Abbreviated ionic form of the equation. Ionic equations

Since electrolytes in solution are in the form of ions, reactions between solutions of salts, bases and acids are reactions between ions, i.e. ion reactions. Some of the ions, participating in the reaction, lead to the formation of new substances (lowly dissociating substances, precipitation, gases, water), while other ions, present in the solution, do not produce new substances, but remain in the solution. In order to show which ions interaction leads to the formation of new substances, molecular, complete and brief ionic equations are drawn up.

IN molecular equations All substances are presented in the form of molecules. Complete ionic equations show the entire list of ions present in the solution during a given reaction. Brief ionic equations are composed only of those ions, the interaction between which leads to the formation of new substances (lowly dissociating substances, sediments, gases, water).

When composing ionic reactions, it should be remembered that substances are slightly dissociated (weak electrolytes), slightly and poorly soluble (precipitated - “ N”, “M”, see appendix, table 4) and gaseous ones are written in the form of molecules. Strong electrolytes, almost completely dissociated, are in the form of ions. The “↓” sign after the formula of a substance indicates that this substance is removed from the reaction sphere in the form of a precipitate, and the “” sign indicates that the substance is removed in the form of a gas.

The procedure for composing ionic equations using known molecular equations Let's look at the example of the reaction between solutions of Na 2 CO 3 and HCl.

1. The reaction equation is written in molecular form:

Na 2 CO 3 + 2HCl → 2NaCl + H 2 CO 3

2. The equation is rewritten in ionic form, with well-dissociating substances written in the form of ions, and poorly dissociating substances (including water), gases or sparingly soluble substances - in the form of molecules. The coefficient in front of the formula of a substance in a molecular equation applies equally to each of the ions that make up the substance, and therefore it is placed in front of the ion in the ionic equation:

2 Na + + CO 3 2- + 2H + + 2Cl -<=>2Na + + 2Cl - + CO 2 + H 2 O

3. From both sides of the equality, ions found in the left and right sides are excluded (reduced):

2Na++ CO 3 2- + 2H + + 2Cl -<=> 2Na+ + 2Cl -+ CO 2 + H 2 O

4. The ionic equation is written in its final form (short ionic equation):

2H + + CO 3 2-<=>CO 2 + H 2 O

If during the reaction, and/or slightly dissociated, and/or sparingly soluble, and/or gaseous substances, and/or water are formed, and such compounds are absent in the starting substances, then the reaction will be practically irreversible (→), and for it it is possible to compose a molecular, complete and brief ionic equation. If such substances are present both in the reagents and in the products, then the reaction will be reversible (<=>):

Molecular equation: CaCO 3 + 2HCl<=>CaCl 2 + H 2 O + CO 2

Complete ionic equation: CaCO 3 + 2H + + 2Cl –<=>Ca 2+ + 2Cl – + H 2 O + CO 2

Quite often, schoolchildren and students have to compose the so-called. ionic reaction equations. In particular, task 31, proposed at the Unified State Exam in Chemistry, is devoted to this topic. In this article we will discuss in detail the algorithm for writing short and complete ionic equations, and will analyze many examples of different levels of complexity.

Why are ionic equations needed?

Let me remind you that when many substances are dissolved in water (and not only in water!), a dissociation process occurs - the substances break up into ions. For example, HCl molecules in aquatic environment dissociate into hydrogen cations (H +, more precisely, H 3 O +) and chlorine anions (Cl -). Sodium bromide (NaBr) is found in an aqueous solution not in the form of molecules, but in the form of hydrated Na + and Br - ions (by the way, solid sodium bromide also contains ions).

When writing “ordinary” (molecular) equations, we do not take into account that it is not molecules that react, but ions. Here, for example, is what the reaction equation looks like between hydrochloric acid and sodium hydroxide:

HCl + NaOH = NaCl + H 2 O. (1)

Of course, this diagram does not describe the process entirely correctly. As we have already said, in an aqueous solution there are practically no HCl molecules, but there are H + and Cl - ions. The same is true with NaOH. It would be more correct 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 of why we wrote H 2 O in molecular form. This will be explained a little later. As you can see, there is nothing complicated: we replaced the molecules with ions that are formed during their dissociation.

