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Photosynthesis

Photosynthesis. Stoichiometry:. Afra Khanani Honors Chemistry Period 6 March 31 st. PROBLEM:.

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Photosynthesis

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  1. Photosynthesis Stoichiometry: Afra Khanani Honors Chemistry Period 6 March 31st

  2. PROBLEM: A solution containing 6720 mg of H20 is added to a solution containing 10.67 Liters of CO2 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.

  3. STEP 1Write and balance the equation

  4. STEP 1Write and balance the equation __ H20 + __ CO2 __ C6H12O6 + __ 02 REACTANTS PRODUCTS

  5. STEP 1Write and balance the equation __ H20 + __ CO2 __ C6H12O6 + __ 02 REACTANTS PRODUCTS

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

  7. STEP 2Start with one of the knowns (convert mg to g) 6720 mg of H20 (Given)

  8. STEP 2Start with one of the knowns (convert mg to g) 6720 mg of H20 (Given) 6720 mg H20 1 gram H20 1000 milligrams H20 1 gram H20 = 1000 mg H20

  9. STEP 3Convert grams to moles

  10. STEP 3Convert grams to moles 6.72 g H20 1 mole H20 18 grams H20 1 mole H20 = 18 g H20

  11. STEP 4Convert mole to moles

  12. STEP 4Convert mole to moles 6 H20 + 6 CO2 C6H12O6 + 6 02

  13. STEP 4Convert mole to moles 6 H20 + 6 CO2 C6H12O6 + 6 02 .373 mole H20 1 mole C6H12O6 6 mole H20 • 6 mole H20 = 1 mole C6H1206

  14. STEP 5Convert moles to grams

  15. STEP 5Convert moles to grams .062 mole C6H1206 180 grams C6H1206 1 mole C6H1206 • 1 mole C6H1206 = 180 g C6H1206

  16. You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6 Now figure out: 10.67 L CO2 = ? grams C6H12O6

  17. STEP 1Convert L at STP to moles 10.67 L of CO2 (Given)

  18. STEP 1Convert L at STP to moles 10.67 L of CO2 (Given) 10.67 L CO2 1 mole CO2 22.4 Liters CO2 • 22.4 L CO2 =1 mole CO2

  19. STEP 2Convert moles to moles 6 H20 + 6 CO2 C6H12O6 + 6 02

  20. STEP 2Convert moles to moles 6 H20 + 6 CO2 C6H12O6 + 6 02 .476 mole CO2 1 mole C6H12O6 6 mole CO2 • 6 mole CO2 = 1 mole C6H1206

  21. STEP 3Convert moles to grams

  22. STEP 3Convert moles to grams .079 mole C6H12O6 180 grams C6H12O6 1 mole C6H12O6 • 1 mole C6H1206 = 180 g C6H1206

  23. You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6 10.67 L of CO2 = 14.22 mol C6H12O6 CO2 is the excess reactant

  24. You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6 10.67 L of CO2 = 14.22 mol C6H12O6 CO2 is the excess reactant How much excess CO2 ? (In grams)

  25. You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6 10.67 L of CO2 = 14.22 mol C6H12O6 CO2 is the excess reactant How much excess CO2 ? (In grams) 14.22 grams – 11.16 grams = 3.06 grams CO2 in excess

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

  27. STEP 1Convert moles to molecules 0.62 mole of C6H12O6 (Found)

  28. STEP 1Convert moles to molecules 0.62 mole of C6H12O6 (Found) 0.62 moles C6H12O6 6.02 x 1023 C6H12O6 1 mole C6H12O6 • 1 mole C6H1206 = 6.02 x 1023 molecules C6H1206

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

  30. RESTATING THE QUESTION: Find the number of molecules of glucose that precipitated. 3.73 E22 molecules C6H12O6

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