Useful Tips

Lesson 13

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A chemical equation is a conditional record of a process that occurs during a chemical reaction. The law of conservation of mass excludes the appearance of new atoms and the destruction of old ones during a chemical reaction. Thus, the number of atoms contained in the reagents should be equal to the number of atoms in the products of a chemical reaction. The guide below will help you understand how to solve chemical equations.

Method 1 Traditional Method

  1. 1 Write down the equation you need to balance. For example, we will use the following equation: C3H8 + O2 -> H2O + CO2This is a propane combustion reaction (C3H8) in the presence of oxygen to form carbon dioxide and water.
  2. 2 Write down the number of atoms on each side of the equation. Pay attention to the index of each of the atoms, it is he who indicates their number.
    • Left side: 3 carbon atoms, 8 hydrogen atoms and 2 oxygen atoms.
    • Right side: 1 carbon atom, 2 hydrogen atoms and 3 oxygen atoms.
  3. 3 Always equalize the hydrogen and oxygen atoms at the end. This means that you need to start with carbon atoms.
  4. 4 Add the coefficient to the carbon atom on the right side of the equation in order to balance it with the three atoms on the left side. C3H8 + O2 -> H2O + 3 CO2
    • The coefficient 3 in front of the carbon atom on the right side of the equation indicates the number of atoms in the same way as the index 3 under the carbohydrate atom on the left side of the equation.
    • In the chemical equation, only the coefficients can be adjusted, while the indices cannot be changed in any case.
  5. 5 Now balance the hydrogen atoms. There are 8 atoms on the right side of the equation, respectively, the same number of hydrogen atoms should be on the left side. C3H8 + O2 -> 4 H2O + 3CO2
    • On the right side of the equation, we added a coefficient of 4 in front of hydrogen, since it already has an index of 2.
    • Thus, multiplying the coefficient 4 by index 2, we get 8 atoms.
  6. 6 other 6 oxygen atoms are derived from 3CO2. (3x2 = 6 oxygen atoms + another 4 = 10)
  7. 7 At the end, balance the oxygen atoms.
    • By adding indices in front of the molecules of substances on the right side of the equation, we also changed the number of oxygen atoms. Now we have 4 oxygen atoms in the molecules of water and 6 oxygen atoms in the molecules of carbon dioxide (carbon dioxide). Together it will be 10 atoms.
    • Then we need to put the coefficient 5 in front of the oxygen molecule on the left side of the equation. Thus, we get 10 oxygen atoms on each side of the equation.

C3H8 + 5 O2 -> 4H2O + 3CO2. Now the number of all carbon, hydrogen and oxygen atoms is the same in both sides of the equation. So your equation is balanced.

Chemical equation

The combustion reaction of methane CH4 in oxygen O2 carbon dioxide CO2 and water H2O. This reaction can be described chemical equation:

Let's try to extract more information from the chemical equation than just an indication products and reagents reactions. The chemical equation (1) is incomplete and therefore does not give any information about how many O molecules2 expended per 1 molecule of CH4 and how many CO molecules2 and H 2 O is the result. But if we write down the numerical coefficients in front of the corresponding molecular formulas that indicate how many molecules of each sort are involved in the reaction, then we get complete chemical equation reactions.

In order to complete the preparation of chemical equation (1), one simple rule must be remembered: the same number of atoms of each sort must be present on the left and right sides of the equation, since during the chemical reaction no new atoms arise and existing ones are not destroyed. This rule is based on the law of conservation of mass, which we examined at the beginning of the chapter.

Chemical equalization

Chemical equalization it is necessary in order to get the full from a simple chemical equation. So, let's move on to directly equalizing reaction (1): once again, look at the chemical equation, exactly at the atoms and molecules in the right and left sides. It is easy to notice that three types of atoms participate in the reaction: carbon C, hydrogen H and oxygen O. Let's calculate and compare the number of atoms of each sort in 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 molecule4, and on the right side one C atom is part of CO2. Thus, the number of carbon atoms coincides in the left and right parts, so we leave it alone. But for clarity, we put a coefficient of 1 in front of molecules with carbon, although this is not necessary:

Then we proceed to the calculation of the hydrogen atoms H. On the left side there are 4 H atoms (in a quantitative sense, H4 = 4H) in the composition of the CH molecule4, and in the right - only 2 H atoms in the H molecule2O, which is two times less than on the left side of the chemical equation (2). We will call! To do this, put the coefficient 2 in front of the molecule H2O. Now we will have 4 hydrogen molecules H in our reagents and products:

Please note that the coefficient 2, which we wrote down in front of the water molecule H2O to equalize hydrogen H, increases by 2 times all the atoms in its composition, i.e. 2H2O is 4H and 2O. Well, it seems to be sorted out, it remains to calculate and compare the number of oxygen atoms O in the chemical equation (3). It is immediately evident that on the left side of the O atoms there are exactly 2 times less than on the right. Now you yourself already know how to equalize the chemical equations, so immediately write down the final result:

As you can see, equalization of 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 observes the law of conservation of mass, i.e. the number of atoms of each variety entering into the reaction exactly coincides with the number of atoms of this variety at the end of the reaction. Each part 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 separate intermediate stages, and therefore, for example, one cannot interpret equation (4) 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 complicated. The second point: the complete reaction equation does not tell us anything about its molecular mechanism, that is, about the sequence of events that occur at the molecular level during its course.

