1. 12.1 STOICHIOMETRY Deals with amounts of reactants used & products formed.

2. What is Stoichiometry? The study of the relationship between amounts of reactants used and products formed in a chemical reaction Based on the Law of Conservation of Mass and Matter –Matter is neither created nor destroyed –Mass of reactants equals the mass of the products

3. MOLE-MASS RELATIONSHIPS Balanced equation: 4Fe (s) + 3O 2 (g)  2Fe 2 O 3 (s) Interpret: 4 atoms 3 molecules 2 formula units (Particles) Mole ratio: 4 moles 3 moles 2 moles (Coefficients) Note: Particles of ionic compounds are called “formula units” Calculate the mass of each reactant and product by multiplying the number of moles by the molar mass

4. SHOW MASS IS CONSERVED 4Fe (s) + 3O 2 (g)  2Fe 2 O 3 (s) Mass reactants: 4 mol Fe (55.8g Fe) = 223.2 g Fe (1 mole Fe) 3 mol O 2 (32.00 g O 2 ) = + 96.0 g O 2 (1 mol O 2 ) ____________ 319.2 g Total

5. MASS OF PRODUCTS: 4Fe (s) + 3O 2 (g)  2Fe 2 O 3 (s) 2 mol Fe 2 O 3 (159.6 g Fe 2 O 3 ) = 319.2 g (1 mol Fe 2 O 3 ) Equals the mass of the reactants (319.2g) Law of Conservation of Matter

6. PROBLEM Interpret in terms of particles, moles, and mass. Show that mass is conserved: (Hint: look at coefficients for particles & moles) 4 Al + 3O 2  2 Al 2 O 3 Particles: Moles: Mass: Conserved?

7. SOLUTION: 4Al + 3O 2  2 Al 2 O 3 Particles: 4 molecules 3 molecule 2 molecule Moles: 4 mole 3 mole 2 moles Mass: 4 (27.0 g) + 3 (26.0 g) = 2 (102.0 g) Conserved? 204.0 g = 204.0 gYES! Law of Conservation of Matter shown.

8. MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced chemical equation There are six mole ratios for the following: Ex. 4 Al(s) + 3 O 2 (g)  2 Al 2 O 3 (s) Note: 4 moles 3 moles 2 moles 4 mol Al and 3mol O 2 3 mol O 2 4 mol Al

9. MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced chemical equation 4 mol Al and 2 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al 3 mol O 2 and 2 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 All stoichiometry calculations begin with a balanced equation and mole ratios!!

10. PROBLEM - FIND MOLE RATIOS FOR: 2 NH 3  N 2 + 3 H 2 2 mol 1 mol 3 mol Ans. 2 mol NH 3 or 1 mol N 2 1 mol N 2 2 mol NH 3 2 mol NH 3 or 3 mol H 2 3 mol H 2 2 mol NH 3

11. Ans. 2 NH 3  N 2 + 3 H 2 1 mole N 2 or 3 mole H 2 3 mole H 2 1 mole N 2

12. 12.2 STOICHIOMETRIC CALCULATIONS There are 3 Basic Stoichiometry Calculations

13. 1. Mole to Mole Conversions A piece of magnesium burns in the presence of oxygen forming magnesium oxide (MgO). How many moles of oxygen are needed to produce 12 moles of magnesium oxide?  Step 1: Write a balanced equation 2 Mg (s)+O 2 (g)  2 MgO (s)

14. Write mole ratios Choose the correct mole ratio needed for this problem

15. Mole to Mole Conversion cont’d  Multiply the known number of moles of MgO by the mole ratio  6 mols of oxygen is needed to produce 12 mols of magnesium oxide

16. 2. Mole to Mass Conversions The following reaction occurs in plants undergoing photosynthesis CO 2 (g) + H 2 O(l)  C 6 H 12 O 6 (s)+ O 2 (g) How many grams of glucose (C 6 H 12 O 6 ) are produced when 24.0 mols CO 2 reacts in excess water?

17. Mole to Mass Conversion cont’d  Write a balanced equation 6 CO 2 (g) +6 H 2 O(l)  C 6 H 12 O 6 (s)+ 6 O 2 (g)  Use mole ratios to determine the number of moles of glucose produced by the given amount of carbon dioxide

18. Mole to Mass Conversion cont’d  Multiply by the molar mass  721 g glucose is produced from 24.0 moles carbon dioxide

19. 3. Mass to Mass Problems The only new step is first step: Convert grams of given substance to moles! Mass given  Moles giv  Moles desired  Mass des ÷ molar mass given X mole ratio X molar mass desired 3 step problem!

20. Label: 25.0 g ? g NH 4 NO 3  N 2 O + 2H 2 O Mole Ratio: 1 mol 1 mol2 mol Ammonium nitrate (NH 4 NO 3 ) produces N 2 O gas and H 2 O when it decomposes. Determine the mass of water produced from the decomposition of 25.0 g of NH 4 NO 3. 1) Find moles NH 4 NO 3 (molar mass): 80.04 g/mol Use the inverse of the molar mass to convert grams of NH 4 NO 3 to moles of NH 4 NO 3

21. Label: 25.0 g ? g NH 4 NO 3  N 2 O + 2H 2 O Mole ratio: 1 mol 1 mol 2 mol Determine the mole ratio of mol H 2 O to mol NH 4 NO 3 from the chemical equation. The desired substance is the numerator. Multiply mol NH 4 NO 3 by the mole ratio. Calculate the mass of H 2 O using the molar mass.

