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Stoichiometry

Stoichiometry. 9.1 Calculating Quantities in Reactions. Determine mole ratios from a balanced chemical equation Explain why mole ratios are central to solving stoichiometry problems Solve stoichiometry problems involving: Mass using molar mass Volume using denisty

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Stoichiometry

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  1. Stoichiometry

  2. 9.1 Calculating Quantities in Reactions • Determine mole ratios from a balanced chemical equation • Explain why mole ratios are central to solving stoichiometry problems • Solve stoichiometry problems involving: • Mass using molar mass • Volume using denisty • Particles using Avogadro’s number

  3. Ham Sandwich How many sandwiches could you make from 24 slices of bread? How many slices of • lettuce would you need? • tomato? • ham? • cheese? This process models the calculations in this chapter.

  4. Equations are like recipes. 2C8H18 + 25O2→ 16CO2 + 18H2O • Coefficients can be read as ratios of particles or of moles. If 2 mols of C8H18 reacts completely how many moles of CO2 would be produced?

  5. Stoichoimetry= • the proportional relationship between 2 or more substances during a chemical reaction • quantitative analysis of the outcomes of a reaction • Predict amount of product you couldmake from starting amounts of reactant

  6. The Mole Ratio is the Key • In stoichiometry problems, the unit that bridges the gap between one substance and another is the mole. • You can use the coefficients in conversion factors called mole ratios

  7. Sample Problem A N2 + 3H2→ 2NH3 • 3 mols of hydrogen are needed to prepare 2 moles of ammonia. 3 mol H2 = 2 mol NH3 PROBLEM: How many moles of hydrogen are needed to prepare 312 moles of ammonia?

  8. 312 mol NH3 x 3 mol H2 = ? mol H2 2 mol NH3 = 468 mol H2 needed

  9. Practice Prob. P. 304 • Calculate the amounts requested if 1.34 mol H2O2 completely react according to the following equation 2H2O2→ 2H2O + O2 • mols of oxygen formed Given: 1.34 mol H2O2 reacts 2 mol H2O2 = 1 mol O2 1.34 mol H2O2 x 1 mol O2 = 0.670 mol O2 produced 2 mol H2O2

  10. Practice Prob. P. 304 • Calculate the amounts requested if 1.34 mol H2O2 completely react according to the following equation 2H2O2→ 2H2O + O2 b. mols of water formed Given: 1.34 mol H2O2 reacts 2 mol H2O2 = 2 mol H2O 1.34 mol H2O2 x 2 mol H2O= 1.34 mol H2O produced 2 mol H2O2

  11. 2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation Fe2O3 + 2Al → 2Fe + Al2O3 Thermite Reaction • mols of aluminum needed Given: 3.30 mol Fe2O3 reacts 1 mol Fe2O3 = 2 mol Al needed 3.30 mol Fe2O3 x 2 mol Al= 6.60 mol Al needed 1 mol Fe2O3

  12. 2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation Fe2O3 + 2Al → 2Fe + Al2O3 b. mols of iron formed Given: 3.30 mol Fe2O3 reacts 1 mol Fe2O3 = 2 mol Fe formed 3.30 mol Fe2O3 x 2 mol Fe= 6.60 mol Fe formed 1 mol Fe2O3

  13. 2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation Fe2O3 + 2Al → 2Fe + Al2O3 c. mols of aluminum oxide formed Given: 3.30 mol Fe2O3 reacts 1 mol Fe2O3 = 1 mol Al2O3 formed 3.30 mol Fe2O3 x 1 mol Al2O3= 3.30 mol Al2O3 formed 1 mol Fe2O3

  14. Application/Homework Worksheet “Stoichiometry: Mole-Mole Problems” – front only

  15. Stoichiometry: Mass-Mass Problems • You must convert to moles using molar mass of known • Then use mole ratio to find moles unknown • Convert back to mass using molar mass of unknown

  16. Solving Mass-Mass Problems Mass Unknown 1 mol ___ grams Periodic Table Mol unknown Mol known Balanced Chemical Equation ___ grams 1 mol Periodic Table

  17. Sample Problem B p. 307 What mass of NH3 can be made from 1221 g H2 and excess N2? N2 + 3H2→ 2 NH3 1221 g H2 X 1 mol H2 x 2 mol NH3x 17.04g NH3 2.02 g 3 mol H2 1 mol NH3 GRAMS MOLAR MOLAR MOLAR KNOWN MASS RATIO MASS KNOWN UNKNOWN = 6867 g NH3 made

  18. Practice p. 307#1-4 Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3 1. How many grams Al needed to completely react with 135 grams Fe2O3 135 g Fe2O3 X 1 mol Fe2O3 x 2 mol Al x 26.98 g Al 159.7 g Fe2O3 1 mol Fe2O3 1 mol Al GRAMS MOLAR MOLAR MOLAR KNOWN MASS RATIO MASS KNOWN UNKNOWN = 45.6 g Al needed

  19. Practice p. 307#1-4 Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3 2. How many grams Al2O3 can form when 23.6g Al react with excess Fe2O3? 23.6 g Al X 1 mol Alx 1 mol Al2O3x 101.96 g Al2O3 26.98 g Al 2 mol Al 1 mol Al2O3 = 44.6 g Al2O3 formed

