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Afra Khanani Period 6 Honors Chemistry March 31 st

Afra Khanani Period 6 Honors Chemistry March 31 st. PROBLEM:. A solution containing 6720 mg of H 2 0 is added to another containing 10.67 Liters of CO 2 at STP.

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Afra Khanani Period 6 Honors Chemistry March 31 st

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  1. Afra Khanani Period 6 Honors Chemistry March 31st

  2. PROBLEM: A solution containing 6720 mg of H20 is added to another 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 as well, as the theoretical and percent yield of glucose if 10.22 g C6H12O6 was obtained.

  3. Products The chemical reaction of water and carbon dioxide will produce oxygen gas, and an unknown element. This unknown element has an empirical formula of CH2O and a molecular mass of 180.15. Its molecular formula is the unknown product needed for your equation.

  4. STEP 1Find the unknown product

  5. STEP 1Find the unknown product Givens: Empirical formula: CH2O Molecular mass: 180.15

  6. STEP 1Find the unknown product Givens: Empirical formula: CH2O Molecular mass: 180.15 1. Find the mass of your empirical formula: CH2O = 30g

  7. STEP 1Find the unknown product Givens: Empirical formula: CH2O Molecular mass: 180.15 1. Find the mass of your empirical formula: CH2O = 30g 2. Divide the molecular mass by the empirical mass: 180/30 = 6

  8. STEP 1Find the unknown product Givens: Empirical formula: CH2O Molecular mass: 180.15 1. Find the mass of your empirical formula: CH2O = 30g 2. Divide the molecular mass by the empirical mass: 180/30 = 6 3. Multiply that number (6) to your empirical formula (CH2O) : C6H12O6

  9. STEP 1Find the unknown product Givens: Empirical formula: CH2O Molecular mass: 180.15 1. Find the mass of your empirical formula: CH2O = 30g 2. Divide the molecular mass by the empirical mass: 180/30 = 6 3. Multiply that number (6) to your empirical formula (CH2O) : C6H12O6 Molecular Formula and Unknown Product= C6H12O6

  10. STEP 1Write and balance the equation

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

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

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

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

  15. 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

  16. STEP 3Convert grams to moles

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

  18. STEP 4Convert mole to moles

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

  20. 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

  21. STEP 5Convert moles to grams

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

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

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

  25. 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

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

  27. 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

  28. STEP 3Convert moles to grams

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

  30. 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

  31. 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)

  32. 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

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

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

  35. 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

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

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

  38. THEORETICAL & PERCENT YIELD Find theoretical percent yield of C6H12O6 (Actual amount of Glucose obtained was 10.22 as stated before)

  39. THEORETICAL & PERCENT YIELD Find theoretical percent yield of C6H12O6 (Actual amount of Glucose obtained was 10.22 as stated before) Theoretical Yield: (Already Found) 11.16 grams

  40. THEORETICAL & PERCENT YIELD Find theoretical percent yield of C6H12O6 (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

  41. THEORETICAL & PERCENT YIELD Find theoretical percent yield of C6H12O6 (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%

  42. Percent Error %Error = (|Your Result - Accepted Value| / Accepted Value) x 100

  43. Percent Error %Error = (|Your Result - Accepted Value| / Accepted Value) x 100 How much should have been made = 11.16 g Glucose How much was made: 10.22 g Glucose

  44. Percent Error %Error = (|Your Result - Accepted Value| / Accepted Value) x 100 How much should have been made = 11.16 g Glucose How much was made: 10.22 g Glucose (|10.22 – 11.16| / 11.16) x 100 = 8%

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