Let's talk about how to create a chemical equation, because they are the main elements of this discipline. Thanks to a deep understanding of all the patterns of interactions and substances, you can control them and apply them in various fields of activity.
Theoretical features
Compiling chemical equations is an important and responsible stage, considered in the eighth grade of secondary schools. What should precede this stage? Before the teacher tells his students how to create a chemical equation, it is important to introduce schoolchildren to the term “valence” and teach them to determine this value for metals and non-metals using the periodic table of elements.
Compilation of binary formulas by valency
In order to understand how to create a chemical equation by valence, you first need to learn how to create formulas for compounds consisting of two elements using valence. We propose an algorithm that will help cope with the task. For example, you need to create a formula for sodium oxide.
First, it is important to take into account that the chemical element that is mentioned last in the name should be in first place in the formula. In our case, sodium will be written first in the formula, oxygen second. Let us recall that oxides are binary compounds in which the last (second) element must be oxygen with an oxidation state of -2 (valency 2). Next, using the periodic table, it is necessary to determine the valence of each of the two elements. To do this we use certain rules.
Since sodium is a metal that is located in the main subgroup of group 1, its valence is a constant value, it is equal to I.
Oxygen is a non-metal, since it is the last one in the oxide; to determine its valency, we subtract 6 from eight (the number of groups) (the group in which oxygen is located), we obtain that the valency of oxygen is II.
Between certain valences we find the least common multiple, then divide it by the valency of each of the elements to obtain their indices. We write down the finished formula Na 2 O.
Instructions for composing an equation
Now let's talk in more detail about how to write a chemical equation. First, let's look at the theoretical aspects, then move on to specific examples. So, composing chemical equations presupposes a certain procedure.
- 1st stage. After reading the proposed task, you need to determine which chemicals should be present on the left side of the equation. A “+” sign is placed between the original components.
- 2nd stage. After the equal sign, you need to create a formula for the reaction product. When performing such actions, you will need the algorithm for composing formulas for binary compounds, which we discussed above.
- 3rd stage. We check the number of atoms of each element before and after chemical interaction, if necessary, we put additional coefficients in front of the formulas.
Example of a combustion reaction
Let's try to figure out how to create a chemical equation for the combustion of magnesium using an algorithm. On the left side of the equation we write the sum of magnesium and oxygen. Do not forget that oxygen is a diatomic molecule, so it must be given an index of 2. After the equal sign, we compose the formula for the product obtained after the reaction. It will be in which magnesium is written first, and oxygen is written second in the formula. Next, using the table of chemical elements, we determine the valencies. Magnesium, which is in group 2 (the main subgroup), has a constant valency II; for oxygen, by subtracting 8 - 6 we also get valency II.
The process record will look like: Mg+O 2 =MgO.
In order for the equation to comply with the law of conservation of mass of substances, it is necessary to arrange the coefficients. First, we check the amount of oxygen before the reaction, after the process is completed. Since there were 2 oxygen atoms, but only one was formed, a coefficient of 2 must be added on the right side before the magnesium oxide formula. Next, we count the number of magnesium atoms before and after the process. As a result of the interaction, 2 magnesium was obtained, therefore, on the left side in front of the simple substance magnesium, a coefficient of 2 is also required.
The final type of reaction: 2Mg+O 2 =2MgO.
Example of a substitution reaction
Any chemistry summary contains a description of different types of interactions.
Unlike a compound, in a substitution there will be two substances on both the left and right sides of the equation. Let's say we need to write the reaction of interaction between zinc and We use the standard writing algorithm. First, on the left side we write zinc and hydrochloric acid through the sum, and on the right side we write the formulas for the resulting reaction products. Since zinc is located before hydrogen in the electrochemical voltage series of metals, in this process it displaces molecular hydrogen from the acid and forms zinc chloride. As a result, we get the following entry: Zn+HCL=ZnCl 2 +H 2.
Now we move on to equalizing the number of atoms of each element. Since there was one atom on the left side of chlorine, and after the interaction there were two, it is necessary to put a factor of 2 in front of the formula of hydrochloric acid.
As a result, we obtain a ready-made reaction equation corresponding to the law of conservation of mass of substances: Zn+2HCL=ZnCl 2 +H 2 .
