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Reactions and Chemical Equations

One of the fundamental principles of chemistry is that matter is neither creates nor destroyed in a chemical process. (We leave it to the physiequationscists to explore the destructions and creation of matter.) The manifestation of this law of conservation of mass or matter is that the number of each type of atoms that one begins with does not change during the course of the reaction. If there were exactly one billion hydrogen atoms at the start of a reaction there will still be exactly one billion hydrogen atoms at the end of that reaction.

Chemists have developed a short hand notation to describe reactions based on the symbols in the periodic table and the idea (borrowed from algebra) of balanced equations. These equations are a powerful way of presenting some of the information that is known about a chemical reaction in a concise and unambiguous way. Each equation has three parts:

It is essential that equations be balanced if you wish to make practical use of them. This should become very clear when we look at unit factor analysis and reaction stoichiometry.

The balanced equation for a very simple reaction is written below to illustrate several points about how the atomic symbols are used in equations. Since we have already begun to talk about molecules and nomenclature, you should be familiar with the notation that is used in writing chemical formulas.


2 H2    +    O2    --->    2 H2O

The "2" that precedes the symbol for the hydrogen molecule, H2, is called a "coefficient" and represents the number of hydrogen molecules involved in the reaction. Since no number precedes the symbol for the oxygen molecule, O2, we assume that the coefficient is "1". This balanced equation says that two hydrogen molecules can react with one oxygen molecule to form two water molecules. We can check to make sure that the law of conservation of mass is obeyed by determining the number of every type of atoms on either side of the equation. For this reaction, if the total number of hydrogens and the total number of oxygens have remained the same, the equation is balanced. On the reactant side, there are four hydrogen atoms. It's easy to calculate this since two (that's the number of hydrogen molecules) times two (that's the number of hydrogen atoms per molecule) equals four. When this process is repeated to find the number of hydrogen atoms on the product side of the equation and of oxygen atoms on both sides of the equation, it is easy to see that the equation is balanced: each side has four hydrogen atoms and two oxygen atoms.

Balancing an Unbalanced Equation

So let's look at another equation. When butane (C4H10) is burned with enough oxygen (O2) around, cabon dioxide (CO2) and water (H2O) are the only compounds produced. The unbalanced equation for this is give below. We'll need to balance the equation before we can use it for most purposes.

C4H10    +    O2    --->    CO2    +    H2O


We use a stepwise process to balance equations like this. First we'll balance the number of one individual element on both sides of the equation. Let's start with carbon. Since there are four carbons in each butane molecule, we'll make four molecules of carbon dioxide on the product side.

C4H10    +    O2    --->    4 CO2    +    H2O


Once we've balanced one of the elements, we balance a second element. Since each butane has ten hydrogens and each water molecule has two hydrogens, there must be five water molecules for each butane molecule.

C4H10    +    O2    --->    4 CO2    +    5 H2O


Now we have balanced both carbon and hydrogen but we haven't balanced oxygen. Since we have already adjusted the number of carbon dioxide and water molecules on the product side, we'll use them to find the number of oxygen molecules needed for the reaction. There are eight oxygen atoms in the four CO2 molecules and five oxygen atoms in the five H2O molecules. That gives us a total of thirteen oxygen atoms. But each oxygen molecule has two oxygen atoms, so we'll need to divide thirteen by two to get the proper number of oxygen atoms.

C4H10    +    13/2 O2    --->    4 CO2    +    5 H2O


If we are talking strictly about the equation on the molecular level, this equation won't work becase we can't have half a molecule. In that case we need to multiply each of the coefficients in the equation by two.

2 C4H10    +    13 O2    --->    8 CO2    +    10 H2O


However, molecules are so small that as a practical matter we nearly never handle compounds or run reactions on the scale of one or two molecules. Therefore, it is logical to talk about a collective number of molecules. A dozen or a score or even a million is still too small an amount to be useful, so chemists use a much larger collective number called a mole to talk about the amounts of the various compounds that they use. In that case, the coefficients can be fractions of moles. In the same way that you could have a half-dozen you might also have a half-mole.


Let's balance one more equation before we move on. Consider the unbalanced equation:

Fe2(SO4)3    +    BaCl2    --->    FeCl3    +    BaSO4


I'll balance iron first.


Fe2(SO4)3    +    BaCl2    --->    2 FeCl3    +    BaSO4


Next I'll balance the sulfate ions. Since the SO4 unit stays intact throughout the reaction and there are no other sulfur or oxygen species present, I don't need to balance sulfur and oxygen separately.


Fe2(SO4)3    +    BaCl2    --->    2 FeCl3    +    3BaSO4


Then I would balance the barium.


Fe2(SO4)3    +    3 BaCl2    --->    2 FeCl3    +    3 BaSO4


And I can check my work by seeing if the number of chlorine atoms is the same on both sides of the equation. Yes, six on both sides.

Practice Problems

Balance the following equations:


a)        Ca3(PO4)2    +    Na2CO3    --->    CaCO3    +    Na3PO4


b)        CrO3    +    NaOH    --->    Na2Cr2O7    +    H2O


c)        C3H8    +    O2    --->    CO2    +    H2O


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