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Stoichiometric Calculations

Stoichiometry – Ch. 8. Stoichiometric Calculations. Proportional Relationships. Ratio of eggs to cookies. I have 5 eggs. How many cookies can I make?. 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

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Stoichiometric Calculations

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  1. Stoichiometry – Ch. 8 Stoichiometric Calculations

  2. Proportional Relationships Ratio of eggs to cookies • I have 5 eggs. How many cookies can I make? 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. 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies

  3. Proportional Relationships • Stoichiometry • mass relationships between substances in a chemical reaction • based on the mole ratio • Mole Ratio • indicated by coefficients in a balanced equation 2 Mg + O2 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 • Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

  5. 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? C5H11OH + CH3COOH  CH3COOC5H11 + H20

  6. Stoichiometry Problems Equation balanced? C5H11OH + CH3COOH  CH3COOC5H11 + H20 YEA!!!

  7. Stoichiometry Problems What do you know? 45 kg of C5H11OH ? kg of isoamyl acetate Molar mass of C5H11OH = 88.17g/mol Molar mass of CH3COOC5H11= 130.21 g/mol 1 mol C5H11OH : 1 mol CH3COOC5H11 1000 g = 1 kg

  8. Stoichiometry Problems 45 kg 1mol C5H11OH1000 g 88.17 g C5H11OH 1kg = 510.mol C5H11OH 510.mol C5H11OH 1 mol CH3COOC5H11 = 1 mol C5H11OH 510. mol CH3COOC5H11

  9. Stoichiometry Problems 510. mol CH3COOC5H11130.21 g1kg 1 mol CH3COOC5H11 1000g =66.4 kg isoamyl acetate 66 kg isoamyl acetate is the maximum amount that can be produced

  10. 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 + O2 → 2MgO

  11. Stoichiometry Problems 189 ml O2 D = 1.429 g/L 1 mol O2 = 2 mol Mg MM O2 = 32.00 g/ mol MM Mg = 24.30 g/ mol

  12. Stoichiometry Problems 189 ml O2 • 1.429 g • 1 L • 1 mol O2 1 L 1000 ml 32.00 g O2 = 8.44 X 10-3 mol O2 8.44 X 10-3 mol O2 • 2 mol Mg 1 mol O2 = 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

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

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

  15. 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.

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

  17. Limiting Reactants Example problem page 286 CO + H2 → CH3OH Not balanced! CO+ 2H2 → CH3OH 1 C 1 C 1 O 1 O 4 H 4 H BALANCED!!!!!!!!!!!

  18. Limiting Reactants Looking for kg of CH3OH produced 152.5 kg CO MM CO = 28.01 g/mol 24.5 kg H2 MM H2 = 2.02 g/mol 1mol CO = 2mol H2 MM CH3OH = 32.05g/mol 1mol CO = 1mol CH3OH 2mol H2 = 1mol CH3OH

  19. 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 H2 x 1mol x 1000 g = 2.02 g H2 1 kg 1.213 x 104 mol H2

  20. Limiting Reactants Mole ratios but this time not for product for reactants 1 mol CO = 2 mol H2 1.213 x 104 mol H2 x 1mol CO = 2 mol H2 6.065 x 103 mol CO Need 6.065 X 103 mol CO Have 5.444 X 103 mol CO

  21. Limiting Reactants CO is the limiting reactant!!!! Not enough to react with all of the H2 present. Use the limiting reactant to solve the problem

  22. Limiting Reactants 5.444 X 103 mol CO 1 mol CH3OH 1 mol CO 32.05 g CH3OH 1 kg 1 mol CH3OH 1000 g = 174.5 kg CH3OH

  23. Percent Yield measured in lab calculated on paper

  24. Percent Yield • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

  25. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: 45.8 g ? g actual: 46.3 g 45.8 g K2CO3 1 mol K2CO3 138.21 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

  26. Percent Yield 46.3 g 49.4 g K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield = 49.4 g KCl 45.8 g 49.4 g actual: 46.3 g  100 = 93.7% % Yield =

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