Today we’ll talk about how to arrange coefficients in chemical equations. This question is of interest not only to high school students of general educational institutions, but also to children who are only acquainted with the basic elements of a complex and interesting science. If at the first stage you understand how to write chemical equations, in the future there will be no problems with solving problems. Let's get it right from the start.
What is an equation?
It is customary to mean a conditional record of a chemical reaction occurring between selected reagents. For such a process, indices, coefficients, formulas are used.
Compilation algorithm
How to make chemical equations? Examples of any interactions can be written by summing up the original compounds. An equal sign indicates that interaction occurs between the reacting substances. Next, the product formula for valency (oxidation state) is compiled.
How to record a reaction
For example, if you need to write chemical equations that confirm the properties of methane, we choose the following options:
- halogenation (radical interaction with the element VIIA of the periodic table of D. I. Mendeleev);
- combustion in oxygen.
For the first case, on the left side we write the starting materials, on the right - the resulting products. After checking the number of atoms of each chemical element, we obtain the final record of the ongoing process. When methane burns in atmospheric oxygen, an exothermic process occurs, resulting in the formation of carbon dioxide and water vapor.
In order to correctly set the coefficients in chemical equations, the law of conservation of mass of substances is used. We begin the adjustment process by determining the number of carbon atoms. Next, we carry out calculations for hydrogen and only after that we check the amount of oxygen.
OVR
Complex chemical equations can be balanced by using the electronic balance method or half-reactions. We offer a sequence of actions designed to arrange the coefficients in the reactions of the following types:
- decomposition;
- substitution.
First, it is important to arrange the oxidation state of each element in the compound. When arranging them, some rules must be taken into account:
- In a simple substance, it is zero.
- In a binary connection, their sum is 0.
- In the compound of three or more elements, the first manifests a positive value, the extreme ion - a negative value of the oxidation state. The central element is calculated mathematically, given that the total must be 0.
Next, those atoms or ions are selected in which the oxidation state indicator has changed. The plus and minus signs indicate the number of electrons (received, given). Next, the least multiple is determined between them. When dividing the NOC by these numbers get numbers. This algorithm will be the answer to the question of how to arrange the coefficients in chemical equations.
First example
Let’s say the task is given: “Set the coefficients in the reaction, supplement the omissions, determine the oxidizing agent and reducing agent.” Such examples are offered to school graduates who have chosen chemistry as the exam.
KMnO 4 + H 2 SO 4 + KBr = MnSO 4 + Br 2 + ... + ...
Let's try to understand how to arrange the coefficients in the chemical equations proposed by future engineers and physicians. After arranging the oxidation states of the elements in the starting materials and available products, we find that the manganese ion acts as an oxidizing agent, and the bromide ion demonstrates the reducing properties.
We conclude that the missing substances do not participate in the redox process. One of the missing products is water, and the second will be potassium sulfate. After compiling the electronic balance, the final stage will be the setting of the coefficients in the equation.
Second example
We give one more example to understand how to arrange the coefficients in the chemical equations of the redox type.
Suppose the following scheme is given:
P + HNO 3 = NO 2 + ... + ...
Phosphorus, which by hypothesis is a simple substance, exhibits reducing properties, increasing the oxidation state to +5. Therefore, one of the missed substances will be phosphoric acid H 3 PO 4. OVR suggests the presence of a reducing agent, which will be nitrogen. It goes into nitric oxide (4), forming NO 2
In order to put the coefficients in this reaction, we will compose an electronic balance.
P 0 gives 5e = P +5
N +5 takes e = N +4
Given that nitric acid and nitric oxide (4) must have a coefficient of 5, we get the finished reaction:
P + 5HNO 3 = 5NO 2 + H 2 O + H 3 PO 4
The stereochemical coefficients in chemistry make it possible to solve various computational problems.
Third example
Given that the arrangement of the coefficients causes difficulties for many high school students, it is necessary to work out a sequence of actions using specific examples. We offer another example of a task, the implementation of which involves the possession of the methodology for arranging the coefficients in the redox reaction.
H 2 S + HMnO 4 = S + MnO 2 + ...
A feature of the proposed task is that it is necessary to supplement the missed reaction product and only after that it is possible to proceed to the formulation of the coefficients.
After arranging the oxidation states of each element in the compounds, it can be concluded that manganese, which reduces valency, exhibits oxidizing properties. Sulfur demonstrates the reducing ability in the proposed reaction, reducing to a simple substance. After compiling the electronic balance, we can only arrange the coefficients in the proposed process scheme. And the job is done.
Fourth example
A chemical equation is called a complete process when it fully observes the law of conservation of mass of substances. How to check this pattern? The number of atoms of one species that have entered into a reaction should correspond to their number in the products of interaction. Only in this case it will be possible to talk about the usefulness of the recorded chemical interaction, the possibility of its use for performing calculations, solving computational problems of different levels of complexity. Here is a variant of the task, involving the arrangement of the missing stereochemical coefficients in the reaction:
Si + ... + HF = H 2 SiF 6 + NO + ...
The complexity of the task is that both the starting materials and the interaction products are missing. After setting all the oxidation state elements, we see that the silicon atom exhibits reducing properties in the proposed task. Among the reaction products, nitrogen (II) is present; nitric acid is one of the starting compounds. Logically, we determine that the missing reaction product is water. The final step will be the arrangement of the obtained stereochemical coefficients in the reaction.
3Si + 4HNO 3 + 18HF = 3H 2 SiF 6 + 4NO + 8 H 2 O
An example of an equation problem
It is necessary to determine the volume of a 10% hydrogen chloride solution, the density of which is 1.05 g / ml, necessary to completely neutralize the calcium hydroxide formed during the hydrolysis of its carbide. It is known that the gas released during hydrolysis occupies a volume of 8.96 liters (n.a.) In order to cope with the task, it is necessary to first compile the equation for the process of hydrolysis of calcium carbide:
CaC 2 + 2H 2 O = Ca (OH) 2 + C 2 H 2
Calcium hydroxide interacts with hydrogen chloride, there is a complete neutralization:
Ca (OH) 2 + 2HCl = CaCl 2 + 2H 2 O
We calculate the mass of acid that will be required for this process. We determine the volume of a solution of hydrogen chloride. All calculations for the problem are carried out taking into account stereochemical coefficients, which confirms their importance.
Finally
An analysis of the results of the unified state exam in chemistry indicates that the tasks associated with stereochemical coefficients in the equations, the preparation of the electronic balance, the determination of the oxidizing agent and reducing agent cause serious difficulties for modern graduates of secondary schools. Unfortunately, the degree of independence of modern graduates is almost minimal, so high school students do not conduct the development of the theoretical base proposed by the teacher.
Among the typical mistakes made by schoolchildren, arranging the coefficients in reactions of various types, there are many mathematical errors. For example, not everyone knows how to find the least common multiple, correctly divide and multiply numbers. The reason for this phenomenon is a decrease in the number of hours allocated in educational schools to study this topic. With the basic chemistry program, teachers have no way to work out with their students questions related to compiling the electronic balance in the redox process.