22. Label: 5.6g ? g PROBLEM: N 2 + 3H 2  2 NH 3 If 5.6 g nitrogen reacts completely with hydrogen, what mass of ammonia is formed? 1) Find moles of nitrogen (given) – molar mass: 5.6 g N 2 |_____ mass N 2 = 2 x 14.0 = 28.0 g/mol 1 | g N 2 5.6 mol N 2 |_1mol N 2 1 | 28.0 g N 2

23. Label: 5.6 g ? g N 2 + 3H 2  2 NH 3 Mole Ratio: 1mol 3mol 2mol 2) Find moles ammonia (desired) - mole ratio: 5.6 g N 2 |1mol N 2 |_2 mol NH 3 ___ 1 | 28 g N 2 | 1 mol N 2 3) Find grams ammonia (desired) – molar mass: NH 3 = 14 + 3 = 17 g/mole 5.6 g N 2 |1mol N 2 |_2 mol NH 3 | 17.0 g NH 3 1 | 28g N 2 | 1 mol N 2 | 1 mol NH 3 given ÷ MM given mole ratio x MM desired = 6.8 g NH 3 Mass H 2 = 6.8 g – 5.6 g = 1.2 g

24. LIMITING REACTANTS

25. Limiting and Excess Reactants When a chemical reaction occurs, the reactants are not always present in the exact ratio indicated by the balanced equation. What usually happens is that a chemical reaction will run until the reactant that is in short supply is used up. Which reactant will be used up first?

26. What is a Limiting Reactant? A limiting reactant is the reactant that limits (stops) a reaction and determines the amount of product An excess reactant is any reactant that is left over after the reaction stops (all reactants except the limiting reactant)

27. IN ORDER FOR CHEMICAL REACTION TO OCCUR, YOU MUST HAVE A COMPLETE SET OF REACTANTS (REAGENTS): You don’t always have the exact amounts. Ex. Let’s make some McBurgers!!! Ingredients: – 2 buns; 1 beef patty; 1cheese slice; 1 tomato slice; 1 lettuce leaf; 3 pickles

28. YOU HAVE AVAILABLE: 6 buns 3 burger patties 5 cheese slices 6 tomato slices 5 lettuce leafs 6 pickles RECIPE CALLS FOR: 2 buns 1 burger patty 1 cheese slice 1 tomato slice 1 lettuce leaf 3 pickles How many McBurgers can you make?

29. Based on the individual ingredients you have available: You have enough: Buns for(3) Burger Patties for(3) Cheese Slices for(5) Tomato Slices for(6) Lettuce Leafs for(5) Pickles for(2)

30. The Limiting Reactant (LR) for the McBurger making process in the _______! This is the ingredient we ran out of first and were unable to continue with the McBurger making process We were only able to make 2 McBurger from the 6 pickles we had available to use –(Remember each McBurger requires 3 pickles) pickles!

31. Excess reactants (XR) are the ingredients not used in the McBurger making process: 2 buns 1 burger patty 3 cheese slices 4 tomato slices 3 lettuce leafs Note: You made 2 McBurgers

32. N 2 + 3H 2  2 NH 3 Using the above equation you are given 3.0 mols N 2 and 5.0 mols H 2 Determine the limiting reactant Determine the excess reactant Determine the amount of NH 3 produced PRACTICE PROBLEM:

33. Draw 2 dimensional analysis problems. Smaller product is your answer. (Always use L.R. to find answer!) 3.0 mol N 2 x 2 mol NH 3 = 6.0 mol NH 3 1 mol N 2 5.0 mol H 2 x 2 mol NH 3 = 3.3 mol NH 3 3 mol H 2. Label: 3.0mol 5.0 mol ?mol N 2 + 3H 2  2 NH 3 M.R. 1 mol 3 mol 2 mol

34. ANSWER: 3.3 mol ammonia formed. Hydrogen (H 2 ) is L.R. Nitrogen (N 2 ) is the XS.

35. IF A PAPER BURNS IN A ROOM: What is the limiting reactant? What is in excess? What would happen if the paper burned in a closed jar? LR? XS? ++ +++

36. 4.0mol 7.0mol PROBLEM: 2 Al + 3 Cl 2  2 AlCl 3 Given 4.0 mol Al and 7.0 mol Cl 2, what is the maximum amount of aluminum chloride formed? Step 1: 2 set ups: 4.0 mol Al _____________ mol Al 7.0 mol Cl 2 ________ mol Cl 2

37. Label: 4.0mol 7.0 mol ? Mol 2 Al + 3 Cl 2  2 AlCl 3 4.0 mol Al 2 mol AlCl 3 = 4.0 mol AlCl 3 2 mol Al 7.0 mol Cl 2 2 mol AlCl 3 = 4.7 mol AlCl 3 3 mol Cl 2 Which is the limiting reactant? *Al is L.R. Ans. 4.0 mol AlCl 3 *Cl 2 is XS

38. THE END