However, even the complete ionic equation is not perfect. Indeed, take a closer look: both the left and right sides of equation (2) contain the same 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 Brief 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 brief ionic equations are written down. If we had solved problem 31 on the Unified State Exam in chemistry, we would have received the maximum score for it - 2 points.

So, once again about the terminology:

HCl + NaOH = NaCl + H 2 O - molecular equation ("ordinary" 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

Let's create a molecular equation for the reaction.
All particles that dissociate in solution to a noticeable extent are written in the form of ions; substances that are not prone to dissociation are left “in the form of molecules.”
We remove the so-called from the two parts of the equation. 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 complete and short ionic equations describing the interaction of aqueous solutions of barium chloride and sodium sulfate.

Solution. We will act in accordance with the proposed algorithm. Let's first create a molecular equation. 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 is formed during the reaction. Let's check:

Exercise 2. Complete the equations for the following reactions:

KOH + H2SO4 =
H 3 PO 4 + Na 2 O=
Ba(OH) 2 + CO 2 =
NaOH + CuBr 2 =
K 2 S + Hg(NO 3) 2 =
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) phosphoric acid and calcium hydroxide, d) silver nitrate and potassium chloride, e) phosphorus oxide (V) and potassium hydroxide.

I sincerely hope that you have no problems completing these three tasks. If this is not the case, you need to return to the topic "Chemical properties of the main classes of inorganic compounds."

How to turn a molecular equation into a complete ionic equation

The fun begins. We must understand which substances should be written as ions and which should be left in “molecular form”. You will have to remember the following.

In the form of ions write:

soluble salts (I emphasize, only salts that are highly soluble in water);
alkalis (let me remind you that alkalis are bases that are soluble in water, 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, remembering this list is not at all difficult: it includes strong acids and bases and all soluble salts. By the way, especially vigilant for young chemists, who may be outraged by the fact that strong electrolytes (insoluble salts) are not included in this list, I can report the following: NOT including insoluble salts in this list does not at all deny 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 “to make public full list"I give the following information.

In the form of molecules write:

all insoluble salts;
All weak grounds(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, looks like I haven't forgotten anything! Although it’s easier, in my opinion, to remember list No. 1. Of the fundamentally important things in list No. 2, I’ll once again mention water.

Let's train!

Example 2. Write a complete ionic equation describing the interaction of copper (II) hydroxide and hydrochloric acid.

Solution. Let's start, naturally, with the molecular equation. Copper(II) hydroxide is an insoluble base. All insoluble bases react with strong acids to form salt and water:

Cu(OH) 2 + 2HCl = CuCl 2 + 2H 2 O.

Now let’s find out which substances should be written down as ions and which ones as 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 is a strong acid; in solution it almost completely dissociates into ions. CuCl 2 is a soluble salt. We write it in ionic form. Water - only in the form of molecules! We get the complete ionic equation:

Сu(OH) 2 + 2H + + 2Cl - = 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.

Solution. Carbon dioxide is a typical acidic oxide, NaOH is an alkali. When interacting acid oxides With aqueous solutions of alkalis, salt and water are formed. Let’s create a molecular equation for the reaction (don’t forget about the coefficients, by the way):

CO 2 + 2NaOH = Na 2 CO 3 + H 2 O.

CO 2 - oxide, gaseous compound; maintaining molecular shape. NaOH - strong base (alkali); We write it in the form of ions. Na 2 CO 3 - soluble salt; we write in the form of ions. Water is a weak electrolyte and practically does not dissociate; leave in molecular form. We get the following:

CO 2 + 2Na + + 2OH - = Na 2+ + CO 3 2- + H 2 O.

