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Quantities in Chemical Reactions

Quantities in Chemical Reactions. Stoichiometry (Greek): Stoicheion = “element” Metron = “to measure”. Why are we learning this?. Stoichiometry can allow you to: determine amount of product from known quantity of reactant determine amount of starting material needed

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Quantities in Chemical Reactions

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  1. Quantities in Chemical Reactions

  2. Stoichiometry (Greek): Stoicheion= “element” Metron = “to measure”

  3. Why are we learning this? • Stoichiometry can allow you to: • determine amount of product from known • quantity of reactant • determine amount of starting material needed • to yield desired quantity of product • identify reaction efficiency (and make amendments • in reaction conditions) • determine purity of starting materials • and more

  4. What is required to perform stoichiometric analysis? Balanced chemical equations!

  5. What is required to perform stoichiometric analysis? Balanced chemical equations! Like a recipe, reaction components must interact with one another in fixed proportions!

  6. What information is conveyed In balanced chemical equation? • Numbers of particles • Numbers of moles • Mass of reaction components • Volumes (gasses)

  7. Example: Production of gaseous ammonia by Haber process † Total values are equal before and after the reaction. * Total values may or may not be equal before and after the reaction.

  8. Overall Strategy for Solving Stoichiometric Problems MASS MASS ÷ M (molar mass) x M (molar mass) Mole Ratio MOLES MOLES ÷ NA(Avogadro’s constant) x NA(Avogadro’s constant) MOLECULES MOLECULES mole ratio – obtained from the balanced chemical equation

  9. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S)

  10. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2?

  11. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2? Answer: 2 Al atoms (directly from balanced equation)

  12. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2? Answer: 2 Al atoms (directly from balanced equation) B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2?

  13. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2? Answer: 2 Al atoms (directly from balanced equation) B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2? Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3 2 atoms Al

  14. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2? Answer: 2 Al atoms (directly from balanced equation) B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2? Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3 2 atoms Al C) How many Cl2 molecules are needed to react with 1.834 x 1024 atoms of Al?

  15. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2? Answer: 2 Al atoms (directly from balanced equation) B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2? Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3 2 atoms Al C) How many Cl2 molecules are needed to react with 1.834 x 1024 atoms of Al? Answer: 1.834 x 1024atoms Al x 3 molecules Cl2 = 2.751 x 1024 molecules Cl2 2 atoms Al

  16. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2? Answer: 2 Al atoms (directly from balanced equation) B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2? Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3 2 atoms Al C) How many Cl2 molecules are needed to react with 1.834 x 1024 atoms of Al? Answer: 1.834 x 1024atoms Al x 3 molecules Cl2 = 2.751 x 1024 molecules Cl2 2 atoms Al D) How many molecules of AlCl3 are formed when 2.751 x 1024 molecules of Cl2 react with sufficient Al?

  17. Relating particles in balanced equations Example (Modelled after textbook, page 237, #2) 2 Al(s) + 3 Cl2(g)→ 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2? Answer: 2 Al atoms (directly from balanced equation) B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2? Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3 2 atoms Al C) How many Cl2 molecules are needed to react with 1.834 x 1024 atoms of Al? Answer: 1.834 x 1024atoms Al x 3 molecules Cl2 = 2.751 x 1024 molecules Cl2 2 atoms Al D) How many molecules of AlCl3 are formed when 2.751 x 1024 molecules of Cl2 react with sufficient Al? Answer: 2.751 x 1024molecules Cl2 x 2 molecules AlCl3= 1.834 x 1024 molecules AlCl3 3 molecules Cl2

  18. Equation coefficients represent Particles and moles In previous example, we saw that coefficients for reactants and products in balanced equation represented the number of particles (atoms or molecules) participating directly in chemical reaction. But…

  19. Equation coefficients represent Particles and moles Multiply ratio of coefficients by NA : e.g. (2 atoms Al(s) : 3 molecules Cl2(g) : 2 molecules AlCl3(s)) x NA 2 NA Al(s) : 3 NACl2(g) : 2 NA AlCl3(s) = 2 mol Al(s) : 3 mol Cl2(g) : 2 mol AlCl3(s)

  20. Equation coefficients represent Particles and moles 2 NA Al(s) : 3 NACl2(g) : 2 NA AlCl3(s) = 2 mol Al(s) : 3 mol Cl2(g) : 2 mol AlCl3(s) Products Total 2 mol Reactants Total 5 mol Note: Because atoms are rearranged in different proportions, there is no law of conservation of moles.

  21. Equation coefficients represent Particles and moles By analogy, imagine leaving 3 moles of puddles of water on the floor, and then using one mole of paper towel sheets to clean it up. 1 mol wet paper towel sheets 1 mol dry paper towel sheets 3 mol puddles of water Reactants total 4 moles Products total 1 mole Think about product as 1 paper towel . 3 H2O puddles. Where have we seen such a formulation recently?

  22. MASS TO MASS STOICHIOMETRIC CALCULATIONS Most likely, situations will involve being given the mass (or moles, or particles) of one participant in a reaction, and being tasked with finding the mass of another reaction species. Use knowledge of the molar mass of species present in the reaction to interconvert between mass and moles, using the molar ratio of species from the balanced chemical equation to identify the number of moles of the desired chemical entity.