Conclusion
A typical chemistry note necessarily contains several chemical transformations. Not a single section of this science is limited to a simple verbal description of transformations, processes of dissolution, evaporation; everything is necessarily confirmed by equations. The specificity of chemistry lies in the fact that all processes that occur between different inorganic or organic substances can be described using coefficients and indices.
How else does chemistry differ from other sciences? Chemical equations help not only to describe the transformations that occur, but also to carry out quantitative calculations based on them, thanks to which it is possible to carry out laboratory and industrial production of various substances.
In lesson 13 "" from the course " Chemistry for dummies» consider why chemical equations are needed; Let's learn how to equalize chemical reactions by correctly arranging coefficients. This lesson will require you to know the basic chemistry from previous lessons. Be sure to read about elemental analysis for an in-depth look at empirical formulas and chemical analysis.
As a result of the combustion reaction of methane CH 4 in oxygen O 2, carbon dioxide CO 2 and water H 2 O are formed. This reaction can be described chemical equation:
- CH 4 + O 2 → CO 2 + H 2 O (1)
Let's try to extract more information from a chemical equation than just an indication products and reagents reactions. Chemical equation (1) is INcomplete and therefore does not provide any information about how many O 2 molecules are consumed per 1 CH 4 molecule and how many CO 2 and H2 O molecules are obtained as a result. But if we write down numerical coefficients in front of the corresponding molecular formulas, which indicate how many molecules of each type take part in the reaction, then we get complete chemical equation reactions.
In order to complete the composition of the chemical equation (1), you need to remember one simple rule: the left and right sides of the equation must contain the same number of atoms of each type, since during the chemical reaction no new atoms are created and existing ones are not destroyed. This rule is based on the law of conservation of mass, which we discussed at the beginning of the chapter.
It is necessary in order to obtain a complete one from a simple chemical equation. So, let's move on to the actual equation of reaction (1): take another look at the chemical equation, exactly at the atoms and molecules on the right and left sides. It is easy to see that the reaction involves three types of atoms: carbon C, hydrogen H and oxygen O. Let's count and compare the number of atoms of each type on the right and left sides of the chemical equation.
Let's start with carbon. On the left side, one C atom is part of the CH 4 molecule, and on the right side, one C atom is part of CO 2. Thus, on the left and right sides the number of carbon atoms is the same, so we leave it alone. But for clarity, let’s put a coefficient of 1 in front of molecules with carbon, although this is not necessary:
- 1CH 4 + O 2 → 1CO 2 + H 2 O (2)
Then we move on to counting the hydrogen atoms H. On the left side there are 4 H atoms (in the quantitative sense, H 4 = 4H) in the CH 4 molecule, and on the right side there are only 2 H atoms in the H 2 O molecule, which is two times less than on the left side of the chemical equation (2). Let's equalize! To do this, let’s put a coefficient of 2 in front of the H 2 O molecule. Now we will have 4 molecules of hydrogen H in both the reactants and the products:
- 1CH 4 + O 2 → 1CO 2 + 2H 2 O (3)
Please note that the coefficient 2, which we wrote in front of the water molecule H 2 O to equalize the hydrogen H, increases by 2 times all the atoms included in its composition, i.e. 2H 2 O means 4H and 2O. Okay, we seem to have sorted this out, all that remains is to count and compare the number of oxygen atoms O in the chemical equation (3). It immediately catches your eye that there are exactly 2 times fewer O atoms on the left side than on the right. Now you already know how to balance chemical equations yourself, so I’ll immediately write down the final result:
- 1CH 4 + 2O 2 → 1CO 2 + 2H 2 O or CH 4 + 2O 2 → CO 2 + 2H 2 O (4)
As you can see, equalizing chemical reactions is not such a tricky thing, and it is not chemistry that is important here, but mathematics. Equation (4) is called complete equation chemical reaction, because it obeys the law of conservation of mass, i.e. the number of atoms of each type that enter into the reaction exactly coincides with the number of atoms of this type upon completion of the reaction. Each side of this complete chemical equation contains 1 carbon atom, 4 hydrogen atoms, and 4 oxygen atoms. However, it is worth understanding a couple of important points: a chemical reaction is a complex sequence of individual intermediate stages, and therefore, for example, equation (4) cannot be interpreted in the sense that 1 methane molecule must simultaneously collide with 2 oxygen molecules. The processes occurring during the formation of reaction products are much more complex. The second point: the complete equation of a reaction does not tell us anything about its molecular mechanism, that is, about the sequence of events that occur at the molecular level during its occurrence.