Example 4. Sodium sulfide in aqueous solution reacts with zinc chloride to form a precipitate. Write a complete ionic equation for this reaction.

Solution. Sodium sulfide and zinc chloride are salts. When these salts interact, a precipitate of zinc sulfide precipitates:

Na 2 S + ZnCl 2 = ZnS↓ + 2NaCl.

I will immediately write down the complete ionic equation, and you will analyze it yourself:

2Na + + S 2- + Zn 2+ + 2Cl - = ZnS↓ + 2Na + + 2Cl - .

I offer you several tasks for independent work and a small test.

Exercise 4. Write molecular and complete ionic equations for the following reactions:

NaOH + HNO3 =
H2SO4 + 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).

When dissolved in water, not all substances have the ability to conduct electricity. Those compounds, water solutions which are capable of conducting electric current are called electrolytes. Electrolytes conduct current due to the so-called ionic conductivity, which many compounds with an ionic structure (salts, acids, bases) possess. There are substances that have highly polar bonds, but in solution they undergo incomplete ionization (for example, mercury chloride II) - these are weak electrolytes. Many organic compounds (carbohydrates, alcohols) dissolved in water do not disintegrate into ions, but retain their molecular structure. Such substances do not conduct electric current and are called non-electrolytes.

Here are some principles that can be used to determine whether a particular compound is a strong or weak electrolyte:

Acids . The most common strong acids include HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4. Almost all other acids are weak electrolytes.
Reasons. The most common strong bases are hydroxides of alkali and alkaline earth metals (excluding Be). Weak electrolyte – NH 3.
Salt. Most common salts, ionic compounds, are strong electrolytes. Exceptions are mainly salts of heavy metals.

Electrolytic dissociation theory

Electrolytes, both strong and weak and even very diluted, do not obey Raoult's law And . Having the ability to conduct electrically, the vapor pressure of the solvent and the melting point of electrolyte solutions will be lower, and the boiling point will be higher compared to similar values of a pure solvent. In 1887, S. Arrhenius, studying these deviations, came to the creation of the theory electrolytic dissociation.

Electrolytic dissociation suggests that electrolyte molecules in solution break down into positively and negatively charged ions, which are called cations and anions, respectively.

The theory puts forward the following postulates:

In solutions, electrolytes break down into ions, i.e. dissociate. The more dilute the electrolyte solution, the greater its degree of dissociation.
Dissociation is a reversible and equilibrium phenomenon.
Solvent molecules interact infinitely weakly (i.e., solutions are close to ideal).

Different electrolytes have different degrees of dissociation, which depends not only on the nature of the electrolyte itself, but the nature of the solvent, as well as the concentration of the electrolyte and temperature.

Degree of dissociation α , shows how many molecules n disintegrated into ions, compared to the total number of dissolved molecules N:

α = n/N

In the absence of dissociation α = 0, with complete dissociation of the electrolyte α = 1.

From the point of view of the degree of dissociation, according to strength, electrolytes are divided into strong (α > 0.7), medium strength (0.3 > α > 0.7), weak (α< 0,3).

More precisely, the process of electrolyte dissociation is characterized by dissociation constant, independent of the concentration of the solution. If we imagine the process of electrolyte dissociation in general form:

A a B b ↔ aA — + bB +

K = a b /

For weak electrolytes the concentration of each ion is equal to the product of α by the total concentration of the electrolyte C, so the expression for the dissociation constant can be transformed:

K = α 2 C/(1-α)

For dilute solutions(1-α) =1, then

K = α2C

It's not hard to find from here degree of dissociation

Ionic-molecular equations

Consider an example of neutralization of a strong acid with a strong base, for example:

HCl + NaOH = NaCl + HOH

The process is presented as molecular equation. It is known that both the starting substances and the reaction products in solution are completely ionized. Therefore, let us represent the process in the form complete ionic equation:

H + + Cl - + Na + + OH - = Na + + Cl - + HOH

After “reducing” identical ions on the left and right sides of the equation, we get abbreviated ionic equation:

H + + OH - = HOH

We see that the neutralization process comes down to the combination of H + and OH - and the formation of water.

