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Stoichiometry

Stoichiometry. This word sounds scary, but it’s really just a big word for the relationships between reactants and products in a chemical reaction. “based on law of conservation of mass” mass reactants = mass products. Ratio of eggs to cookies. Proportional Relationships. 2 1/4 c. flour

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Stoichiometry

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  1. Stoichiometry This word sounds scary, but it’s really just a big word for the relationships between reactants and products in a chemical reaction. “based on law of conservation of mass” mass reactants = mass products

  2. Ratio of eggs to cookies 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. Assuming I have plenty of the other ingredients, how many cookies can I make? 5 dozen cookies 2 eggs 5 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 + 1O2 2MgO

  4. Mole Ratios 2 Mg + 1 O2 2MgO • Relationships between coefficients are used as conversion factors in stoichiometry! • A mole ratio is a ratio between the numbers of moles of any 2 substances.

  5. Mole Ratios 2 Mg + 1 O2 2MgO • What are the mole ratios in this equation? 2 molMg 2 molMg 2 mol MgO 1 mol O2 1 mol O2 1 mol O2 2 mol Mg 2 mol MgO 2 mol MgO 2 mol MgO 2 mol Mg 1 mol O2 Total of 6 mole ratios!

  6. Mole Ratios 2KClO3→2KCl + 3O2 • What are the mole ratios in this equation? 2 molKClO3 2 molKClO3 3 mol O2 2 mol KCl 2 mol KCl 2 mol KCl 2 mol KClO3 3 mol O2 3 mol O2 3 mol O2 2 mol KClO3 2 mol KCl Total of 6 mole ratios!

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

  8. Mole ratios • 1CO2 + 2LiOH → 1Li2CO3 + 1H2O • What is the ratio of • CO2 molecules to LiOH molecules? • CO2 molecules to Li2CO3 molecules • Li2CO3 molecules to LiOH molecules

  9. Mole Map 1 Moles 1 step (1 conversion factor) 1 step (1 conversion factor) Molar Mass 6.02 × 1023 Mass (g) Particles 2 steps (2 conversion factors) Atoms, Ions, Molecules, or Formula Units

  10. Chemical Equations and the Mole 1P4O10+6H2O4H3PO4 1 molecule of P4O10 reacts with 6 molecules of H2O to give 4 molecules of H3PO4 1 mole of P4O10 reacts with 6 moles of H2O to give 4 moles of H3PO4

  11. Mole to Mole (1 step1 or you need to use 1 conversion factor)

  12. Mole to Mole conversions -use mole ratios from last notes Formula used: Moles of known Moles of unknown Moles of known = moles of unknown

  13. Mole to Mole conversions • Steps in Solving • Determine moles given in problem • Find mole ratio showing unknown on top and known on bottom • Fill into formula above and calculate moles of unknown

  14. 3 mol H2 2 mol NH3 5.46 mol N2 5.46 mol N2 1 mol N2 1 mol N2 Stoichiometry and Chemical Equations The coefficients in a chemical equation not only describe the number of atoms/molecules being reacted/produced, they also describe the number of moles of atoms/molecules being reacted/produced. N2 + 3H2 2NH3 Suppose 5.46 mol N2 completely react in this fashion: = 16.4 mol H2 are needed to react = 10.9 mol NH3 must be produced

  15. 6 mol HCl 15.0 mol H2S 3 mol H2S Stoichiometry Practice Bismuth(III) chloride will react with hydrogen sulfide to form bismuth(III) sulfide and hydrochloric acid. Write the balanced equation for this reaction, then calculate how many moles of acid would be formed if 15.0 mol of hydrogen sulfide react. 2 BiCl3 + 3 H2S 1 Bi2S3 + 6 HCl = 30.0 mol HCl

  16. Practice ProblemsMole to Mole • P. 375 # 11-12 • Stoichiometry WS #1 #1-3

  17. Mole to Mass&Mass to Mole(2 steps or you need to use 2 conversion factors)

  18. Mole to Mass conversions • -use if you know # moles of one thing and you want to know mass of another thing • Steps in solving • Convert moles of known to moles of unknown. • Convert moles of unknown to grams of unknown

  19. Stoichiometry Practice Sodium nitride can be formed by reacting sodium metal with nitrogen gas. Write the balanced equation for this reaction, then calculate how many moles of sodium nitride can be produced from 25.0 g of sodium. 6 Na + 1 N2 2 Na3N 25.0 g Na 1 mol Na 2 mol Na3N 23.0 g Na 6 mol Na = 0.362 mol Na3N

