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Stoichiometry Lesson # 1

Stoichiometry Lesson # 1. Balanced chemical equations can be used to relate the amounts of reactants and products in chemical reactions. 2 Fe 2 O 3 + 3 C → 4 Fe + 3 CO 2

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Stoichiometry Lesson # 1

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  1. Stoichiometry Lesson # 1

  2. Balanced chemical equations can be used to relate the amounts of reactants and products in chemical reactions. 2Fe2O3 + 3C → 4Fe + 3CO2 The coefficients mean that 2 moles of Fe2O3react with 3 moles of C to produce 4 moles of Fe and 3 moles of CO2. The coefficients are a chemical recipe that describes the exact amounts of reactants required to make exact amounts of products in moles.

  3. 1. How many grams of Fe2O3 are required to produce 105 g of Fe? 2Fe2O3 + 3C → 4Fe + 3CO2 ?g 105 g

  4. 1. How many grams Fe2O3 of are required to produce 105 g of Fe? 2Fe2O3 + 3C → 4Fe + 3CO2 ?g 105 g 105 g Fe

  5. 1. How many grams Fe2O3 of are required to produce 105 g of Fe? 2Fe2O3 + 3C → 4Fe + 3CO2 ?g 105 g 105 g Fe x 1 mole 55.8 g

  6. 1. How many grams Fe2O3 of are required to produce 105 g of Fe? 2Fe2O3 + 3C → 4Fe + 3CO2 ?g 105 g 105 g Fe x 1 mole x 2 mole Fe2O3 55.8 g 4 mole Fe

  7. 1. How many grams Fe2O3 of are required to produce 105 g of Fe? 2Fe2O3 + 3C → 4Fe + 3CO2 ?g 105 g 105 g Fe x 1 mole x 2 mole Fe2O3 x 159.6 g 55.8 g 4 mole Fe 1 mole

  8. 1. How many grams Fe2O3 of are required to produce 105 g of Fe? 2Fe2O3 + 3C → 4Fe + 3CO2 ?g 105 g 105 g Fe x 1 mole x 2 mole Fe2O3 x 159.6 g = 150. g 55.8 g 4 mole Fe 1 mole

  9. 2. How many grams of C are require to consume 155 g of Fe2O3? 2Fe2O3 + 3C → 4Fe + 3CO2 155g ? g

  10. 2. How many grams of C are require to consume 155 g of Fe2O3? 2Fe2O3 + 3C → 4Fe + 3CO2 155g ? g 155 g Fe2O3

  11. 2. How many grams of C are require to consume 155 g of Fe2O3? 2Fe2O3 + 3C → 4Fe + 3CO2 155g ? g 155 g Fe2O3 x 1 mole 159.6 g

  12. 2. How many grams of C are require to consume 155 g of Fe2O3? 2Fe2O3 + 3C → 4Fe + 3CO2 155g ? g 155 g Fe2O3 x 1 mole x 3 mole C 159.6 g 2 mole Fe2O3

  13. 2. How many grams of C are require to consume 155 g of Fe2O3? 2Fe2O3 + 3C → 4Fe + 3CO2 155g ? g 155 g Fe2O3 x 1 mole x 3 mole C x 12.0 g 159.6 g 2 mole Fe2O3 1 mole

  14. 2. How many grams of C are require to consume 155 g of Fe2O3? 2Fe2O3 + 3C → 4Fe + 3CO2 155g ? g 155 g Fe2O3 x 1 mole x 3 mole C x 12.0 g = 17.5 g 159.6 g 2 mole Fe2O3 1 mole

  15. 3. How many grams of Al2(CO3)3 are produced by the complete reaction of 452 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 452 g ? g

  16. 3. How many grams of Al2(CO3)3 are produced by the complete reaction of 452 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO32 452 g ? g 452 g Al(NO3)3

  17. 3. How many grams of Al2(CO3)3 are produced by the complete reaction of 452 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO32 452 g ? g 452 g Al(NO3)3 x 1 mole 213.0 g

