Stoichiometry - Chemical equations are the recipes that tell the manufacturer how much of each chemical to use in making a product. - Stoichiometry is the calculation of quantities in chemical reactions. - Stoichiometric calculations are calculations done using balanced chemical equations.

Proportional Relationships 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. I have 5 eggs. How many cookies can I make? I have 5 eggs. How many cookies can I make? 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

2 cups mix + 2 egg cup oil + 1 cup milk → 8 pancakes How many cups of mix will you need to make 24 pancakes? How many cups of oil will you need to make 28 pancakes? How many pancakes can you make with 13 cups of mix? Proportional Relationships

N2N2 + H2H2 NH 3 23 What does this equation tell us? Interpreting Chemical Reactions

Proportional Relationships in Equations N 2 + 3H 2 → 2NH 3 What is the relationship between N 2 and H 2 ? What is the relationship between N 2 and H 2 ? What is the relationship between N 2 and NH 3 ? What is the relationship between N 2 and NH 3 ? What is the relationship between H 2 and NH 3 ? What is the relationship between H 2 and NH 3 ?

Interpreting Chemical Equations Interpreting Chemical Equations Lets describe this reaction in terms of the number of particles. Lets describe this reaction in terms of the number of particles. 2H 2 + O 2 → 2H 2 O 2H 2 + O 2 → 2H 2 O 2 molecules H 2 +1 molecule O 2 → 2 molecules H 2 O

Interpreting Chemical Equations Interpreting Chemical Equations Lets describe this reaction in terms of the number of moles. Lets describe this reaction in terms of the number of moles. 2H 2 + O 2 → 2H 2 O 2H 2 + O 2 → 2H 2 O 2 moles H moles O 2 → 2 moles H 2 O

Interpreting Chemical Equations Interpreting Chemical Equations Lets describe this reaction in terms of mass. Lets describe this reaction in terms of mass. 2H 2 + O 2 → 2H 2 O 2H 2 + O 2 → 2H 2 O g H g O 2 → g H 2 O 2 mol H 2 x g 1 mol 1 mol O 2 x g 1 mol 2 mol H 2 Ox g 1 mol

Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. a. Mole ratio (moles to moles) is the core step in all stoichiometry problems.

How many moles of ammonia (NH 3 ) are produced when 0.60 moles Nitrogen reacts with Hydrogen? How many moles of ammonia (NH 3 ) are produced when 0.60 moles Nitrogen reacts with Hydrogen? Mole to Mole Calculations Mole to Mole Calculations First you need the balanced equation.N 2 +3H 2 →2NH 3 Now according to the equation: When 1 mole of nitrogen reacts with hydrogen, 2 mole of ammonia are produced. Conversion Factor: 1 mol N 2 = 2 mol NH mol N 2 x 1 mol N 2 2 mol NH 3 =1.2 mol NH 3

How many moles of hydrogen are needed to produce 2.3 moles of ammonia (NH 3 )? How many moles of hydrogen are needed to produce 2.3 moles of ammonia (NH 3 )? Mole to Mole Calculations Mole to Mole Calculations N 2 +3H 2 →2NH 3 Now according to the equation: When 3 mole of hydrogen reacts with nitrogen, 2 mole of ammonia are produced. Conversion Factor: 2 mol NH 3 = 3 mol H mol NH 3 x 2 mol NH 3 3 mol H 2 =3.45 mol H 2

How many grams of hydrogen are needed to react with 23.2 grams nitrogen? How many grams of hydrogen are needed to react with 23.2 grams nitrogen? Mass to Mass Calculations Mass to Mass Calculations N 2 +3H 2 →2NH 3 Now according to the equation: 3 moles of hydrogen are required for every 1 mole of nitrogen. Conversion Factor: 1 mol N 2 = 3 mol H g N 2 x g N 2 1 mol N 2 =5.01 g H 2 x 3 mol H 2 1 mol N 2 x 1 mol H g H 2

How many liters of hydrogen are needed to react with 54.3 liters of nitrogen? How many liters of hydrogen are needed to react with 54.3 liters of nitrogen? Volume to Volume Calculations Volume to Volume Calculations N 2 +3H 2 →2NH 3 Now according to the equation: 3 moles of hydrogen are required for every 1 mole of nitrogen. Conversion Factor: 1 mol N 2 = 3 mol H L N 2 x 22.4 L N 2 1 mol N 2 =162.9 L H 2 x 3 mol H 2 1 mol N 2 x 1 mol H L H 2

How many grams of hydrogen are needed to react with 54.3 liters of nitrogen? How many grams of hydrogen are needed to react with 54.3 liters of nitrogen? Mass to Volume Calculations Mass to Volume Calculations N 2 +3H 2 →2NH 3 Now according to the equation: 3 moles of hydrogen are required for every 1 mole of nitrogen. Conversion Factor: 1 mol N 2 = 3 mol H L N 2 x 22.4 L N 2 1 mol N 2 = g H 2 x 3 mol H 2 1 mol N 2 x 1 mol H g H 2

How many grams of hydrogen are needed to react with 54.3 liters of nitrogen? 54.3 L N 2 x 22.4 L N 2 1 mol N 2 = g H 2 x 3 mol H 2 1 mol N 2 x 1 mol H g H 2 How many liters of hydrogen are needed to react with 54.3 liters of nitrogen? 54.3 L N 2 x 22.4 L N 2 1 mol N 2 = L H 2 x 3 mol H 2 1 mol N 2 x 1 mol H L H 2 Difference between the last 2 equations

- The Stoichiometry Map is the same as the Mole Map with 1 step added in the middle. Stoichiometry Map