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Bell Ringer

Bell Ringer. When 6.58 g SO 3 and 1.64 g H 2 O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield?. SO 3. +. H 2 O. H 2 SO 4. 1 mol SO 3. 1 mol H 2 SO 4. 98.09 g H 2 SO 4. 6.58 g SO 3. x. x. x. =. 80.07 g SO 3.

richard-gentry
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Bell Ringer

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  1. Bell Ringer When 6.58 g SO3 and 1.64 g H2O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield? SO3 + H2O H2SO4 1 mol SO3 1 mol H2SO4 98.09 g H2SO4 6.58 g SO3 x x x = 80.07 g SO3 1 mol SO3 1 mol H2SO4 8.06 g H2SO4 1 mol H2O 1 mol H2SO4 98.09 g H2SO4 1.64 g H2O x x x = 18.02 g H2O 1 mol H2O 1 mol H2SO4 8.93 g H2SO4

  2. % Yield = Actual Yield x 100 % Theoretical Yield Bell Ringer When 6.58 g SO3 and 1.64 g H2O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield? SO3 + H2O H2SO4 Info we’ve learned: Theoretical Yield = 8.06 g H2SO4 7.99 g H2SO4 = 99.1 % 8.06 g H2SO4

  3. Limiting Factors & Percent Yield Quiz Good Luck!!

  4. Homework Answers • D • A • C • B • A • C • C • A • B • D • 11. Mass-mass • 12. Mass-volume • 13. Mole-mole • 14. Limiting reactant • 15. Volume-volume • 0.52 mol PBr3 • 0.13 mol Na • 50 g NaClO3 • 23. 5000 g HCl • 4770 g H2O • 27. 98 g AgCl; 120 g AgNO3 • 29. 700 g CO2 ; 500 g O2 • 31. 89.6 L H2 • 33. 234 g ZnSO4 • 35. 7.75 L O2 • 39. 8.06 g H2SO4 ; 99.1 %

  5. Stoichiometry Review Ms. Besal 3/7/2006

  6. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  7. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  8. 2 H2O 2 H2 + O2 Mole-Mole Problems • 1 conversion step • Given: moles “A” • Required: moles “B” • Convert moles “A” to moles “B” using mole ratio. • The mole ratio is used in EVERY STOICHIOMETRY PROBLEM. EVER. I PROMISE. How many moles of water can be formed from 0.5 mol H2? 2 mol H2O 0.5 mol H2 0.5 mol H2O x = 2 mol H2

  9. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  10. 2 H2O 2 H2 + O2 Mass-Mole Problems • 2 conversion steps • Given: mass “A” • Required: moles “B” • Step 1: convert grams “A” to moles “A” using Periodic Table • Step 2: convert moles “A” to moles “B” using mole ratio How many moles of water can be formed from 48.0 g O2? 1 mol O2 2 mol H2O 48.0 g O2 3.00 mol H2O x x = 32.00 g O2 1 mol O2

  11. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  12. 2 H2O 2 H2 + O2 Mass-Mass Problems • 3 conversion steps • Given: mass “A” • Required: mass “B” • Step 1: convert grams “A” to moles “A” using Periodic Table • Step 2: convert moles “A” to moles “B” using mole ratio • Step 3: convert moles “B” to grams “B” using Periodic Table How many grams of water can be formed from 48.0 g O2? 1 mol O2 2 mol H2O 18.02 g H2O 48.0 g O2 54.1 g H2O x x x = 32.00 g O2 1 mol O2 1 mol H2O

  13. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  14. 2 H2O 2 H2 + O2 Mass-Volume Problems • 3 – 4 conversion steps • Given: mass “A” • Required: volume “B” • Step 1: convert grams “A” to moles “A” using Periodic Table • Step 2: convert moles “A” to moles “B” using mole ratio • Step 3: convert moles “B” to liters “B” How many liters of oxygen are necessary to create 48.0 g H2O? 48.0 g H2O 1 mol H2O 1 mol O2 22.4 L O2 29.8 L O2 x x x = 18.02 g H2O 2 mol H2O 1 mol O2

