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Stoichiometric Calculations

Stoichiometry – Ch. 11. Stoichiometric Calculations. Background on things you NEED to know how to do:. Name/write correct chemical formula Write chemical equations Balance chemical equations Predict Products Mole/mass conversions. Stoichiometry.

colette-hammond
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Stoichiometric Calculations

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  1. Stoichiometry – Ch. 11 Stoichiometric Calculations

  2. Background on things you NEED to know how to do: • Name/write correct chemical formula • Write chemical equations • Balance chemical equations • Predict Products • Mole/mass conversions

  3. Stoichiometry • Stoichiometry uses ratios to determine relative amounts of reactants or products. • For example If you were to make a bicycle, you would need one frame and two tires. • 1 frame + 2 tires  1 bicycle • If I had 74 tires, what is the most # of bicycles I could make? 74 tires 1 bicycle = 37 bicycles 2 tires

  4. 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

  5. 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 • can be used to determine expected amounts of products given amounts of reactants. 2 Mg + O2 2 MgO

  6. 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.

  7. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS NUMBER OF PARTICLES MOLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

  8. Mole – Mole Conversions • The first type of problems we encounter will go between moles and moles. For this we need to use mole ratios. • Ex: Write and balance the reaction between lead (II) nitrate and potasium iodide. Pb(NO3)2 + 2KI  2 KNO3 + PbI2 Mole ratio of potasium iodide to lead (II) iodide: 2 moles KI 1 mole PbI2

  9. Mole to Mole Problems • How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3 2KCl + 3O2 ? mol 9 mol 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3

  10. Mole to Mass • We can also convert from moles to mass, and mass to moles • For Example: • 4 Al + 3 O2 2Al2O3 • If you know how many grams of Al you start with, we can write a flow chart to show how to calculate the # of moles of oxygen need to fully react with the Al. • g Al  moles Al  moles of oxygen

  11. Mass to Moles: 4 Al + 3 O2 2Al2O3 • If the reaction starts with .84 moles of aluminum, how many grams of aluminum oxide are produced? 2 mol Al2O3 .84 mol Al 101.9 grams Al2O3 1 mol Al2O3 4 mol Al = 42.8 grams Al2O3 • 0.92 g of Aluminum oxide are produced from the reaction. How much aluminum was used up? 26.9 g Al 4 mol Al 1 mol Al2O3 .92 g Al2O3 1 mol Al 2 mol Al2O3 101.9 g Al2O3 = .49 grams Al

  12. Mass to Mass • How many grams of silver will be formed from 12.0 g copper reacting with silver nitrate? Cu + 2AgNO3 2Ag + Cu(NO3)2 12.0 g ? g 12.0 g Cu 1 mol Cu 63.55 g Cu 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag = 40.7 g Ag

  13. Stoichiometry Problems – Mole/Mass • In photosynthesis, carbon dioxide and water react to form glucose, C6H12O6 and oxygen gas. ___CO2 + ___H2O  ___C6H12O6 + ___O2 If 15.6 grams of carbon dioxide react, how many moles of glucose will be produced? How many grams of carbon dioxide must react to produce 0.25 moles of glucose? 6 6 6 15.6 g CO2 1 mol CO2 1 mol C2 H12O6 = 0.0591 mol C2H12O6 6 mol CO2 44.01 g CO2 0.25 mol C2 H12O6 44.01 g CO2 6 mol CO2 = 66 g CO2 1 mol C2 H12O6 1 mol CO2

  14. Stoichiometry with Gases • If the pressure and temperature are constant, the ratio of moles in the balanced equation is the ratio of liters in an all gas reaction.

  15. Molar Volume at STP StandardTemperature&Pressure 0°C and 1 atm At STP 1 mol of a gas=22.4 L

  16. Molar Volume • Hydrogen and chlorine gas react to produce hydrochloric acid. If 7.00 L of hydrogen gas react, how many liters of HCl gas are formed? H2 (g) + Cl2 (g) 2 HCl(g) 2.0 L HCl 7.00 L H2 = 14.0 L HCl *only in all gas reactions! 1.0 L H2

  17. Molar Volume • In the following reaction, if 17 g of Mg react, how many L of H2 forms? Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g) 17.0 g Mg 1 mol Mg 24.31 g Mg 1 mol H2 1 mol Mg 22.4 L H2 1 mol H2 = 15.7 L H2

  18. Molar Volume Problems • How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3 2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L O2 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3

  19. Molar Volume Problems • How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3 2Ag + Cu(NO3)2 1.5L 0.10M ? g 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu = 4.8 g Cu

  20. Molarity • Molarity is the number of moles of solute dissolved in one liter of solution. • Units are moles per liter or moles of solute per liter of solution. • Molarity abbreviated by a capital M • Molarity = moles of solute liter of solution

  21. Molarity Problems • As an example, suppose we dissolve 23 g of ammonium chloride (NH4Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution? 1 mole NH4Cl 23 g NH4Cl = .43 mol NH4Cl 53.5 g NH4Cl 145 mL 1 L = .145 L 1000 mL .43 mol NH4Cl = 2.97 M NH4Cl .145 L