  20. Practice p. 307#1-4 Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3 3. How many grams of Fe2O3 react with excess Al to make 475 g Fe? 475 g Fe X 1 mol Fex 1 mol Fe2O3x 159.7 g Fe2O3 55.85 g Fe 2 mol Fe 1 mol Fe2O3 = 679 g Fe2O3 react

  21. Practice p. 307#1-4 Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3 4. How many grams of Fe will form when 97.6 g Al2O3 form? 97.6 g Al2O3 X 1 mol Al2O3 x 2 mol Fex 55.85 g Fe 101.96 g Al2O3 1 mol Al2O3 1 mol Fe = 107 g Fe formed

  22. Homework Worksheet “Stoichiometry: Mass-Mass Problems” – back of mol-mol worksheet Lab: Mass Relations in Chemical Reactions (Baking soda & HCl reaction)

  23. Stoichiometry:Volume-Volume Problems • Liquid amounts often measured in volumes. You might use density and molar mass for these problems p. 308 Toolkit • Density unit g/mL means ____g = 1 mL • Gases 22.41 L = 1 mol of any gas • Basics the same: Change to moles, use mole ratio, and then change to desired units.

  24. Solving Volume-Volume Problems Volume Unknown Density Volume Known Density Mass Unknown

  25. Sample Problem C p. 309 What volume of H3PO4 forms when 56 mL POCl3 completely react? (density of POCl3 = 1.67 g/mL; density of H3PO4 = 1.83 g/mL) POCl3(l) + 3H2O(l) → H3PO4 (l) + 3HCl (g) 56mL POCl3 x 1.67 g POCl3 x 1 mol POCl3 x 1 mol H3PO4 1 mL153.32g 1 mol POCl3 X 98.00 g H3PO4 x 1 mL H3PO4 = 33 mL H3PO4 1 mol 1.83 g

  26. Practice p. 309 # 1-4 C5H12 (l) → C5H8(l) + 2H2(g) Densities: C5H12 = 0.620 g/mL C5H8 = 0.681 g/mL H2 = 0.0899 g/L = 0.0899 g/1000mL = 8.99 x 10-5 g/mL Calculate Molar Masses: C5H12 = 72.17 g/mol C5H8 = 68.13 g/mol H2 = 2.02 g/mol

  27. How many mL of C5H8can be made from 366mL C5H12? 366mL C5H12 x 0.620 g x 1 mol C5H12x 1 molC5H8 1 mL72.17 g 1 mol C5H12 X 68.13 g x 1 mL 1 mol C5H8 0.681 g = 315 mL C5H8

  28. 2. How many liters of H2 can form when 4.53 x 103 mL C5H8 form? 4.53 x 103 mLC5H8 x 0.681 g x 1 mol C5H8x 2 mol H2 1 mL 68.13 g 1 mol C5H8 X 2.02 g x 1 L 1 mol H2 0.0899 g = 2030 L H2

  29. 3. How many mL of C5H12 are needed to make 97.3 mL of C5H8? 97.3mLC5H8 x 0.681 g x 1 mol C5H8x 1 mol C5H12 1 mL 68.13 g 1 mol C5H8 X 72.17 g x 1 mL 1 mol C5H120.620 g = 113 mLC5H12

  30. 4. How many milliliters of H2 can be made from 1.98 x 103 mL C5H12? 1.98 x 103 mLC5H12 x 0.620 g x 1 mol C5H12x 2 molH2 1 mL72.17 g 1 mol C5H12 X 2.02 g x 1 mL 1 mol H2 0.0000899 g = 7.64 x 105mL H2

  31. Homework • WS “Stoichiometry: Volume-Volume Problems” front only • WS “Stoichiometry: MixedProblems” back – classwork 9.1 QUIZ will be on __________________

  32. 9.2 Limiting Reactants and Percentage Yield In this section you will: • Identify the limiting reactant for a reaction and use it to calculate theoretical yield. • Perform calculations involving percentage yield.

  33. Limiting Reactant • To drive a car you need gasoline and oxygen from the air. • When the gas runs out, you can’t go any farther even though there is still plenty of oxygen. • The gasoline limits the distance you can travel because it runs out first. Page 312/313 Figures 3&4 Which of the starting supplies limited the number of mums that could be made? Which items where in excess/left over?

  34. Limiting reactant = substance that controls the quantity of product that can form in a chemical reaction; runs out first Excess reactant = substance that is not used up completely in a reaction. Identify the Limiting Reactant from Lab Data • Calculate the amount of product that each could form. • Whichever reactant would produce the least amount of product is the limiting reactant.

  35. Theoretical yield = maximum amount of product that can be made if everything about the reaction works out perfectly; determined by the limiting reactant. • Whenever a problem gives you quantities of 2 or more reactants, you must • Determine the limiting reactant (makes less product) • Use the limiting reactant quantity to determine the theoretical yield.