When composing ionic equations, it should be remembered that only strong electrolytes are written in ionic form. Weak electrolytes solids and gases are written in their molecular form.

The deposition process is reduced to the interaction of only Ag + and I - and the formation of water-insoluble AgI.

To find out whether the substance we are interested in is able to dissolve in water, we need to use the insolubility table.

Let's consider the third type of reaction, which results in the formation of a volatile compound. These are reactions involving carbonates, sulfites or sulfides with acids. For example,

When mixing some solutions of ionic compounds, interactions between them may not occur, for example

So, to summarize, we note that chemical transformations observed when one of the following conditions is met:

Non-electrolyte formation. Water can act as a non-electrolyte.
Formation of sediment.
Gas release.
Formation of a weak electrolyte for example acetic acid.
Transfer of one or more electrons. This is realized in redox reactions.
Formation or rupture of one or more.

Categories ,

In electrolyte solutions, reactions occur between hydrated ions, which is why they are called ionic reactions. towards them important have the nature and strength of the chemical bond in the reaction products. Typically, exchange in electrolyte solutions results in the formation of a compound with a stronger chemical bond. Thus, when solutions of barium chloride salts BaCl 2 and potassium sulfate K 2 SO 4 interact, the mixture will contain four types of hydrated ions Ba 2 + (H 2 O)n, Cl - (H 2 O)m, K + (H 2 O) p, SO 2 -4 (H 2 O)q, between which the reaction will occur according to the equation:

BaCl 2 +K 2 SO 4 =BaSO 4 +2КCl

Barium sulfate will precipitate in the form of a precipitate, in the crystals of which chemical bond between the Ba 2+ and SO 2- 4 ions is stronger than the bond with the water molecules hydrating them. The connection between the K+ and Cl - ions only slightly exceeds the sum of their hydration energies, so the collision of these ions will not lead to the formation of a precipitate.

Therefore, we can draw the following conclusion. Exchange reactions occur during the interaction of such ions, the binding energy between which in the reaction product is much greater than the sum of their hydration energies.

Ion exchange reactions are described by ionic equations. Sparingly soluble, volatile and slightly dissociated compounds are written in molecular form. If, during the interaction of electrolyte solutions, none of the indicated types of compounds are formed, this means that practically no reaction occurs.

Formation of sparingly soluble compounds

For example, the interaction between sodium carbonate and barium chloride in the form of a molecular equation will be written as follows:

Na 2 CO 3 + BaCl 2 = BaCO 3 + 2NaCl or in the form:

2Na + +CO 2- 3 +Ba 2+ +2Сl - = BaCO 3 + 2Na + +2Сl -

Only the Ba 2+ and CO -2 ions reacted, the state of the remaining ions did not change, so the short ionic equation will take the form:

CO 2- 3 +Ba 2+ =BaCO 3

Formation of Volatile Substances

The molecular equation for the interaction of calcium carbonate and hydrochloric acid will be written as follows:

CaCO 3 +2HCl=CaCl 2 +H 2 O+CO 2

One of the reaction products - carbon dioxide CO 2 - was released from the reaction sphere in the form of a gas. The expanded ionic equation is:

CaCO 3 +2H + +2Cl - = Ca 2+ +2Cl - +H 2 O+CO 2

The result of the reaction is described by the following short ionic equation:

CaCO 3 +2H + =Ca 2+ +H 2 O+CO 2

Formation of a slightly dissociated compound

An example of such a reaction is any neutralization reaction, resulting in the formation of water, a slightly dissociated compound:

NaOH+HCl=NaCl+H 2 O

Na + +OH-+H + +Cl - = Na + +Cl - +H 2 O

OH-+H+=H 2 O

From the brief ionic equation it follows that the process is expressed in the interaction of H+ and OH- ions.