  23. Example #1: (Mass of one reactant to another) Find the mass of solid Cadmium required to fully react with 0.45 grams Nickel (III) oxide hydroxide in a rechargeable nickel-cadmium battery. The overall reaction is: 2 NiO(OH)(s) + 2 H2O(l) + Cd(s)→ 2 Ni(OH)2(s) + Cd(OH)2(s)

  24. Example #1: (Mass of one reactant to another) Find the mass of solid Cadmium required to fully react with 0.45 grams Nickel (III) oxide hydroxide in a rechargeable nickel-cadmium battery. The overall reaction is: 2 NiO(OH)(s) + 2 H2O(l) + Cd(s)→ 2 Ni(OH)2(s) + Cd(OH)2(s) 0.45 g NiO(OH) x 1 molNiO(OH) = 0.0049073 molNiO(OH) 91.70 g NiO(OH) Mass → Moles

  25. Example #1: (Mass of one reactant to another) Find the mass of solid Cadmium required to fully react with 0.45 grams Nickel (III) oxide hydroxide in a rechargeable nickel-cadmium battery. The overall reaction is: 2 NiO(OH)(s) + 2 H2O(l) + Cd(s)→ 2 Ni(OH)2(s) + Cd(OH)2(s) 0.45 g NiO(OH) x 1 molNiO(OH) = 0.0049073 molNiO(OH) 91.70 g NiO(OH) Mass → Moles 0.0049073 molNiO(OH) x 1 mol Cd = 0.0024537 mol Cd 2 molNiO(OH) Moles→ Moles

  26. Example #1: (Mass of one reactant to another) Find the mass of solid Cadmium required to fully react with 0.450 grams Nickel (III) oxide hydroxide in a rechargeable nickel-cadmium battery. The overall reaction is: 2 NiO(OH)(s) + 2 H2O(l) + Cd(s)→ 2 Ni(OH)2(s) + Cd(OH)2(s) 0.450 g NiO(OH) x 1 molNiO(OH) = 0.0049073 molNiO(OH) 91.70 g NiO(OH) Mass → Moles 0.0049073 molNiO(OH) x 1 mol Cd = 0.0024537 mol Cd 2 molNiO(OH) Moles→ Moles 0.0024537 mol Cd x 112.41 g Cd = 0.2758 g Cd(s) = 0.276 g Cd(s) 1 mol Cd Moles → Mass

  27. Example #2: (Mass of reactant to product) (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) +4 H2O(g) Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

  28. Example #2: (Mass of reactant to product) (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) +4 H2O(g) Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams. 1.75 g (NH4)2Cr2O7 x 1 mol(NH4)2Cr2O7 = 0.0069417 mol(NH4)2Cr2O7 252.10 g (NH4)2Cr2O7 Mass → Moles

  29. Example #2: (Mass of reactant to product) (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) +4 H2O(g) Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams. 1.75 g (NH4)2Cr2O7 x 1 mol(NH4)2Cr2O7 = 0.0069417 mol(NH4)2Cr2O7 252.10 g (NH4)2Cr2O7 Mass → Moles • 0.0069417 mol(NH4)2Cr2O7 x 1 mol Cr2O3 = 0.0069417 molCr2O3 • 1 mol(NH4)2Cr2O7 Moles→ Moles

  30. Example #2: (Mass of reactant to product) (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) +4 H2O(g) Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams. 1.75 g (NH4)2Cr2O7 x 1 mol(NH4)2Cr2O7 = 0.0069417 mol(NH4)2Cr2O7 252.10 g (NH4)2Cr2O7 Mass → Moles • 0.0069417 mol(NH4)2Cr2O7 x 1 mol Cr2O3 = 0.0069417 molCr2O3 • 1 mol(NH4)2Cr2O7 Moles→ Moles 0.0069417 molCr2O3 x 152 g Cr2O3 = 1.055 g Cr2O3(s) = 1.06 g Cr2O3(s) 1 molCr2O3 Moles → Mass

  31. Example #2: (Mass of reactant to product) (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) +4 H2O(g) Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams. 1.75 g (NH4)2Cr2O7 x 1 mol(NH4)2Cr2O7 = 0.0069417 mol(NH4)2Cr2O7 252.10 g (NH4)2Cr2O7 Mass → Moles • b)0.0069417 mol(NH4)2Cr2O7 x 1 mol N2 = 0.0069417 mol N2 • 1 mol(NH4)2Cr2O7 Moles→ Moles

  32. Example #2: (Mass of reactant to product) (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) +4 H2O(g) Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams. 1.75 g (NH4)2Cr2O7 x 1 mol(NH4)2Cr2O7 = 0.0069417 mol(NH4)2Cr2O7 252.10 g (NH4)2Cr2O7 Mass → Moles 0.0069417 mol(NH4)2Cr2O7 x 1 mol N2 = 0.0069417 mol N2 1 mol(NH4)2Cr2O7 Moles→ Moles 0.0069417 molN2 x 28.02 g N2 = 0.1945 g N2(g) = 0.195 g N2(g) 1 mol N2 Moles → Mass

  33. Example #2: (Mass of reactant to product) (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) +4 H2O(g) Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams. 1.75 g (NH4)2Cr2O7 x 1 mol(NH4)2Cr2O7 = 0.0069417 mol(NH4)2Cr2O7 252.10 g (NH4)2Cr2O7 Mass → Moles 0.0069417 mol(NH4)2Cr2O7 x 4 mol H2O = 0.0277668 mol H2O 1 mol(NH4)2Cr2O7 Moles→ Moles

  34. Example #2: (Mass of reactant to product) (NH4)2Cr2O7(s)→ Cr2O3(s) + N2(g) +4 H2O(g) Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams. 1.75 g (NH4)2Cr2O7 x 1 mol(NH4)2Cr2O7 = 0.0069417 mol(NH4)2Cr2O7 252.10 g (NH4)2Cr2O7 Mass → Moles 0.0069417 mol(NH4)2Cr2O7 x 4 mol H2O = 0.0277668 mol H2O 1 mol(NH4)2Cr2O7 Moles→ Moles 0.0277668 molH2O x 18.02 g H2O = 0.5004 g H2O(g) = 0.500 g H2O(g) 1 mol H2O Moles → Mass

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