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Chapter 9 – STOICHIOMETRY

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Presentation on theme: "Chapter 9 – STOICHIOMETRY"— Presentation transcript:

1 Chapter 9 – STOICHIOMETRY
The MATH of a CHEMICAL REACTION

2 9.1 – Reaction Stoichiometry
-Ratio of compounds in a reaction -YOU NEED A BALANCED CHEMICAL REACTION!! Examples:       Al +       O2 ®       Al2O3 What this means: 4 MOLES Al react with 3 MOLES O2 to form 2 MOLES Al2O3

3 9.1 – Reaction Stoichiometry
Example:       C2H6 +       O2 ®       CO2 +       H2O 2 moles C2H6 reacts with 7 moles O2 to form 4 mole CO2 and 6 moles H2O

4 9.2 – Stoichiometry Calculations
-You need to be sure your units are canceling -CAN ONLY SWITCH FROM ONE SUBSTANCE TO ANOTHER BY MOLES!!

5 9.2 – Stoichiometry Calculations
Types of Calculations: 1. mole of one substance to mole of another Ratio from Balanced Equation Moles A Moles B

6 Example: 4 NH3 + 3 O2 -> 2 N2 + 6 H2O 9.4 mole NH3 = ? mole N2
Rewrite the information given Multiply Draw a fraction line DENOMINATOR IS THE SAME UNIT AS GIVEN Set up numerator Plug in known fraction numbers Solve Work:

7 More Practice:

8 9.2 – Stoichiometry Calculations
Types of Calculations: 2. mole of one substance to gram of another Ratio from Balanced Equation Molar Mass from Periodic Table Moles A Moles B Grams B

9 Example: 4 NH3 + 3 O2 -> 2 N2 + 6 H2O 4.35 mole NH3 = ? g H2O Work:

10 More Practice:

11 9.2 – Stoichiometry Calculations
Types of Calculations: 3. gram of one substance to mole of another Ratio from Balanced Equation Molar Mass from Periodic Table Grams A Moles A Moles B

12 Example: 4 NH3 + 3 O2 -> 2 N2 + 6 H2O 19.3 g NH3 = ? mol O2 Work:

13 More Practice:

14 9.2 – Stoichiometry Calculations
Types of Calculations: 4. gram of one substance to gram of another Ratio from Balanced Equation Molar Mass from Periodic Table Molar Mass from Periodic Table Grams A Moles A Moles B Grams B

15 Example: 4 NH3 + 3 O2 -> 2 N2 + 6 H2O 18.6 g O2 = ? g N2 Work:

16 More Practice:

17 More Practice:

18 Limiting Reactants -Need a balanced equation!
-You will be given information about more than one starting material -It is nearly impossible to add the perfect amount of both reactants -Limiting Reactant = The reactant which will run out first -Excess Reactant = The reactant which you have “more than enough of”

19 Limiting Reactants Examples: + + + + 10 18 13 1 1 1 2 1 2 1 1
-Making a Bicycle: 1 1 1 + 2 + + 1 + 2 1 frame tires handlebars 1 bike Stockroom Inventory: 10 18 13

20 Limiting Reactants + 10 18 13 Examples:
-Making a Bicycle: How many bikes can we make??? 1 2 + frame tires handlebars bike 10 18 13 How many bikes COULD we make from each component? 13 10 9

21 9 Limiting Reactants How many maximum could be made??
*We picked the smallest value of the “could make” answers!! Which was the limiting reactant?? tires What are the excess reactants?? frames and handlebars

22 Now a Chemistry Example:
Limiting Reactants Now a Chemistry Example: 3Fe + 4H2O -> Fe3O4 + 4H2 Starting materials: 24 g Fe and 20 g H2O How many moles / How many grams Fe3O4 will be produced? Work:

23 Limiting Reactants Fe2O3 + 3CO -> 2Fe + 3CO g Fe2O3 and 21.5 g CO How many grams of iron will be formed?

24 Percent Yield Compares the amount actually received in lab to the amount that was expected. Theoretical Yield = The amount expected. The number you calculate with pen and paper. Actual Yield = The amount actually obtained % Yield =

25

26 Percent Yield Example: 3Fe + 4H2O -> Fe3O4 + 4H2
Starting materials: 14.2 g Fe and 27.8 g H2O After the experiment is complete: 17.3 g Fe3O4 are actually obtained

27 Percent Yield Work: 3Fe + 4H2O -> Fe3O4 + 4H2 14.2 g Fe 27.8 g H2O


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