Coefficients in chemical reaction equations
Another clear example of how to correctly arrange odds in chemical reaction equations: Trinitrotoluene (TNT) C 7 H 5 N 3 O 6 combines vigorously with oxygen to form H 2 O, CO 2 and N 2. Let us write down the reaction equation that we will equalize:
- C 7 H 5 N 3 O 6 + O 2 → CO 2 + H 2 O + N 2 (5)
It is easier to construct the complete equation based on two TNT molecules, since the left side contains an odd number of hydrogen and nitrogen atoms, and the right side contains an even number:
- 2C 7 H 5 N 3 O 6 + O 2 → CO 2 + H 2 O + N 2 (6)
Then it is clear that 14 carbon atoms, 10 hydrogen atoms and 6 nitrogen atoms must turn into 14 molecules of carbon dioxide, 5 molecules of water and 3 molecules of nitrogen:
- 2C 7 H 5 N 3 O 6 + O 2 → 14CO 2 + 5H 2 O + 3N 2 (7)
Now both parts contain the same number of all atoms except oxygen. Of the 33 oxygen atoms present on the right side of the equation, 12 are supplied by the two original TNT molecules, and the remaining 21 must be supplied by 10.5 O 2 molecules. Thus the complete chemical equation will look like:
- 2C 7 H 5 N 3 O 6 + 10.5O 2 → 14CO 2 + 5H 2 O + 3N 2 (8)
You can multiply both sides by 2 and get rid of the non-integer coefficient 10.5:
- 4C 7 H 5 N 3 O 6 + 21O 2 → 28CO 2 + 10H 2 O + 6N 2 (9)
But you don’t have to do this, since all the coefficients of the equation do not have to be integers. It would be even more correct to create an equation based on one TNT molecule:
- C 7 H 5 N 3 O 6 + 5.25O 2 → 7CO 2 + 2.5H 2 O + 1.5N 2 (10)
The complete chemical equation (9) contains a lot of information. First of all, it indicates the starting substances - reagents, and products reactions. In addition, it shows that during the reaction all atoms of each type are individually preserved. If we multiply both sides of equation (9) by Avogadro's number N A = 6.022 10 23, we can state that 4 moles of TNT react with 21 moles of O 2 to form 28 moles of CO 2, 10 moles of H 2 O and 6 moles of N 2.
There is one more trick. Using the periodic table, we determine the molecular masses of all these substances:
- C 7 H 5 N 3 O 6 = 227.13 g/mol
- O2 = 31.999 g/mol
- CO2 = 44.010 g/mol
- H2O = 18.015 g/mol
- N2 = 28.013 g/mol
Now equation 9 will also indicate that 4 227.13 g = 908.52 g of TNT require 21 31.999 g = 671.98 g of oxygen to complete the reaction and as a result 28 44.010 g = 1232.3 g of CO 2 are formed, 10·18.015 g = 180.15 g H2O and 6·28.013 g = 168.08 g N2. Let's check whether the law of conservation of mass is satisfied in this reaction:
| Reagents | Products | |
| 908.52 g TNT | 1232.3 g CO2 | |
| 671.98 g CO2 | 180.15 g H2O | |
| 168.08 g N2 | ||
| Total | 1580.5 g | 1580.5 g |
But individual molecules do not necessarily have to participate in a chemical reaction. For example, the reaction of limestone CaCO3 and hydrochloric acid HCl to form an aqueous solution of calcium chloride CaCl2 and carbon dioxide CO2:
- CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2 O (11)
Chemical equation (11) describes the reaction of calcium carbonate CaCO 3 (limestone) and hydrochloric acid HCl to form an aqueous solution of calcium chloride CaCl 2 and carbon dioxide CO 2. This equation is complete, since the number of atoms of each type in its left and right sides is the same.
The meaning of this equation is macroscopic (molar) level is as follows: 1 mole or 100.09 g of CaCO 3 requires 2 moles or 72.92 g of HCl to complete the reaction, resulting in 1 mole of CaCl 2 (110.99 g/mol), CO 2 (44.01 g /mol) and H 2 O (18.02 g/mol). From these numerical data it is easy to verify that the law of conservation of mass is satisfied in this reaction.