  36. Sample E p. 314 Identify the limiting reactant and the theoretical yield of phosphorous acid, H3PO3 if 225g of PCl3 is mixed with 123 g of H2O. PCl3 + 3H2O → H3PO3 + 3HCl • Determine how many g of H3PO4 that each reactant could make. • Need to calculate molar masses to use as conversion factors.

  37. 225g PCl3x 1mol PCl3 X 1mol H3PO4 x 82.00g H3PO4 137.32g 1 mol PCl3 1 mol = 134 g H3PO4 123g H2O x 1mol PCl3 X 1mol H3PO4 x 82.00g H3PO4 18.02g 3 mol H2O 1 mol = 187 g H3PO4 • PCl3 is the limiting reactant because 134 g is less than 187 g. • The theoretical yield is 134 grams of H3PO3.

  38. Practice p. 314 #1-3PCl3 + 3H2O → H3PO3 + 3HClIdentify the limiting reactant and the theoretical yield (in grams) of HCl for each pair • 3.00 mol PCl3 and 3.00 mol H2O 3.00 mol PCl3 x 3 mol HCl = 9.00 mol HCl 1 mol PCl3 3.00 mol H2Ox 3 mol HCl = 3.00 mol HCl 3 mol H2O H2O is limiting reactant. 3.00 mol HCl x 36.51 g = 109 grams HCl 1 mol theoretical yield (made)

  39. Practice p. 314 #1-3PCl3 + 3H2O → H3PO3 + 3HClIdentify the limiting reactant and the theoretical yield (in grams) of HCl for each pair • 75.0 g PCl3 and 75.0 g H2O 75.0g PCl3 x 1 mol PCl3x 3 mol HCl = 1.65 mol HCl 136.5 g 1 mol PCl3 75.0 g H2O x 1 mol H2O x 3 mol HCl = 4.16 mol HCl 18.02 3 mol H2O PCl3 is limiting reactant. 1.65 mol HClx 36.51 g = 60.2 grams HCl 1 mol theoretical yield (made)

  40. Practice p. 314 #1-3PCl3 + 3H2O → H3PO3 + 3HClIdentify the limiting reactant and the theoretical yield (in grams) of HCl for each pair 3. 1.00 mol PCl3 and 50.0 g H2O 1.00 mol PCl3 x 3 mol HCl = 3.00 mol HCl 1 mol PCl3 50.0 g H2O x 1 mol H2O x 3 mol HCl = 2.77 mol H3PO3 18.02 3 mol H2O H2O is limiting reactant. 2.77 mol HClx 36.51 g = 101 grams HCl 1 mol theoretical yield (made)

  41. Limiting Reactants & Industry • The most expensive chemicals are chosen as the limiting reactants • Less expense reactants can be used in excess to ensure all of the expensive chemicals are completely used up (none wasted).

  42. Actual Yield and Percentage Yield • Although equations tell you what should happen, they cannot always tell you what will happen in real life/in the lab. • Some reactions do not make all of the product predicted by the theoretical yield. ACTUAL YIELD = the mass of product actually formed, measured in lab, often less than expected (theoretical).

  43. Examples of things that could reduce yield: • Incomplete distillation/purification needed to separate product from a mixture. • Side reactions that can use up reactants without making desired products

  44. PERCENTAGE YIELD = ratio relating the actual to the theoretical yield; describes how efficient the reaction was. Percentage Yield = actual x 100 theoretical SHOULD ALWAYS BE LESS THAN 100%. If not, you switched the actual & theoretical in the formula!!!

  45. Sample Problem F p. 317 N2 + 3 H2→ 2NH3 Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form. 14.0g N2 x 1 mol N2x 2 mol NH3x 17.04 g NH3= 17.0g NH3 28.02 g 1 mol N2 1 mol NH3 9 .0g H2 x 1 mol N2x 2 mol NH3x 17.04 g NH3= 51 g NH3 2.02 g 3 mol H2 1 mol NH3 N2 makes the smaller amount and is the limiting reactant.

  46. Sample Problem F p. 317 N2 + 3 H2→ 2NH3 Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form. • N2 makes the smaller amount and is the limiting reactant. • The theoretical amount made is 17.0g NH3. • The actual amount made is 16.1 g NH3 PERCENTAGE YIELD = 16.1 g x 100 = 94.7% 17.0g

  47. Practice p. 317 #1 N2 + 3 H2→ 2NH3 1. Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 , 3.15g H2 and actual is 14.5 g NH3. 14.0g N2 x 1 mol N2x 2 mol NH3x 17.04 g NH3= 17.0g NH3 28.02 g 1 mol N2 1 mol NH3 3.15g H2 x 1 mol N2x 2 mol NH3x 17.04 g NH3= 17.7 g NH3 2.02 g 3 mol H2 1 mol NH3 N2 makes the smaller amount and is the limiting reactant.

  48. N2 + 3 H2→ 2NH3 Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form. • N2 makes the smaller amount and is the limiting reactant. • The theoretical amount made is 17.0g NH3. • The actual amount made is 14.5 g NH3 PERCENTAGE YIELD = 14.5 g x 100 = 85.3% 17.0g

  49. Homework Section 9.2 Review p. 319 # 1,3,4,5,6,8,10,12,14 Quiz on 9.2 will be on __________________.

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