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Entry Task: March 13 th- 14 th Block 1

Entry Task: March 13 th- 14 th Block 1. Entry task question: Question : 1.5 grams of zinc was added with 0.5 grams of iodine to produce zinc II iodide. Calculate the limited reactant and the amount of excess. You have ~10 minutes. Agenda:. Sign off and Discuss Limited Reactant #1 ws

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Entry Task: March 13 th- 14 th Block 1

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  1. Entry Task: March 13th-14th Block 1 Entry task question: Question: 1.5 grams of zinc was added with 0.5 grams of iodine to produce zinc II iodide. Calculate the limited reactant and the amount of excess. You have ~10 minutes

  2. Agenda: • Sign off and Discuss Limited Reactant #1 ws • Notes on Theoretical and % yield • HW: Limited and yields practice ws

  3. Limited and Excess #1 ws

  4. 25.0 grams of gold III sulfide reacts with 15.0 grams of hydrogen gas to create gold metal and dihydrogenmonosulfide. Which reactant is limiting, which is excess, and how much product is produced? Au2S3+ 3H2 2Au + 3H2S 25.0 g Au2S3 1 moleAu2S3 2 molAu 196.97 g Au 20.1g Au 490.14 g Au2S3 1 mole Au2S3 1 molAu 15.0g H2 1 mole H2 2 mole Au 196.97 g Au 977.13 g Au 3 mole H2 2.0 g H2 1 mole Au

  5. Which reactant is Limited and which reactant is Excess? Au2S3+ 3H2 2Au + 3H2S Limited= Au2S3 25.0 g Au2S3 1 moleAu2S3 2 molAu 196.97 g Au 20.1g Au 490.14 g Au2S3 1 mole Au2S3 1 molAu Excess = H2 15.0g H2 1 mole H2 2 mole Au 196.97 g Au 977.13 g Au 3 mole H2 2.0 g H2 1 mole Au

  6. 25.0 grams of gold III sulfide reacts with 15.0 grams of hydrogen gas to create gold metal and dihydrogenmonosulfide. Which reactant is limiting, which is excess, and how much product is produced? Au2S3+ 3H2 2Au + 3H2S Start with the limited reactant!! Used In reaction 25.0g Au2S3 1 mole Au2S3 3mole H2 2.0 g H2 0.31 g H2 1 mole Au2S3 490.14 g Au2S3 1 mole H2 Given Used In reaction 14.7 g EXCESS 15.0 grams H2 minus 0.31 g H2=

  7. 45.0 g Aluminum hydroxide reacts 15.0 grams of dihydrogenmonosulfurtetraoxide to create aluminum sulfate and water. Which reactant is limiting, which is excess, and how much product is produced? 2Al(OH)3+ 3 H2SO4 Al2(SO4)3 + 6 H2O 45.0 gAl(OH)3 1 mole Al(OH)3 18.0g H2O 6 molH2O 31.1g H2O 77.98 g Al(OH)3 2 mole Al(OH)3 1 mol H2O 15.0 g H2SO4 1 molH2SO4 6 molH2O 18.0 g H2O 5.51 g H2O 3 molH2SO4 97.996 g H2SO4 1 mol H2O

  8. Which reactant is Limited and which reactant is Excess? 2Al(OH)3+ 3 H2SO4 Al2(SO4)3 + 6 H2O Excess = Al(OH)3 45.0 gAl(OH)3 1 mole Al(OH)3 18.0g H2O 6 molH2O 31.1g H2O 77.98 g Al(OH)3 2 mole Al(OH)3 1 mol H2O Limited= H2SO4 15.0 g H2SO4 1 molH2SO4 6 molH2O 18.0 g H2O 5.51 g H2O 3 molH2SO4 97.996 g H2SO4 1 mol H2O

  9. 45.0 g Aluminum hydroxide reacts 15.0 grams of dihydrogenmonosulfurtetraoxide to create aluminum sulfate and water. Which reactant is limiting, which is excess, and how much product is produced? 2Al(OH)3+ 3 H2SO4 Al2(SO4)3 + 6 H2O Start with the limited reactant!! Used In reaction 15.0 g H2SO4 1 molH2SO4 2 molAl(OH)3 77.98 g Al(OH)3 7.98 g Al(OH)3 3 molH2SO4 1 molAl(OH)3 97.996 g H2SO4 Given Used In reaction 37.0 g EXCESS 45.0 gramsAl(OH)3minus 7.98 g Al(OH)3=

  10. I can… • Calculate the theoretical yield of a chemical reaction from data. • Determine the percent yield for a chemical reaction

  11. Theoretical yield • Is the maximum amount of product that can be produced from the given amount of reactants. • Calculated by stoichiometry

  12. Actual yield • Is the maximum amount of product ACTUALLY produced from the given amount of reactants in a lab experiment. • FOR REAL

  13. Percent Yield • Percent yield of a product is the ratio of the actual amount of product to the theoretical amount of product expressed as a percent.

  14. Percent Yield Percent yield = Actual (for real) Theoretical (calculated) X 100 =

  15. NaBr + KCl NaCl + KBr • This is a double replacement reaction

  16. NaBr + KCl NaCl + KBr • If we mixed 25 grams of sodium bromide with a large amount of potassium chloride, what would our theoretical yield of sodium chloride be? So a gram to gram stoich set up. 25g NaBr 1 mole NaCl 1 mole NaBr 58.44 g NaCl 14 g NaCl 1 mole NaCl 1 mole NaBr 102.98g NaBr 1461 102.98

  17. NaBr + KCl NaCl + KBr If our actual yield from this reaction were 18 grams of sodium chloride, what would our percent yield be for this reaction? Actual (for real) Theoretical (calculated) Percent yield = 18 g NaCl (actual) 128% X 100 = 14 g NaCl (calculated)

  18. NaBr + KCl NaCl + KBr 5. Is your answer in question 2 reasonable? If so, explain how this could happen in a real lab. If not, explain what is wrong with it and discuss possible reasons you might have gotten this answer in a real lab. 18 g NaCl (actual) 128% X 100 = 14 g NaCl (calculated) There is NO WAY to have over 100% percent yield. It has to be experimental error- procedural or math error.

  19. NaBr + KCl NaCl + KBr 6. What are some factors that might cause percent yield to really be greater than 100%? If there were added substances, dirty glassware or something on the scale when massing- experimental procedure mess up

  20. NaBr + KCl NaCl + KBr 6. What are some factors that might cause percent yield to really be less than 100%? If the reactants were not finished reacting, or not enough reactants to move the reaction forward for more product.

  21. Limited reactants % yields practice

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