# Stoichiometry

## What is a Chemical Equation?

In chemistry, we use symbols to represent the various chemicals. Success in chemistry depends upon developing a strong familiarity with these basic symbols. For example, the symbol "C"represents an atom of carbon, and "H" represents an atom of hydrogen. To represent a molecule of table salt, sodium chloride, we would use the notation "NaCl", where "Na" represents sodium and "Cl" represents chlorine. We call chlorine "chloride" in this case because of its connection to sodium. You should have reviewed naming schemes, or nomenclature. in earlier readings.

A chemical equation is an expression of a chemical process. For example:

AgNO3 (aq) + NaCl(aq) ---> AgCl(s) + NaNO3 (aq)

In this equation, AgNO3 is mixed with NaCl. The equation shows that the reactants (AgNO3 and NaCl) react through some process (--->) to form the products (AgCl and NaNO3 ). Since they undergo a chemical process, they are changed fundamentally.

Often chemical equations are written showing the state that each substance is in. The (s) sign means that the compound is a solid. The (l) sign means the substance is a liquid. The (aq) sign stands for aqueous in water and means the compound is dissolved in water. Finally, the (g) sign means that the compound is a gas.

Coefficients are used in all chemical equations to show the relative amounts of each substance present. This amount can represent either the relative number of molecules. or the relative number of moles (described below). If no coefficient is shown, a one (1) is assumed.

On some occasions, a variety of information will be written above or below the arrows. This information, such as a value for temperature, shows what conditions need to be present for a reaction to occur. For example, in the graphic below, the notation above and below the arrows shows that we need a chemical Fe2 O3. a temperature of 1000° C, and a pressure of 500 atmospheres for this reaction to occur.

The graphic below works to capture most of the concepts described above:

## The Mole

Given the equation above, we can tell the number of moles of reactants and products. A mole simply represents Avogadro's number (6.022 x 10 23 ) of molecules. A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve of them. Similarly, if you have a mole of carrots, you have 6.022 x 10 23 carrots. In the equation above there are no numbers in front of the terms, so each coefficient is assumed to be one (1). Thus, you have the same number of moles of AgNO3. NaCl, AgCl, NaNO3 .

Converting between moles and grams of a substance is often important. This conversion can be easily done when the atomic and/or molecular mass of the substance(s) are known. Given the atomic or molecular mass of a substance, that mass in grams makes a mole of the substance. For example, calcium has an atomic mass of 40 atomic mass units. So, 40 grams of calcium makes one mole, 80 grams makes two moles, etc.

## Balancing Chemical Equations

Sometimes, however, we have to do some work before using the coefficients of the terms to represent the relative number of molecules of each compound. This is the case when the equations are not properly balanced. We will consider the following equation:

+ Fe

Since no coefficients are in front of any of the terms, it is easy to assume that one (1) mole of Al and one (1) mole of Fe3 O4 react to form one (1) mole of Al2 O3. If this were the case, the reaction would be quite spectacular: an aluminum atom would appear out of nowhere, and two (2) iron

atoms and one (1) oxygen atom would magically disappear. We know from the Law of Conservation of Mass (which states that matter can neither be created nor destroyed) that this simply cannot occur. We have to make sure that the number of atoms of each particular element in the reactants equals the number of atoms of that same element in the products. To do this we have to figure out the relative number of molecules of each term expressed by the term's coefficient.

Balancing a simple chemical equation is essentially done by trial and error. There are many different ways and systems of doing this, but for all methods, it is important to know how to count the number of atoms in an equation. For example we will look at the following term.

This term expresses two (2) molecules of Fe3 O4. In each molecule of this substance there are three (3) Fe atoms. Therefore in two (2) molecules of the substance there must be six (6) Fe atoms. Similarly there are four (4) oxygen atoms in one (1) molecule of the substance so there must be eight (8) oxygen atoms in two (2) molecules.

Now let's try balancing the equation mentioned earlier:

Developing a strategy can be difficult, but here is one way of approaching a problem like this.

1. Count the number of each atom on the reactant and on the product side.
• Determine a term to balance first. When looking at this problem, it appears that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplest way to balance the oxygen terms is:

Be sure to notice that the subscript times the coefficient will give the number of atoms of that element. On the reactant side, we have a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. On the product side, we have a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. Now, the oxygens are balanced.

• Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have:

• Balance the last term. In this case, since we had eight (8) aluminum atoms on the product side we need to have eight (8) on the reactant side so we add an eight (8) in front of the Al term on the reactant side.

Now, we're done, and the balanced equation is:

• ## Limiting Reagents

Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the "limiting reagent". Often, it is necessary to identify the limiting reagent in a problem.

Example: A chemist only has 6.0 grams of C2 H2 and an unlimited supply of oxygen and he desires to produce as much CO2 as possible. If she uses the equation below, how much oxygen should she add to the reaction?

To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up (this is the way to produce the maximum amount of CO2 ).

First, we calculate the number of moles of C2 H2 in 6.0 g of C2 H2. To be able to calculate the moles we need to look at a periodic table and see that 1 mole of C weighs 12.0 g and H weighs 1.0 g. Therefore we know that 1 mole of C2 H2 weighs 26 g (2 × 12 grams + 2 × 1 gram).

Source: www.shodor.org

Category: Forex