All three types of reactions proceed irreversibly to completion.

If you merge solutions of, for example, sodium chloride and calcium nitrate, then, as the ionic equation shows, no reaction will occur, since no precipitate, no gas, or low-dissociating compound is formed:

Using the solubility table, we establish that AgNO 3, KCl, KNO 3 are soluble compounds, AgCl is an insoluble substance.

We create an ionic equation for the reaction taking into account the solubility of the compounds:

A brief ionic equation reveals the essence of the chemical transformation taking place. It can be seen that only Ag+ and Cl - ions actually took part in the reaction. The remaining ions remained unchanged.

Example 2. Make up a molecular and ionic equation for the reaction between: a) iron (III) chloride and potassium hydroxide; b) potassium sulfate and zinc iodide.

a) We compose the molecular equation for the reaction between FeCl 3 and KOH:

Using the solubility table, we establish that of the resulting compounds, only iron hydroxide Fe(OH) 3 is insoluble. We compose the ionic equation of the reaction:

The ionic equation shows that the coefficients of 3 in the molecular equation apply equally to ions. This general rule drawing up ionic equations. Let us represent the reaction equation in short ionic form:

This equation shows that only Fe3+ and OH- ions took part in the reaction.

b) Let's create a molecular equation for the second reaction:

K 2 SO 4 + ZnI 2 = 2KI + ZnSO 4

From the solubility table it follows that the starting and resulting compounds are soluble, therefore the reaction is reversible and does not reach completion. Indeed, no precipitate, no gaseous compound, or slightly dissociated compound is formed here. Let's create a complete ionic equation for the reaction:

2K + +SO 2- 4 +Zn 2+ +2I - + 2K + + 2I - +Zn 2+ +SO 2- 4

Example 3. Using the ionic equation: Cu 2+ +S 2- -= CuS, create a molecular equation for the reaction.

The ionic equation shows that on the left side of the equation there must be molecules of compounds containing Cu 2+ and S 2- ions. These substances must be soluble in water.

According to the solubility table, we will select two soluble compounds, which include the Cu 2+ cation and the S 2- anion. Let's create a molecular equation for the reaction between these compounds:

CuSO 4 +Na 2 S CuS+Na 2 SO 4

Instructions

Before you begin ionic equations, you need to understand some rules. Insoluble in water, gaseous and poorly dissociating substances (for example, water) do not disintegrate into ions, which means write them in molecular form. This also includes weak electrolytes such as H2S, H2CO3, H2SO3, NH4OH. The solubility of compounds can be determined from the solubility table, which is an approved reference material for all types of control. All charges that are inherent in cations and anions are also indicated there. To fully complete the task, you must write molecular, complete and ionic abbreviated equations.

Example No. 1. neutralization reaction between sulfuric acid and potassium hydroxide, consider it from the point of view of ED (electrolytic dissociation theory). First, write down the reaction equation in molecular form and .H2SO4 + 2KOH = K2SO4 + 2H2O Analyze the resulting substances for their solubility and dissociation. All compounds are soluble in water, which means they are ions. The only exception is water, which does not disintegrate into ions and therefore remains in molecular form. Write the complete ionic equation, find the same ions on the left and right sides and . To cancel identical ions, cross them out.2H+ +SO4 2- +2K+ +2OH- = 2K+ +SO4 2- + 2H2OThe result is an ionic abbreviation equation:2H+ +2OH- = 2H2OCoefficients in the form of twos can also be abbreviated:H+ +OH- = H2O

