Chapter 9 Notes



Chapter 9 Notes

STOICHIOMETRY deals with mass relationships. COPOSITION STOICHIOMETRY is the mass relationship between the elements in a compound. REACTION STOICHIOMETRY is the mass relationship between the reactants and products of a chemical reaction.

You have been studying composition stoichiometry in the previous chapters. Our study of chemical reactions introduces mass relationships between the elements or compounds found in chemical reactions. Stoichiometry in regards to a chemical reaction can be compared to a recipe used in cooking. In a recipe, you can find a list of ingredients. These ingredients can be compared to the reactants in a chemical reaction. The product would be comparable to the final dish prepared from the recipe. Anytime the recipe is adjusted due to the lack of ingredients or to produce larger amounts of a product is the same as stoichiometry.

With a chemical reaction, it is very important that the chemical equation is balanced. A proper balanced equation is our recipe. From this balanced chemical equation we find MOLAR RATIOS. Molar ratios are conversion factors between the numbers of moles of any two substances in a chemical reaction.

For the equation:

2Al2O3 ( 4Al + 3O2

The following molar ratios can be derived.

2 moles Al2O3 4 moles Al

4 moles Al 2 moles Al2O3

4 moles Al 3 moles O2

3 moles O2 4 moles Al

2 moles Al2O3 3 moles O2

3 moles O2 2 moles Al2O3

So if you have 13 moles of Al2O3, how many moles of Al can be produced?

113 moles Al2O3 x 4 moles Al = 26 moles of Al

2 moles Al2O3

Using the molar mass of Al, we can convert to grams.

26 moles Al x 26.98 grams Al = 701 grams Al

1 mole Al

There are 4 basic types of stoichiometric problems; mole to mole, gram to mole, mole to gram, and gram to gram. However, volumes of a reactant or product can be introduced using the conversion factors of density for liquids or solids and 22.4 liters/1 mole for gases.

Common to all these problems is a “given” and an “unknown”. Identifying these in the problem is very important. The stoichiometric map provided outlines the steps involved in solving these problems.

Let’s look at each of these types of problems.

MOLE TO MOLE

In a space craft, the carbon dioxide exhaled by astronauts can be removed by the reaction with lithium hydroxide, LiOH, according to the following chemical equation.

CO2(g) + 2LiOH(s) ( Li2CO3(s) +H2O(l)

How many moles of LiOH are required to react with 20 moles of CO2, the average amount exhaled by a person per day?

Given: 20 moles CO2 Unknown: ? moles LiOH

The molar ratio is 2 moles LiOH to 1 mole CO2. This is found in the balanced chemical equation.

20 moles CO2 x 2 moles LiOH = 40 moles LiOH

1 mole CO2

MOLES TO GRAMS

In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6 from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 moles of water react with carbon dioxide?

6CO2(g) + 6H2O(l) ( C6H12O6(s) + 6O2(g)

Given: 3.00 moles water Unknown: ? grams of glucose

From the balanced chemical equation, the molar ratio is 6 moles H2O to 1 mole C6H12O6. Using the periodic table to calculate, the molar mass of C6H12O6 is 180.18 grams.

3.00 moles H2O x 1 mole C6H12O6 x 180.18 grams = 90.1 grams C6H12O6

6 moles H2O 1 mole C6H12O6

GRAMS TO MOLES

The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.

4NH3(g) + 5O2(g) ( 4NO(g) + 6H2O(g)

Using 824 grams of NH3 and excess oxygen, how many moles of NO is formed?

Given: 824 grams NH3 Unknown: ? moles of NO

From the balanced chemical equation, the molar ratio is 4 moles NH3 to 4 moles NO. Using the periodic table to calculate, the molar mass of NH3 is 17.04 grams.

824g NH3 x 1 mole NH3 x 4 moles NO = 48.4 moles NO

17.04 grams 4 moles NH3

GRAMS TO GRAMS

Tin (II) fluoride, SnF2, is used in some toothpaste. It is made by reacting tin with hydrogen fluoride according to the following equation.

Sn(s) + 2HF(g) (SnF2(s) + H2(g)

How many grams of SnF2 are produced from 30.00 grams HF?

Given: 30.00 grams HF Unknown: ? grams of SnF2

From the balanced chemical equation, the molar ratio is 1 mole SnF2 to 2 moles HF. Using the periodic table to calculate, the molar mass of HF is 20.01 grams and the molar mass of SnF2 is 156.71 grams.

30.00g HF x 1 mole HF x 1 mol SnF2 x 156.71g SnF2 = 117.5 grams SnF2

20.01g HF 2 mol HF 1 mol SnF2

In most reactions, one of the reactants is completely used up in the reaction and the other reactant will have some amount remaining. The reactant that is completely consumed in the reaction is known as the LIMITTING REACTANT. The limiting reactant determines how much product is produced. The other reactant will be in excess and is called an EXCESS REACTANT. Limiting reactant problems use stoichiometric solutions.

EXAMPLE

Silicon dioxide is usually unreactive, but reacts readily with hydrogen fluoride according to the following equation.

SiO2(s) + 4HF(g) ( SiF4(g) + 2H2O(l)

If 6.0 moles of HF is added to 4.5 moles of SiO2, which is the limiting reactant?

Given: 6.0 moles HF and 4.5 moles SiO2 Unknown: ? Limiting reactant

You will have to select a product. It is not suggested that water is chosen as a product unless it is the only product in the equation. So in this case, we will choose SiO4. We now work a mole to mole stoichiometric problem with both reactants.

6.0 moles HF x 1 mole SiO4 = 1.5 moles SiO4

4 moles HF

4.5 moles SiO2 x 1 mole SiO4 = 4.5 moles SiO4

1 mole SiO2

Because 6.0 moles of HF produce the lesser amount of product, SiO4, it is considered to be the limiting reactant.

Stoichiometric problems can also be used to calculate the amount of product is expected to be produced in a reaction. This calculated amount of product can be compared to the amount actually produced in a laboratory setting. In order to make this type of comparison, the calculated amount of product expected to be produced is known as the THEORETICAL YIELD. Using the following equation,

Percentage yield = experimental yield x 100

theoretical yield

EXAMPLE

Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C6H6, with chlorine.

C6H6(l) + Cl2(g) ( C6H5Cl(l) + HCl(g)

When 36.8 grams of benzene react with excess chlorine, the actual yield of chlorobenzene is 38.8 grams. What is the percentage yield?

Given: 36.8 grams C6H6 and actual yield is 38.8 grams

Unknown: Percentage yield of chlorobezene

From the balanced chemical equation, the molar ratio is 1 mole of C6H6 to 1 mole of C6H5Cl. Using the periodic table to calculate, the molar mass of benzene is 78.12 grams and the molar mass of chlorobenzene is 112.56 grams.

36.8g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl x 112.56g = 53.0 grams C6H5Cl

78.12 grams 1 mol C6H6 1 mol C6H5Cl

Percentage yield = 38.8 grams x 100 = 73.2 %

53.0 grams

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