  15. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  16. 2 H2O 2 H2 + O2 Volume-Mass Problems • 3 – 4 conversion steps • Given: volume “A” • Required: mass “B” • Step 1: convert liters “A” to moles “A” • Step 2: convert moles “A” to moles “B” using mole ratio • Step 3: convert moles “B” to grams “B” using Periodic Table How many grams of water are formed by reacting 36.0 L O2? 36.0 L O2 1 mol O2 2 mol H2O 18.02 g H2O 58.7 g H2O = x x x 22.4L O2 1 mol O2 1 mol H2O

  17. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  18. 2 H2O 2 H2 + O2 Volume-Volume Problems • 3 – 5 conversion steps • Given: volume “A” • Required: volume “B” • Step 1: convert liters “A” to moles “A” • Step 2: convert moles “A” to moles “B” using mole ratio • Step 3: convert moles “B” to liters “B” How many liters of H2 are required to react with 5.0 L O2? 5.0 L O2 1 mol O2 2 mol H2 22.4 L H2 10. L H2 x x x = 22.4 L O2 1 mol O2 1 mol H2

  19. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  20. Limiting Reactant Problems • Quantities are given for each reactant. • 2 parallel equations • Solve each equation for product desired and determine limiting reactant. • Use Limiting Reactant to solve for amount or excess reactant used. • Subtract amount excess reactant used from amount given to determine how much is left over.

  21. 2 H2 + O2 2 H2O Limiting Reactant Problems If you start with 10.0 g of O2 and 5.00 g H2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be? 1 mol O2 2 mol H2O 18.02 g H2O 10.0 g O2 x x x = 32.00 g O2 1 mol O2 1 mol H2O LIMITING REACTANT 11.3 g H2O THEORETICAL YIELD 5.00 g H2 1 mol H2 2 mol H2O 18.02 g H2O x x x = EXCESS REACTANT 2.02 g H2 2 mol H2 1 mol H2O 44.06 g H2O

  22. Limiting Reactant Problems 2 H2 + O2 2 H2O If you start with 10.0 g of O2 and 5.00 g H2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be? Info we know so far: Limiting Reactant = O2 Excess Reactant = H2 1 mol O2 2 mol H2 2.02 g H2 10.0 g O2 x x = x 32.00 g O2 1 mol O2 1 mol H2 1.26 g H2 USED 5.00 g H2 – 1.26 g H2 = 3.74 g H2 LEFT OVER

  23. Types of Stoichiometry Problems • Mole-Mole • Mass-Mole • Mass-Mass • Mass-Volume • Volume-Mass • Volume-Volume • Limiting Reactant • Percent Yield

  24. May or may not be given You will be given one of these 2 H2 + O2 2 H2O Percent Yield Problems • Critical Information: • Theoretical Yield • Actual Yield • Percent Yield Determine the actual yield of a reaction between 6.25 g H2 and excess O2 that has a 85% percent yield. 6.25 g H2 1 mol H2 2 mol H2O 18.02 g H2O x x x = 2.02 g H2 2 mol H2 1 mol H2O THEORETICAL YIELD 55.6 g H2O ? 85 % = 100 % x 55.6 g H2O ACTUAL YIELD = 47.3 g H2O

  25. Homework Answers • D • A • C • B • A • C • C • A • B • D • 11. Mass-mass • 12. Mass-volume • 13. Mole-mole • 14. Limiting reactant • 15. Volume-volume • 0.52 mol PBr3 • 0.13 mol Na • 50 g NaClO3 • 23. 5000 g HCl • 4770 g H2O • 27. 98 g AgCl; 120 g AgNO3 • 29. 700 g CO2 ; 500 g O2 • 31. 89.6 L H2 • 33. 234 g ZnSO4 • 35. 7.75 L O2 • 39. 8.06 g H2SO4 ; 99.1 %

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