  22. Molarity Problems • Now, suppose we have a beaker with 175 mL of a 5.5 M HCl solution. How many moles of HCl is in this beaker? 5.5 mol HCl = .96 mol HCl 175 mL 1 L 1000 mL 1 L • Suppose you had 70 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution? 1.2 mol NaCl 70 g NaCl 1 mol NaCl 2.0 L = 1.2 mol NaCl 58.44 g NaCl = 0.6 M NaCl

  23. Volume of Solutions • A 10.% HCl solution (soln) means: 10 g HCl (pure) 100 g HCl soln • A solution with a density of 1.5 g/mL means: 1.5 g soln 1 mL soln

  24. Impure Substances • To say a substance is 75% NaCl by mass means: 75 g NaCl (pure) 100 g of NaCl solution or 100 g of NaCl solution 75 g of NaCL (pure) • Or, Iron ore that is 15% iron by mass means: 15 g Fe 100 g ore or 100 g of ore 15 g of Fe

  25. Energy & Stoichiometry

  26. Exothermic and Endothermic HEAT HEAT • Exothermic process – heat is released into the surroundings • Exo = Exit • Endothermic Process – heat is absorbed from the surroundings • Endo = Into

  27. Thermochemical Equations • In a thermochemical equation, the energy of change for the reaction can be written as either a reactant or a product • Enthalpy: the heat content of a system at constant pressure (ΔH) • Endothermic (positive ΔH) 2NaHCO3 + 129kJ Na2CO3 + H2O + CO2 • Exothermic (negative ΔH) CaO + H2O Ca(OH)2 + 65.2kJ

  28. Exo 4 3 2 1 mol Fe 10.00g Fe 1652 kJ =73.97 kJ of heat 4 mol Fe 55.85g Fe Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= -1652 kJ Fe(s) + O2(g)→ Fe2O3(s) + 1652 kJ How much heat is evolved when 10.00g of Iron is reacted with excess oxygen?

  29. Endo 1 mol NaHCO3 50.0 g NaHCO3 129 kJ 83.9 g NaHCO3 2 mol NaHCO3 =38.4 kJ of heat Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = + 129 kJ: 2 NaHCO3 + 129kJ → Na2CO3(s) + H2O + CO2 How much heat is required to break down 50.0g of sodium bicarbonate?

  30. Exo 100 g CaO 1 mol CaO 65.2 kJ 56.07 g CaO 1 mol CaO =116 kJ of heat Write the thermochemical equation for the synthesis of calcium oxide and water with a ΔH= - 65.2 kJ: CaO + H2O → Ca(OH)2 + 65.2kJ How much energy is released when 100 g of calcium oxide reacts?

  31. Endo 420 kJ 1 mol O2 31.98 g O2 =218 g of O2 61.5 kJ 1 mol O2 Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= + 61.5 kJ: 2 MgO + 61.5 → 2 Mg + O2 How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kJ of Energy?

  32. Stoichiometry – Ch. 11 Stoichiometry in the Real World

  33. 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

  34. Limiting Reactants • Available Ingredients • 24 graham cracker squares • 1 bag of marshmallows • 12 pieces of chocolate • Limiting Reactant • chocolate • Excess Reactants • Marshmallows and graham crackers

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

  36. Limiting Reactant Steps 1. Write the balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: • limiting reactant • amount of product actually possible

  37. Limiting Reactants • 79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed? Zn + 2HCl  ZnCl2 + H2 ? g 79.1 g 68.1 g

  38. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? g 79.1 g 68.1 g 68.1 g HCl 1 mol HCl 36.46 g HCl 1 mol H2 2 mol HCl 2.02 g H2 1 mol H2 = 1.89 g H2

  39. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? g 79.1 g 68.1 g 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 1 mol Zn 2.02 g H2 1 mol H2 = 2.44 g H2

  40. Limiting Reactants left over zinc Zn: 2.44 g H2 HCl: 1.89 g H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 1.89 g H2

  41. Limiting Reactants #2 • 5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed? 2Mg + O2 2MgO 4.00 g 5.42 g ? g

  42. Limiting Reactants #2 2Mg + O2 2MgO ? g 4.00 g 5.42 g 5.42 g Mg 1 mol Mg 24.31 g Mg 2 mol MgO 2 mol Mg 40.31 g MgO 1 mol MgO = 8.99 g MgO

  43. Limiting Reactants #2 2Mg + O2 2MgO ? g 4.00 g 5.42 g 4.00 g O2 1 mol O2 32.00 g O2 2 mol MgO 1 mol O2 40.31 g MgO 1 mol MgO = 10.1 g MgO

  44. A. Limiting Reactants #2 Excess oxygen Mg: 8.99 g MgO O2: 10.1 g MgO Limiting reactant: Mg Excess reactant: O2 Product Formed: 8.99 g MgO

  45. Limiting Reactants • What other information could you find in these problems? • How much of each reactant is used – in grams, liters, moles • How much of excess reactant is left over – in grams, liters, moles

  46. Percent Yield measured in lab calculated on paper

  47. 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 + 2 HCl  2 KCl + H2CO3 1 mol K2CO3 138.2 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl 45.8 g K2CO3 = 49.4 grams KCl

  48. 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. 46.3 grams KCl x 100 49.4 grams KCl % yield = 93.7 %

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