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Stoichiometry

Stoichiometry. Calculations based on chemical reactions. Stoichiometry. Stoichiometry is a Greek word that means using chemical reactions to calculate the amount of reactants needed and the amount of products formed.

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Stoichiometry

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  1. Stoichiometry Calculations based on chemical reactions

  2. Stoichiometry • Stoichiometry is a Greek word that means using chemical reactions to calculate the amount of reactants needed and the amount of products formed. • Amounts are typically calculated in grams (or kg), but there are other ways to specify the quantities of matter involved in a reaction.

  3. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen.

  4. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H2O(l ) 

  5. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H2O(l )  NaOH(aq) + H2(g)

  6. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H2O(l )  NaOH(aq) + H2(g) • The equation is not yet balanced. Hydrogens come in twos on the left, and three hydrogens are on the right side of the equation.

  7. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H2O(l )  NaOH(aq) + H2(g) • Try a “2” in front of the water.

  8. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + 2 H2O(l )  NaOH(aq) + H2(g) • We now have two O atoms on the left, so we need to put a 2 before NaOH.

  9. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) • The two sodium atoms on the right require that we put a 2 in front of Na on the left.

  10. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) • The two sodium atoms on the right require that we put a 2 in front of Na on the left. The equation is now balanced.

  11. Balancing Chemical Equations- Problem • Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) Left SideRight Side Na- 2 Na- 2 H- 4 H- 4 O- 2 O- 2

  12. Chemical Equations 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) The balanced chemical equation can be interpreted in a variety of ways. It could say that 2 atoms of sodium react with 2 molecules of water to produce 2 molecules of sodium hydroxide and a molecule of hydrogen.

  13. Chemical Equations 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) The balanced chemical equation can be interpreted in a variety of ways. It could say that 200 atoms of sodium react with 200 molecules of water to produce 200 molecules of sodium hydroxide and 100 molecules of hydrogen.

  14. Chemical Equations 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) The balanced chemical equation can be interpreted in a variety of ways. • It is usually interpreted as 2 moles of sodium will react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen. • The balanced equation tells us nothing about the masses of reactants or products.

  15. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) • How many grams of sodium are needed to produce 50.0g of hydrogen? • Since the balanced chemical equation gives us information about moles of reactants and products, we cannot answer the question until we convert the mass of hydrogen into moles.

  16. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) • How many grams of sodium are needed to produce 50.0g of hydrogen? 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g

  17. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g • Although the question doesn’t state it, you can assume enough water is present for complete reaction. • We can map out the problem: g H2  moles H2 moles Na  grams Na

  18. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g • We can map out the problem: g H2  moles H2 moles Na  grams Na We use the molar mass of H2 to go from grams of H2 to moles of H2.

  19. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g • We can map out the problem: g H2  moles H2 moles Na  grams Na molar mass of H2

  20. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g • We can map out the problem: g H2  moles H2 moles Na  grams Na molar mass of H2 • We use the coefficients from the balanced equation to go from moles of H2 to moles of Na.

  21. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g • We can map out the problem: g H2  moles H2 moles Na  grams Na molar mass of H2 coefficients

  22. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g • We can map out the problem: g H2  moles H2 moles Na  grams Na molar mass of H2 coefficients • We use the molar mass of Na to go from moles of Na to grams of Na.

  23. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g • We can map out the problem: g H2  moles H2 moles Na  grams Na molar mass of H2 coefficients molar mass of Na

  24. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) ? grams 50.0 g g H2  moles H2 moles Na  grams Na molar mass of H2 coefficients molar mass of Na (50.0 g H2) (1 mol H2)(2 moles Na)( 22.99 g Na) (2.02 g H2) (1 mol H2) (1 mol Na)

  25. Stoichiometry Problem 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) (50.0 g H2) (1 mol H2)(2 moles Na)( 22.99 g Na) = (2.02 g H2) (1 mol H2) (1 mol Na) = 1,138 grams Na = 1.14 x 103 g Na = 1.14 kg Na

  26. Limiting Reagent Problems • Sometimes you are given quantities of more than one reactant, and asked to calculate the amount of product formed. The quantities of reactants might be such that both react completely, or one might react completely, and the other(s) might be in excess. These are called limiting reagent problems, since the quantity of one of the reacts will limit the amount of product that can be formed.

  27. Limiting Reagent - Problem • Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 1. First write the formulas for reactants and products. Al + Br2 AlBr3

  28. Limiting Reagent - Problem • Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 2. Now balance the equation by adding coefficients. 2 Al +3 Br22 AlBr3

  29. Limiting Reagent - Problem • Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 2 Al +3 Br22 AlBr3 The theoretical yield is the maximum amount of product that can be formed, given the amount of reactants. It is usually expressed in grams.