  18. 3. How many grams of Al2(CO3)3 are produced by the complete reaction of 452 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO32 452 g ? g 452 g Al(NO3)3 x 1 mole x 1 mole Al2(CO3)3 213.0 g 2 mole Al(NO3)3

  19. 3. How many grams of Al2(CO3)3 are produced by the complete reaction of 452 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO32 452 g ? g 452 g Al(NO3)3 x 1 mole x 1 mole Al2(CO3)3 x 234.0 g = 248 g 213.0 g 2 mole Al(NO3)3 1 mole

  20. 4. How many moles of Na2CO3 are required to completely consume 152 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 152 g ? moles

  21. 4. How many moles of Na2CO3 are required to completely consume 152 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 152 g ? moles 152 g Al(NO3)3

  22. 4. How many moles of Na2CO3 are required to completely consume 152 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 152 g ? moles 152 g Al(NO3)3 x 1 mole 213.0 g

  23. 4. How many moles of Na2CO3 are required to completely consume 152 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 152 g ? moles 152 g Al(NO3)3 x 1 mole x 3 mole Na2CO3 213.0 g 2 mole Al(NO3)3

  24. 4. How many moles of Na2CO3 are required to completely consume 152 g of Al(NO3)3? 2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 152 g ? moles 152 g Al(NO3)3 x 1 mole x 3 mole Na2CO3 = 1.07 moles 213.0 g 2 mole Al(NO3)3

  25. Molar Volume of a Gas at STP Lesson # 2

  26. The molar volume of any gas a STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! Calculate the volume of 10.0 lbs of CO2 at STP.

  27. The molar volume of any gas a STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! Calculate the volume of 10.0 lbs of CO2 at STP. 10.0 lbs

  28. The molar volume of any gas a STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! Calculate the volume of 10.0 lbs of CO2 at STP. 10.0 lbs x 1.00 kg 2.21 lbs

  29. The molar volume of any gas a STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! Calculate the volume of 10.0 lbs of CO2 at STP. 10.0 lbs x 1.00 kg x 1000 g 2.21 lbs 1 kg

  30. The molar volume of any gas a STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! Calculate the volume of 10.0 lbs of CO2 at STP. 10.0 lbs x 1.00 kg x 1000 g x 1 mole 2.21 lbs 1 kg 44.0 g

  31. The molar volume of any gas a STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! Calculate the volume of 10.0 lbs of CO2 at STP. 10.0 lbs x 1.00 kg x 1000 g x 1 mole x 22.4 L 2.21 lbs 1 kg 44.0 g 1 mole

  32. The molar volume of any gas a STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! 1. Calculate the volume of 10.0 lbs of CO2 at STP. 10.0 lbs x 1.00 kg x 1000 g x 1 mole x 22.4 L = 2.30 x 103 L 2.21 lbs 1 kg 44.0 g 1 mole

  33. 2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO. 2HgO → 2Hg + O2 12.5 g ? L 12.5 g HgO

  34. 2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO. 2HgO → 2Hg + O2 12.5 g ? L 12.5 g HgO x 1mole 216.6g

  35. 2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO. 2HgO → 2Hg + O2 12.5 g ? L 12.5 g HgO x 1mole x 1 mole O2 216.6g 2 mole HgO

  36. 2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO. 2HgO → 2Hg + O2 12.5 g ? L 12.5 g HgO x 1mole x 1 mole O2 x 22.4 L 216.6g 2 mole HgO 1 mole

  37. 2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO. 2HgO → 2Hg + O2 12.5 g ? L 12.5 g HgO x 1mole x 1 mole O2 x 22.4 L = 0.646 L 216.6g 2 mole HgO 1 mole

  38. 3. 10.62 g of a common gas occupies 8.50 L at STP. Calculate the molar mass of the gas and determine the gas. 8.50 L x 1 mole=0.37946 moles 22.4 L Molar Mass = grams moles = 10.62 g = 28.0 g/mole 0.37946 moles The gas is N2

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