Interpretation of equation (11) on microscopic (molecular) level is not so obvious, since calcium carbonate is a salt, not a molecular compound, and therefore chemical equation (11) cannot be understood in the sense that 1 molecule of calcium carbonate CaCO 3 reacts with 2 molecules of HCl. Moreover, the HCl molecule in solution generally dissociates (breaks up) into H + and Cl - ions. Thus, a more correct description of what happens in this reaction at the molecular level is given by the equation:
- CaCO 3 (sol.) + 2H + (aq.) → Ca 2+ (aq.) + CO 2 (g.) + H 2 O (l.) (12)
Here, the physical state of each type of particle is briefly indicated in parentheses ( TV- hard, aq.- hydrated ion in aqueous solution, G.- gas, and.- liquid).
Equation (12) shows that solid CaCO 3 reacts with two hydrated H + ions, forming the positive ion Ca 2+, CO 2 and H 2 O. Equation (12), like other complete chemical equations, does not give an idea of the molecular mechanism reactions and is less convenient for counting the amount of substances, however, it gives a better description of what is happening at the microscopic level.
Reinforce your knowledge of composing chemical equations by working through an example with a solution yourself:
I hope from lesson 13" Writing Chemical Equations"You learned something new for yourself. If you have any questions, write them in the comments.
In order to figure out how to balance a chemical equation, you first need to know the purpose of this science.
Definition
Chemistry studies substances, their properties, and transformations. If there is no change in color, precipitation, or release of a gaseous substance, then no chemical interaction occurs.
For example, when filing an iron nail, the metal simply turns into powder. In this case, no chemical reaction occurs.
Calcination of potassium permanganate is accompanied by the formation of manganese oxide (4), the release of oxygen, that is, an interaction is observed. In this case, a completely natural question arises about how to correctly equalize chemical equations. Let's look at all the nuances associated with such a procedure.
Specifics of chemical transformations
Any phenomena that are accompanied by a change in the qualitative and quantitative composition of substances are classified as chemical transformations. In molecular form, the process of burning iron in the atmosphere can be expressed using signs and symbols.
Methodology for setting coefficients
How to equalize coefficients in chemical equations? The high school chemistry course covers the electronic balance method. Let's look at the process in more detail. To begin with, in the initial reaction it is necessary to arrange the oxidation states of each chemical element.
There are certain rules by which they can be determined for each element. In simple substances, the oxidation states will be zero. In binary compounds, the first element has a positive value, corresponding to the highest valency. For the latter, this parameter is determined by subtracting the group number from eight and has a minus sign. Formulas consisting of three elements have their own nuances in calculating oxidation states.
For the first and last element, the order is similar to the definition in binary compounds, and an equation is drawn up to calculate the central element. The sum of all indicators must be equal to zero, based on this, the indicator for the middle element of the formula is calculated.
Let's continue the conversation about how to equalize chemical equations using the electronic balance method. After the oxidation states have been established, it is possible to determine those ions or substances that changed their value during chemical interaction.
The plus and minus signs must indicate the number of electrons that were accepted (donated) during the chemical interaction. The least common multiple is found between the resulting numbers.
When dividing it into received and donated electrons, the coefficients are obtained. How to balance a chemical equation? The figures obtained in the balance sheet must be placed before the corresponding formulas. A prerequisite is to check the quantity of each element on the left and right sides. If the coefficients are placed correctly, their number should be the same.
Law of conservation of mass of substances
When discussing how to balance a chemical equation, it is this law that must be used. Considering that the mass of those substances that entered into a chemical reaction is equal to the mass of the resulting products, it becomes possible to set coefficients in front of the formulas. For example, how to balance a chemical equation if the simple substances calcium and oxygen interact, and after the process is completed, an oxide is obtained?
To cope with the task, it is necessary to take into account that oxygen is a diatomic molecule with a covalent nonpolar bond, therefore its formula is written in the following form - O2. On the right side, when composing calcium oxide (CaO), the valence of each element is taken into account.
First you need to check the amount of oxygen in each side of the equation as it is different. According to the law of conservation of mass of substances, a coefficient of 2 must be placed in front of the product formula. Next, calcium is checked. In order for it to be equalized, we put a coefficient of 2 in front of the original substance. As a result, we get the entry:
- 2Ca+O2=2CaO.
Analysis of the reaction using the electronic balance method
How to balance chemical equations? Examples of OVR will help answer this question. Let us assume that it is necessary to arrange the coefficients in the proposed scheme using the electronic balance method:
- CuO + H2=Cu + H2O.