  20. Stoichiometry Practice When magnesium burns in air, it combines with oxygen to form magnesium oxide according to the following equation: • What mass (in grams) of magnesium oxide is produced from 2.00 mol of magnesium? 2Mg + 1O2 2MgO 16.5 mol Mg 2 mol MgO 40.3g MgO 2 mol Mg 1 mol MgO = 665 g MgO

  21. Practice ProblemsMole to MassMass to Mole • P. 376 #13-14 • Stoichiometry WS#1 #4-7

  22. Mass to Mass(3 steps or you need to use 3 conversion factors)

  23. Mass to Mass conversions • Use if you know the mass of 1 thing and want to find mass of another • Steps in solving • Convert mass of known to moles of known • Do mole-mole conversion • Convert moles of unknown to mass of unknown

  24. 2KClO3→ 2KCl + 3O2Calculate the mass of O2 produced if 2.5 g of KClO3 is completely decomposed when burned 3 mol O2 32 g O2 2.5 g KClO3 1 mol KClO3 122.6g KClO3 2 mol KClO3 1 mol O2 = .98g O2 Mole ratio

  25. 1 NH4NO3→ 1 N2O + 2 H2ODetermine the mass of water produced from the decomposition of 25.0g of solid ammonium nitrate 2 mol H2O 18.02 g H2O 25.0 g NH4NO3 1 mol NH4NO3 80.04 g NH4NO3 1 mol NH4NO3 1 mol H2O = 11.26g H2O Mole ratio

  26. Summary of Conversions Mole to Mole: 1 step Mass to Mole: 2 steps Mass to Mass: 3 steps What setup do all conversions require? Mole:mole ratio

  27. Practice ProblemsMass to Mass • P. 377 #15-16 • Stoichiometry WS#1 #8-10

  28. Limiting Reactants

  29. How many bicycles can be assembled from the parts shown? From eight wheels four bikes can be constructed. From three pedal assemblies three bikes can be constructed. From four frames four bikes can be constructed. The limiting part is the number of pedal assemblies.

  30. Likewise, in a chemical reaction, it is possible for one reactant to be used up before the other(s). A limiting reactant is a substance that is completely used up in a chemical reaction and causes the reaction to stop. Therefore, it is the limiting reactant that also determines how much product can actually be formed.

  31. An excess reactant are the left-over reactants that are in abundance in a reaction. The bicycle wheels and frames were in excess in our example

  32. Limiting Reactant Bikes 8 wheels 1 bike = 4 bikes Wheels and Frames and excess reactants 2 wheels 1 bike 4 frames = 4 bikes 1 frame 3 Pedal A 1 bike = 3 bikes LR or Limiting Reactant 1 Pedal A

  33. The limiting reactant does 3 things: • LR “limits” or determines the number of reactions that can take place because when the LR runs out, the reaction stops! • FEWEST REACTIONS = LR • The LR determines how much product can be made. • The LR determines how much of the other reactant is left over.

  34. The excess reactant (XS) is the reactant that is left over unused after the LR runs out and the reaction stops. MOST REACTIONS = XS

  35. A candle is lit and placed inside a jar and the lid is closed securely. What are the limiting and excess reactants and why? LR is Oxygen (O2) because once the oxygen runs out, the combustion (burning) of the candle will stop. XS is candle wax because once the candle stops burning, no more of the candle wax can be used so some of the candle wax will be left over. Example # 2

  36. Example # 3 • You are making 1 batch of chocolate chip cookies. Here’s what you find in your pantry: 1 egg 1 bag chocolate chips 2 tbsp butter

  37. Example # 3 • Which ingredients are the Limiting and Excess Reactants?

  38. Remember . . . • Fewest Rxns = Limiting Reactant • Most Rxns = Excess Reactant

  39. Example # 3 • Which ingredients are the Limiting and Excess Reactants?

  40. Example # 3 • Which ingredients are the Limiting and Excess Reactants? XS LR

  41. Percent Yield

  42. Theoretical, Actual & Percent Yields Theoretical yield: the amount of product you would get if the reaction occurs with complete efficiency. Actual yield: what you “actually” get when you perform the reaction or experiment. Percent yield the actual yield divided by the theoretical yield multiplied by 100. % Yield = (Actual/Theoretical) x 100

  43. measured in lab calculated on paper Percent Yield

  44. 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

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

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

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