1. Afra Khanani Honors Chemistry Period 6 March 31 st

2. A solution containing 6720 mg of H 2 0 is added to another containing 10.67 Liters of CO 2 at STP. Determine which reactant was in excess, as well as the number of grams over the amount required by the limiting species. Also, find the number of molecules of glucose that precipitated as well, as the theoretical and percent yield of glucose if 10.22 g C 6 H 12 O 6 was obtained.

3. STEP 1 Write and balance the equation

4. __ H 2 0 + __ CO 2 __ C 6 H 12 O 6 + __ 0 2 REACTANTS PRODUCTS

5. STEP 1 Write and balance the equation __ H 2 0 + __ CO 2 __ C 6 H 12 O 6 + __ 0 2 REACTANTS PRODUCTS

6. Water + Carbon Dioxide (+ energy) = Glucose + Oxygen

7. STEP 2 Start with one of the knowns (convert mg to g) 6720 mg of H 2 0 (Given)

8. STEP 2 Start with one of the knowns (convert mg to g) 6720 mg of H 2 0 (Given) 6720 mg H 2 0 1 gram H 2 0 1000 milligrams H 2 0

9. STEP 3 Convert grams to moles

10. 6.72 g H 2 0 1 mole H 2 0 18 grams H 2 0

11. STEP 4 Convert mole to moles

12. 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2 STEP 4 Convert mole to moles

13. 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2.373 mole H 2 0 1 mole C 6 H 12 O 6 6 mole H 2 0

14. STEP 5 Convert moles to grams

15. .062 mole C 6 H 12 0 6 180 grams C 6 H 12 0 6 1 mole C 6 H 12 0 6 STEP 5 Convert moles to grams

16. You have figured out that: 6720 mg of H 2 0 = 11.16 grams C 6 H 12 O 6 Now figure out: 10.67 L CO 2 = ? grams C 6 H 12 O 6

17. STEP 1 Convert L at STP to moles 10.67 L of CO 2 (Given)

18. STEP 1 Convert L at STP to moles 10.67 L of CO 2 (Given) 10.67 L CO 2 1 mole CO 2 22.4 Liters CO 2

19. STEP 2 Convert moles to moles 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2

20. STEP 2 Convert moles to moles 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2.476 mole CO 2 1 mole C 6 H 12 O 6 6 mole CO 2

21. STEP 3 Convert moles to grams

22. .079 mole C 6 H 12 O 6 180 grams C 6 H 12 O 6 1 mole C 6 H 12 O 6

23. You have figured out that: 6720 mg of H 2 0 = 11.16 grams C 6 H 12 O 6 10.67 L of CO 2 = 14.22 mol C 6 H 12 O 6 CO 2 is the excess reactant

24. You have figured out that: 6720 mg of H 2 0 = 11.16 grams C 6 H 12 O 6 10.67 L of CO 2 = 14.22 mol C 6 H 12 O 6 CO 2 is the excess reactant How much excess CO 2 ? (In grams)

25. CO 2 is the excess reactant How much excess CO 2 ? (In grams) 14.22 grams – 11.16 grams = You have figured out that: 6720 mg of H 2 0 = 11.16 grams C 6 H 12 O 6 10.67 L of CO 2 = 14.22 mol C 6 H 12 O 6 3.06 grams CO 2 in excess

26. Find the number of molecules of glucose that precipitated. What’s Next?

27. STEP 1 Convert moles to molecules 0.62 mole of C 6 H 12 O 6 (Found)

28. STEP 1 Convert moles to molecules 0.62 mole of C 6 H 12 O 6 (Found) 0.62 moles C 6 H 12 O 6 6.02 x 10 23 C 6 H 12 O 6 1 mole C 6 H 12 O 6

29. Find the number of molecules of glucose that precipitated. RESTATING THE QUESTION:

30. Find the number of molecules of glucose that precipitated. RESTATING THE QUESTION:

31. THEORETICAL & PERCENT YIELD Find theoretical percent yield of C 6 H 12 O 6 (Actual amount of Glucose obtained was 10.22 as stated before)

32. THEORETICAL & PERCENT YIELD Theoretical Yield: (Already Found) 11.16 grams Find theoretical percent yield of C 6 H 12 O 6 (Actual amount of Glucose obtained was 10.22 as stated before)

33. THEORETICAL & PERCENT YIELD Theoretical Yield: (Already Found) 11.16 grams Percent Yield: ACTUAL YIELD/THEORETICAL x 100 Find theoretical percent yield of C 6 H 12 O 6 (Actual amount of Glucose obtained was 10.22 as stated before)

34. THEORETICAL & PERCENT YIELD Find theoretical percent yield of C 6 H 12 O 6 (Actual amount of Glucose obtained was 10.22 as stated before) Theoretical Yield: (Already Found) 11.16 grams Percent Yield: ACTUAL YIELD/THEORETICAL x 100 PERCENT YIELD : 10.22/11.16 x 100 = 92%