To begin with, we will assign oxidation states for each of the elements in the starting substances and reaction products. We get the following form of the equation:
- Cu(+2)O(-2)+H2(0)=Cu(0)+H2(+)O(-2).
The indicators have changed for copper and hydrogen. It is on their basis that we will draw up an electronic balance:
- Cu(+2)+2е=Cu(0) 1 reducing agent, oxidation;
- H2(0)-2e=2H(+) 1 oxidizing agent, reduction.
Based on the coefficients obtained in the electronic balance, we obtain the following entry for the proposed chemical equation:
- CuO+H2=Cu+H2O.
Let's take another example that involves setting coefficients:
- H2+O2=H2O.
In order to equalize this scheme based on the law of conservation of substances, it is necessary to start with oxygen. Considering that a diatomic molecule reacted, a coefficient of 2 must be placed in front of the formula of the reaction product.
- 2H2+O2=2H2O.
Conclusion
Based on the electronic balance, you can place coefficients in any chemical equations. Graduates of the ninth and eleventh grades of educational institutions who choose an exam in chemistry are offered similar tasks in one of the tasks of the final tests.
It has a valence of two, but in some compounds it can exhibit a higher valency. If it is written incorrectly, it may not equalize.
After correctly writing the resulting formulas, we arrange the coefficients. They are for the equation of elements. The essence of equalization is that the number of elements before the reaction is equal to the number of elements after the reaction. You should always start equalizing with . We arrange the coefficients according to the indices in the formulas. If on one side the reaction has an index of two, but on the other it does not (takes on the value of one), then in the second case we put a two in front of the formula.
Once a coefficient is placed in front of a substance, the values of all elements in it increase to the value of the coefficient. If the element has an index, then the resulting sum will be equal to the product of the index and the coefficient.
After equalizing the metals, we move on to non-metals. Then we move on to acidic residues and hydroxyl groups. Next we equalize the hydrogen. At the very end we check reaction according to equalized oxygen.
Chemical reactions are the interaction of substances, accompanied by a change in their composition. In other words, the substances entering into do not correspond to the substances resulting from the reaction. A person encounters such interactions hourly, every minute. After all, the processes occurring in his body (respiration, protein synthesis, digestion, etc.) are also chemical reactions.
Instructions
So, write down the starting materials on the left side of the reaction: CH4 + O2.
On the right, accordingly, there will be reaction products: CO2 + H2O.
The preliminary notation for this chemical reaction will be: CH4 + O2 = CO2 + H2O.
Equalize the above reaction, that is, ensure that the basic rule is fulfilled: the number of atoms of each element in the left and right sides of the chemical reaction must be the same.
You see that the number of carbon atoms is the same, but the number of oxygen and hydrogen atoms is different. There are 4 hydrogen atoms on the left side, and only 2 on the right side. Therefore, put the coefficient 2 in front of the water formula. Get: CH4 + O2 = CO2 + 2H2O.
The carbon and hydrogen atoms are equalized, now it remains to do the same with oxygen. On the left side there are 2 oxygen atoms, and on the right - 4. By placing a coefficient of 2 in front of the oxygen molecule, you get the final record of the methane oxidation reaction: CH4 + 2O2 = CO2 + 2H2O.
How unsurprising nature is for humans: in winter it envelops the earth in a blanket of snow, in spring it reveals all living things like popcorn flakes, in summer it rages with a riot of colors, in autumn it sets plants on fire with red fire... And only if you think about it and look closely, you can see that behind all these so familiar changes are complex physical processes and CHEMICAL REACTIONS. And in order to study all living things, you need to be able to solve chemical equations. The main requirement when balancing chemical equations is knowledge of the law of conservation of the amount of substance: 1) the amount of substance before the reaction is equal to the amount of substance after the reaction; 2) the total amount of substance before the reaction is equal to the total amount of substance after the reaction.
Instructions
To equalize the "example" you need to perform several steps.
Write down the equation reactions in general. To do this, the unknown coefficients are denoted by Latin letters (x, y, z, t, etc.). Let it be necessary to equalize the reaction of the combination of hydrogen and , as a result of which water is obtained. Before the molecules of hydrogen, oxygen and water put the Latin letters
Solving chemical reaction equations causes difficulties for a considerable number of secondary school students, largely due to the wide variety of elements involved in them and the ambiguity of their interactions. But since the main part of the general chemistry course at school examines the interaction of substances based on their reaction equations, students must necessarily fill gaps in this area and learn to solve chemical equations in order to avoid problems with the subject in the future.