Coefficients in the equations of chemical reactions

Another good example of how to arrange odds in the equations of chemical reactions: Trinitrotoluene (TNT) C7H5N3O6 vigorously combines with oxygen to form H2O CO2 and N2. We write the reaction equation, which we will equalize:

It is easier to draw up a complete equation based on two TNT molecules, since the left side contains an odd number of hydrogen and nitrogen atoms, and the even part is even on the right:

  • 2C7H5N3O6 + O2 → CO2 + H2O + N2 (6)

Then it’s clear that 14 carbon atoms, 10 hydrogen atoms and 6 nitrogen atoms should turn into 14 carbon dioxide molecules, 5 water molecules and 3 nitrogen molecules:

Now both parts contain the same number of all atoms except oxygen. Of the 33 oxygen atoms on the right side of the equation, 12 are delivered by two initial TNT molecules, and the remaining 21 must be supplied by 10.5 O molecules2. Thus, the complete chemical equation will look like:

You can multiply both parts by 2 and get rid of an integer coefficient of 10.5:

But this can not be done, since all the coefficients of the equation do not have to be integer. It’s even more correct to make an equation based on one TNT molecule:

The complete chemical equation (9) carries a lot of information. First of all, it indicates the starting materials - reagents, as well as the products reactions. In addition, it shows that in the course of the reaction all atoms of each sort are individually preserved. If we multiply both sides of equation (9) by the Avogadro number NA= 6.022 · 10 23, we can state that 4 moles of TNT react with 21 moles of O2 with the formation of 28 moles of CO210 moles H2O and 6 moles N2.

There is one more chip. Using the periodic table, we determine the molecular weights of all these substances:

  • C 7 H 5 N 3 O 6 = 227.13 g / mol
  • O 2 = 31.999 g / mol
  • CO 2 = 44.010 g / mol
  • H 2 O = 18.015 g / mol
  • N 2 = 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 for the complete reaction and as a result 28 · 44.010 g = 1232.3 g of CO are formed210 · 18.015 g = 180.15 g H2O and 6 · 28.013 g = 168.08 g N2. Check if the mass conservation law holds in this reaction:

ReagentsProducts
908.52 g TNT1232.3 g CO 2
671.98 g CO 2180.15 g H 2 O
168.08 g N 2
Total1580.5 g1580.5 g

But not necessarily individual molecules must be involved in the chemical reaction. For example, the reaction of limestone CaCO 3 and hydrochloric acid HCl, with the formation of an aqueous solution of calcium chloride CaCl 2 and carbon dioxide CO 2:

Chemical equation (11) describes the reaction of calcium carbonate CaCO3 (limestone) and hydrochloric acid HCl to form an aqueous solution of calcium chloride CaCl2 and carbon dioxide CO2. This equation is complete, since the number of atoms of each sort in its left and right parts is the same.

The meaning of this equation at the macroscopic (molar) level is as follows: 1 mol or 100.09 g of CaCO3 requires 2 moles or 72.92 g of HCl to complete the reaction, resulting in 1 mole of CaCl2 (110.99 g / mol), CO2 (44.01 g / mol) and H2O (18.02 g / mol). From these numerical data, it is easy to verify that the mass conservation law is satisfied in this reaction.

The interpretation of equation (11) at the 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 CaCO3 reacts with 2 molecules of HCl. Moreover, the HCl molecule in solution generally dissociates (decomposes) into H + and Cl - ions. Thus, a more accurate description of what happens in this reaction at the molecular level gives the equation:

Here in parentheses the physical state of each kind of particles is abbreviated (tv - hard aq. - hydrated ion in an aqueous solution, g. - gas g. - liquid).

Equation (12) shows that solid CaCO3 reacts with two hydrated H + ions, forming a positive ion Ca 2+, CO2 and H2O. Equation (12), like other complete chemical equations, does not give an idea of ​​the molecular mechanism of the reaction and is less convenient for calculating the amount of substances, however, it gives a better description of what is happening at the microscopic level.

Fix the knowledge gained in the preparation of chemical equations by independently examining an example with a solution:

I hope from lesson 13 "Drawing up chemical equations»You have learned something new for yourself. If you have any questions, write them in the comments.

Instructions

To equalize a chemical reaction, enter the reaction equation and click the Equate button. The solved equation appears at the top.

  • Use uppercase characters for the initial character of the element and lowercase characters for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.
  • Ion charges are not yet supported and will not be taken into account.
  • Move immutable groups in the joints to avoid ambiguity. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not equalize, but XC2H5 + O2 = XOH + CO2 + H2O will equalize.
  • Intermediate distances [such as (s) (aq) or (g)] are not required.
  • You can use parentheses () and square brackets [].

Chemical equation balancing example

Step 1.

Compare items on the left and right sides

ElementLeft sideRight part
N21
H23

Step 2

To balance the chemical equation, add 2 to the right side.

ElementLeft sideRight part
N22
H26

Step 3

Add 3 to the H element on the left side of the equation.
Equation: N2 + 3H2 = 2NH3

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