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

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. Au 2 S 3 + 3H 2  2Au + 3H 2 S 25.0 grams of gold III sulfide reacts with 15.0 grams of hydrogen gas to create gold metal and dihydrogen monosulfide. Which reactant is limiting, which is excess, and how much product is produced? 25.0 g Au 2 S 3 20.1 g Au 490.14 g Au 2 S 3 1 mole Au 2 S 3 196.97 g Au 1 mol Au 1 mole Au 2 S 3 2 mol Au 15.0g H 2 977.13 g Au 2.0 g H 2 1 mole H 2 196.97 g Au 1 mole Au 3 mole H 2 2 mole Au

5. Au 2 S 3 + 3H 2  2Au + 3H 2 S Which reactant is Limited and which reactant is Excess? 25.0 g Au 2 S 3 20.1 g Au 490.14 g Au 2 S 3 1 mole Au 2 S 3 196.97 g Au 1 mol Au 1 mole Au 2 S 3 2 mol Au 15.0g H 2 977.13 g Au 2.0 g H 2 1 mole H 2 196.97 g Au 1 mole Au 3 mole H 2 2 mole Au Limited = Au 2 S 3 Excess = H 2

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

7. 2Al(OH) 3 + 3 H 2 SO 4  Al 2 (SO 4 ) 3 + 6 H 2 O 45.0 g Aluminum hydroxide reacts 15.0 grams of dihydrogen monosulfur tetraoxide to create aluminum sulfate and water. Which reactant is limiting, which is excess, and how much product is produced? 45.0 g Al(OH) 3 31.1 g H 2 O 77.98 g Al(OH) 3 1 mole Al(OH) 3 18.0g H 2 O 1 mol H 2 O 2 mole Al(OH) 3 6 mol H 2 O 15.0 g H 2 SO 4 5.51 g H 2 O 97.996 g H 2 SO 4 1 mol H 2 SO 4 18.0 g H 2 O 1 mol H 2 O 3 mol H 2 SO 4 6 mol H 2 O

8. 2Al(OH) 3 + 3 H 2 SO 4  Al 2 (SO 4 ) 3 + 6 H 2 O Which reactant is Limited and which reactant is Excess? 45.0 g Al(OH) 3 31.1 g H 2 O 77.98 g Al(OH) 3 1 mole Al(OH) 3 18.0g H 2 O 1 mol H 2 O 2 mole Al(OH) 3 6 mol H 2 O 15.0 g H 2 SO 4 5.51 g H 2 O 97.996 g H 2 SO 4 1 mol H 2 SO 4 18.0 g H 2 O 1 mol H 2 O 3 mol H 2 SO 4 6 mol H 2 O Limited = H 2 SO 4 Excess = Al(OH) 3

9. 45.0 g Aluminum hydroxide reacts 15.0 grams of dihydrogen monosulfur tetraoxide to create aluminum sulfate and water. Which reactant is limiting, which is excess, and how much product is produced? Start with the limited reactant!! Used In reaction 45.0 grams Al(OH) 3 minus 7.98 g Al(OH) 3 = Given Used In reaction 37.0 g EXCESS 15.0 g H 2 SO 4 7.98 g Al(OH) 3 97.996 g H 2 SO 4 1 mol H 2 SO 4 77.98 g Al(OH) 3 1 mol Al(OH) 3 3 mol H 2 SO 4 2 mo l Al(OH) 3 2Al(OH) 3 + 3 H 2 SO 4  Al 2 (SO 4 ) 3 + 6 H 2 O

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

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

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

14. 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.

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

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

17. 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 NaCl 102.98g NaBr 1 mole NaBr 1 mole NaCl 58.44 g NaCl 1461 102.98 14 g

18. 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? Percent yield = Actual (for real) Theoretical (calculated) 18 g NaCl (actual) 14 g NaCl (calculated) X 100 = 128%

19. 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) 14 g NaCl (calculated) X 100 = 128% There is NO WAY to have over 100% percent yield. It has to be experimental error- procedural or math error.

20. 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

21. 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.

22. Limited reactants % yields practice