The equation of a chemical reaction is a symbolic notation that displays the interacting chemical elements, their quantitative ratio and the substances resulting from the interaction. These equations reflect the essence of the interaction of substances from the point of view of atomic-molecular or electronic interaction.
- At the very beginning of the school chemistry course, they are taught to solve equations based on the concept of valence of elements of the periodic table. Based on this simplification, let us consider the solution of a chemical equation using the example of the oxidation of aluminum with oxygen. Aluminum reacts with oxygen to form aluminum oxide. Having the specified initial data, we will draw up an equation diagram.
Al + O 2 → AlO
In this case, we have written down an approximate diagram of a chemical reaction, which only partially reflects its essence. The substances involved in the reaction are written on the left side of the diagram, and the result of their interaction is written on the right. Additionally, oxygen and other typical oxidizing agents are usually written to the right of metals and other reducing agents on both sides of the equation. The arrow shows the direction of the reaction.
- In order for this compiled reaction scheme to acquire a complete form and comply with the law of conservation of mass of substances, it is necessary:
- Place indices on the right side of the equation for the substance resulting from the interaction.
- Level the amount of elements participating in the reaction with the amount of the resulting substance in accordance with the law of conservation of mass of substances.
- Place indices on the right side of the equation for the substance resulting from the interaction.
- Let's start by suspending the subscripts in the chemical formula of the finished substance. Indices are set in accordance with the valence of chemical elements. Valence is the ability of atoms to form compounds with other atoms due to the combination of their unpaired electrons, when some atoms give up their electrons, while others add them to themselves at an external energy level. It is generally accepted that the valency of a chemical element is determined by its group (column) in the periodic table. However, in practice, the interaction of chemical elements is much more complex and varied. For example, the oxygen atom has a valence of Ⅱ in all reactions, despite the fact that it is in the sixth group in the periodic table.
- To help you navigate this diversity, we offer you the following small reference assistant that will help you determine the valency of a chemical element. Select the element you are interested in and you will see the possible values of its valency. Rare valencies for the selected element are indicated in brackets.
- Let's return to our example. Let us write down its valence on the right side of the reaction diagram above each element.
For aluminum Al the valence will be equal to Ⅲ, and for the oxygen molecule O 2 the valency will be equal to Ⅱ. Find the least common multiple of these numbers. It will be equal to six. We divide the least common multiple by the valence of each element and get the indices. For aluminum, divide six by valence to obtain an index of 2, for oxygen 6/2 = 3. The chemical formula of aluminum oxide obtained as a result of the reaction will take the form Al 2 O 3.
Al + O 2 → Al 2 O 3
- After obtaining the correct formula of the finished substance, it is necessary to check and, in most cases, equalize the right and left parts of the diagram according to the law of conservation of mass, since the reaction products are formed from the same atoms that were originally part of the starting substances participating in the reaction.
- Law of conservation of mass states that the number of atoms that entered into the reaction must be equal to the number of atoms resulting from the interaction. In our scheme, the interaction involves one aluminum atom and two oxygen atoms. As a result of the reaction, we obtain two aluminum atoms and three oxygen atoms. Obviously, the diagram must be leveled using coefficients for elements and matter in order for the law of conservation of mass to be observed.
- Equalization is also performed by finding the least common multiple, which is located between the elements with the largest indices. In our example, this will be oxygen with an index on the right side equal to 3 and on the left side equal to 2. The least common multiple in this case will also be equal to 6. Now we divide the least common multiple by the value of the largest index on the left and right sides of the equation and get the following indices for oxygen.
Al + 3∙O 2 → 2∙Al 2 O 3
- Now all that remains is to equalize the aluminum on the right side. To do this, put a coefficient of 4 on the left side.
4∙Al + 3∙O 2 = 2∙Al 2 O 3
- After arranging the coefficients, the equation of a chemical reaction corresponds to the law of conservation of mass and an equal sign can be placed between its left and right sides. The coefficients placed in the equation indicate the number of molecules of substances participating in the reaction and resulting from it, or the ratio of these substances in moles.