Example No. 2. Write the exchange reaction between copper chloride and , consider it from the point of view of TED. Write the reaction equation in molecular form and assign the coefficients. As a result, the resulting copper hydroxide precipitated a color. CuCl2 + 2NaOH = Cu(OH) 2↓ + 2NaCl Analyze all substances for their solubility in water - everything is soluble except copper hydroxide, which will not be ionized. Write down the complete ionic equation, underline and abbreviate the identical ions: Cu2+ +2Cl- + 2Na+ +2OH- = Cu(OH) 2↓+2Na+ +2Cl- The ionic abbreviated equation remains: Cu2+ +2OH- = Cu(OH) 2↓

Example No. 3. Write the exchange reaction between sodium carbonate and hydrochloric acid, consider it from the point of view of TED. Write the reaction equation in molecular form and assign the coefficients. Sodium chloride is formed and CO2 gas (carbon dioxide or carbon monoxide (IV)) is released. It is formed due to the decomposition of weak, breaking up into oxide and water. Na2CO3 + 2HCl = 2NaCl + CO2+H2OAnalyze all substances for their solubility in water and dissociation. Carbon dioxide leaves the system as a gaseous compound, water is a poorly dissociating substance. All other substances disintegrate into ions. Write down the complete ionic equation, underline and abbreviate the identical ions: 2Na+ +CO3 2- +2H+ +2Cl- =2Na+ +2Cl- +CO2+H2O The ionic abbreviated equation remains: CO3 2- +2H+ =CO2+H2O

Video on the topic

note

To correctly determine the number of ions, you need to multiply the coefficient in front of the formula by the index.

Helpful advice

Be sure to check the coefficients in reaction equations.

Sources:

how to write equations for ion exchange reactions

Reaction equation - conventional notation chemical process, in which some substances are transformed into others with a change in properties. To record chemical reactions, formulas of substances and knowledge of chemical properties connections.

Instructions

Write the formulas correctly according to them. For example, put aluminum oxide Al₂O₃, index 3 from aluminum (corresponding to its oxidation state in this compound) near oxygen, and index 2 (oxidation state of oxygen) near aluminum.
If the oxidation state is +1 or -1, then the index is not given. For example, you need to write down the formula. Nitrate is an acidic residue of nitric acid (-NO₃, d.o. -1), ammonium (-NH₄, d.o. +1). Thus, ammonium nitrate is NH₄ NO₃. Sometimes the oxidation number is indicated in the name of the compound. Sulfur oxide (VI) - SO₃, silicon oxide (II) SiO. Some (gases) are written with index 2: Cl₂, J₂, F₂, O₂, H₂, etc.

It is necessary to know which substances react. Visible reactions: gas evolution, color change and precipitation. Very often reactions pass without visible changes.
Example 1: neutralization reaction
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
Sodium hydroxide reacts with sulfuric acid to form the soluble salt sodium sulfate and water. The sodium ion is split off and combines with the acidic one, replacing the hydrogen. The reaction takes place without external signs.
Example 2: iodoform test
С₂H₅OH + 4 J₂ + 6 NaOH→CHJ₃↓ + 5 NaJ + HCOONa + 5 H₂O
The reaction occurs in several stages. The end result is the precipitation of iodoform crystals yellow color(qualitative reaction to).
Example 3:
Zn + K₂SO₄ ≠
The reaction is impossible, because In the series of metal stresses, zinc comes after potassium and cannot displace it from compounds.

The law of conservation of mass states: the mass of substances that react is equal to the mass of the substances formed. Competent recording chemical reaction– half. It is necessary to set the coefficients. Start equalizing with those compounds whose formulas contain large indices.
K₂Cr₂O₇ + 14 HCl → 2 CrCl₃ + 2 KCl + 3 Cl₂ + 7 H₂O
Start setting the coefficients with potassium dichromate, because its formula contains the largest index (7).
Such accuracy in recording is necessary for calculating mass, volume, concentration, released energy and other quantities. Be careful. Memorize the most common formulas and bases, as well as acid residues.

Sources:

chemistry equation

Working with formulas and equations in the Word office application included in the package Microsoft Office, is provided by a special utility “Formula Editor”, which is part of the Math Type program.