  30. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given: 50.0g 500.g ? grams • There are several ways to solve this problem. One method is to solve the problem twice. Once, assuming that all of the aluminum reacts, the other assuming that all of the bromine reacts. The correct answer is whichever assumption provides the smallest amount of product.

  31. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given: 50.0g 500.g ? grams • The problem can be mapped: Al: grams Al  moles Al moles AlBr3  g AlBr3 molar mass Al coefficients molar mass AlBr3

  32. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given:50.0g 500.g ? grams • The problem can be mapped: Al: grams Al  moles Al moles AlBr3  g AlBr3 molar mass Al coefficients molar mass AlBr3 (50.0 g Al) ( 1 mol Al )( 2 mol AlBr3)(266.7 g AlBr3) (26.98 g Al) ( 2 mol Al) (mol AlBr3)

  33. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given:50.0g 500.g ? grams Al: (50.0 g Al) ( 1 mol Al )( 2 mol AlBr3)(266.7 g AlBr3) (26.98 g Al)( 2 mol Al) (mol AlBr3) = 494. g AlBr3 (if all of the Al reacts)

  34. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given:50.0g 500.g ? grams The calculation is repeated for Br2. g Br2 moles Br2  moles AlBr3  g AlBr3 molar mass Br2 coefficients molar mass AlBr3

  35. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given:50.0g 500.g ? grams The calculation is repeated for Br2. g Br2 moles Br2  moles AlBr3  g AlBr3 molar mass Br2 coefficients molar mass AlBr3 (500. g Br2) (1 mol Br2) (2 moles AlBr3)(266.7 g AlBr3) (159.8 g Br2) (3 moles Br2) (1 mol AlBr3) = 556. grams of AlBr3

  36. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given: 50.0g 500.g ? grams • Summary: We have enough Al to produce 494. g AlBr3 We have enough Br2 to produce 556. grams of AlBr3 The theoretical yield is 494. grams of AlBr3.

  37. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given: 50.0g 500.g 494. g Summary: • All of the Al reacts, so Al is limiting. • Bromine is in excess. • Additional questions: 1. How much bromine is left over? 2. If 418 grams of AlBr3 is obtained, what is the % yield?

  38. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given: 50.0g 500.g 494. g 1. How much bromine is left over? Since all 50.0 g of the Al reacts, the product must contain 494.g -50.g = 444. g of bromine. Therefore, 500.g- 444.g = 56. g of Br2 are left over.

  39. Limiting Reagent - Problem 2 Al +3 Br22 AlBr3 Given: 50.0g 500.g 494. g • If 418 grams of AlBr3 is obtained, what is the % yield? The percent yield is (actual yield) (100%) (theoretical yield) % yield = (418 g) (100%) = 84.6 % (494g)

  40. Concentration of Solutions There are many ways to express the concentration of a solution. The one most commonly used for solution stoichiometry is molarity (M). Molarity = M = moles of solute liter of solution

  41. Concentration of Solutions Solutions are often used to determine how much of a certain substance is present in a sample. As a result, it is important to know how to perform calculations using solution stoichiometry.

  42. Concentration of Solutions • 40.0 grams of sodium hydroxide are dissolved in enough water to make 250. mL of solution. Calculate the molarity of the solution. grams NaOH moles NaOH M NaOH

  43. Concentration of Solutions • 40.0 grams of sodium hydroxide are dissolved in enough water to make 250. mL of solution. Calculate the molarity of the solution. grams NaOH moles NaOH M NaOH molar mass of NaOH M=moles/volume

  44. Concentration of Solutions • What volume of the previous solution contains 0.125 moles of NaOH?

  45. Concentration of Solutions • How would you make 100. mL of a 0.075M solution of NaCl? moles solute molarity = _________________ (liter of solution)

  46. Concentration of Solutions • How would you make 100. mL of a 0.075M solution of NaCl? moles solute = M(V) moles NaCl = (0.075mol/L) (.100L) moles NaCl = 0.0075 mol mass NaCl = 0.0075 mol (58.5 g/mol) mass NaCl = 0.439 g = 0.44 g NaCl

  47. The Preparation of a Standard Aqueous Solution

  48. Dilution Most solutions are purchased in highly concentrated form. They often need to be diluted before they can be used. When a solution is diluted, the moles of solute remain unchanged. Dilution involves the addition of more solvent (water), while the amount of solute stays the same.

  49. Dilution Dilution problems can be easily solved using the following relationships. mols concentrated solute = mols diluted solute Mconc(Vconc) = Mdil(Vdil)

  50. Dilution • How many milliliters of 6.00M H2SO4 are needed to make 30.0 mL of 1.25M H2SO4? Mconc (Vconc) = Mdil(Vdil) ________

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