exercise 4. Write the molecular and full ionic equations for the following reactions:
- NaOH + HNO3 =
- H 2 SO 4 + MgO =
- Ca(NO 3) 2 + Na 3 PO 4 =
- CoBr 2 + Ca(OH) 2 =
Exercise 5. Write complete ionic equations describing the interaction of: a) nitric oxide (V) with an aqueous solution of barium hydroxide, b) a solution of cesium hydroxide with hydroiodic acid, c) aqueous solutions of copper sulfate and potassium sulfide, d) calcium hydroxide and an aqueous solution of iron nitrate ( III).