1. Stoichiometric Calculations Stoichiometry – Ch. 8

2. Proportional Relationships b I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

3. Proportional Relationships b Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio b Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

4. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass -moles  grams Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

5. Stoichiometry Problems b How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

6. Stoichiometry Problems If you have 45 kg of isoamyl alcohol and enough acetic acid to react with all of the isoamyl alcohol, what is the maximum number of kg of isoamyl acetate that can be made? C 5 H 11 OH + CH 3 COOH  CH 3 COOC 5 H 11 + H 2 0

7. Stoichiometry Problems Equation balanced? C 5 H 11 OH + CH 3 COOH  CH 3 COOC 5 H 11 + H 2 0 YEA!!!

8. Stoichiometry Problems What do you know? 45 kg of C 5 H 11 OH ? kg of isoamyl acetate Molar mass of C 5 H 11 OH = 88.17g/mol Molar mass of CH 3 COOC 5 H 11 = 130.21 g/mol 1 mol C 5 H 11 OH : 1 mol CH 3 COOC 5 H 11 1000 g = 1 kg

9. Stoichiometry Problems 45 kg  1mol C 5 H 11 OH  1000 g 88.17 g C 5 H 11 OH 1kg = 510.mol C 5 H 11 OH 510.mol C 5 H 11 OH 1 mol CH 3 COOC 5 H 11 = 1 mol C 5 H 11 OH 510. mol CH 3 COOC 5 H 11

10. Stoichiometry Problems 510. mol CH 3 COOC 5 H 11  130.21 g  1kg 1 mol CH 3 COOC 5 H 11 1000g =66.4 kg isoamyl acetate 66 kg isoamyl acetate is the maximum amount that can be produced

11. Stoichiometry Problems b How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g? g

12. Stoichiometry Problems Magnesium burns in oxygen to produce magnesium oxide. How much magnesium will burn in the presence of 189 ml of oxygen? The density of oxygen is 1.429 g/L. 2Mg + O 2 → 2MgO

13. Stoichiometry Problems 189 ml O 2 D = 1.429 g/L 1 mol O 2 = 2 mol Mg MM O 2 = 32.00 g/ mol MM Mg = 24.30 g/ mol

14. Stoichiometry Problems 189 ml O 2 1.429 g 1 L 1 mol O2 1 L 1000 ml 32.00 g O 2 = 8.44 X 10 -3 mol O 2 8.44 X 10-3 mol O2 2 mol Mg 1 mol O 2 = 16.88 X 10 -3 mol Mg 16.88 X 10 -3 mol Mg 24.30 g Mg = 1 mol Mg 0.410 g Mg

15. Limiting Reactants b Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly

16. Limiting Reactants b Limiting Reactant used up in a reaction determines the amount of product b Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

17. Limiting Reactants- Steps 1. Write a balanced equation. 2. Determine # of moles present for each reactant. 3. Mole ratios but this time not for what you are looking for, but for reactants.

18. Limiting Reactants- Steps 4. Use the limiting reactant to solve the problem as usual.

19. Limiting Reactants Example problem page 286 CO + H 2 → CH 3 OH Not balanced! CO+ 2H 2 → CH 3 OH 1 C 1 C 1 O 1 O 4 H 4 H BALANCED!!!!!!!!!!!

20. Limiting Reactants Looking for kg of CH 3 OH produced 152.5 kg CO MM CO = 28.01 g/mol 24.5 kg H 2 MM H 2 = 2.02 g/mol 1mol CO = 2mol H 2 MM CH 3 OH = 32.05g/mol 1mol CO = 1mol CH 3 OH 2mol H 2 = 1mol CH 3 OH

21. Limiting Reactants Determine # of moles present for each reactant. 152.5 kg CO x 1mol CO x 1000g = 28.01 g CO 1 kg 5.444 x 103 mol CO 24.5 kg H 2 x 1mol x 1000 g = 2.02 g H 2 1 kg 1.213 x 10 4 mol H 2

22. Limiting Reactants Mole ratios but this time not for product for reactants 1 mol CO = 2 mol H 2 1.213 x 10 4 mol H 2 x 1mol CO = 2 mol H 2 6.065 x 10 3 mol CO Need 6.065 X 10 3 mol CO Have 5.444 X 10 3 mol CO

23. Limiting Reactants CO is the limiting reactant!!!! Not enough to react with all of the H 2 present. Use the limiting reactant to solve the problem

24. Limiting Reactants 5.444 X 10 3 mol CO  1 mol CH 3 OH  1 mol CO 32.05 g CH 3 OH  1 kg 1 mol CH 3 OH 1000 g = 174.5 kg CH 3 OH

25. Percent Yield calculated on paper measured in lab

26. Percent Yield b When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g

27. Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g Theoretical Yield:

28. Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g