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Mole/ Stoichiometry

Mole/ Stoichiometry. By: Tyler Lewis, Michael Stylc and Erich Johnson. Definition of Stoichiometry. The relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers. Avogadro’s Number. 6.02 x 10^23.

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Mole/ Stoichiometry

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  1. Mole/ Stoichiometry By: Tyler Lewis, Michael Stylc and Erich Johnson

  2. Definition of Stoichiometry • The relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers.

  3. Avogadro’s Number 6.02 x 10^23

  4. What is a mole? • The mole is a unit of measurement used in chemistry to express amounts of a chemical substance.

  5. Molar Mass Definition: The mass, in grams, of one mole of a substance

  6. Mass to Mole conversion 1 mol = g-formula-mass (periodic table) Problem #1: How many moles in 28 grams of CO2 ? Gram-formula-mass of CO2 1 C = 1 x 12.01 g = 12.01 g 2 O = 2 x 16.00 g =32.00 g 44.01 g/mol 28 g CO2x 1 mole CO2= 0.64 mole CO2 44.01 g CO2

  7. Problem #2:How many moles in 480. grams of Fe2O3 ? 2 Fe = 2 x 55.85 g = 111.7 g 3 O = 3 x 16.00 g = 48.0 g 159.7 g/mol 480. gram Fe2O3x 1 moleFe2O3 = 3.01 mole Fe2O3 159.7 gram Fe2O3

  8. Problem #3:Find the number of moles of argon in 452 g of argon. Ar= 1 x 39.95 g =39.95 g 39.95g/mol 452 gram Arx 1 mole Ar= 11.3 mole Ar 39.95 gram Ar

  9. Particle to Mole conversion 1 mol = 6.02 x 1023 particles Problem #1: How many moles in 39.0 x 1028 particles of CO2? 39.0 x 1028particles CO2x 1 mole CO2= 6.48 x 1028 mole CO2 6.02 x 1023 particles CO2

  10. Problem #2:How many moles are 1.20 x 1025 particles of phosphorous? 1.20 x 1025particles P x 1 mole P = 19.9 mole P 6.02 x 1023 particles P

  11. Problem #3:How many moles in 48.0 x 1023 particles of Cs ? 48.0 x 1023particles Cs x 1 mole Cs = 7.97 mole CS 6.02 x 1023 particles Cs

  12. Volume to Mole conversion 1 mol = 22.4 L for a gas at STP Problem #1: How many moles in 22.4 Liters of CO3 ? 22.4 L CO3 x 1 mole CO3= 1 mole CO3 22.4 L CO3

  13. Problem #2: How many moles of argon atoms are present in 11.2 L of argon gas at STP? 11.2 L Arx 1 mole Ar= .500 mole CO3 22.4 L Ar

  14. Problem #3:How many moles of argon atoms are present in 403.2 L of Xenon gas at STP? 403.2 L Xex 1 mole Xe= 18.00 mole Xe 22.4 L Xe

  15. Mol-Mol calculations N2 + 3 H2 ---> 2 NH3 Problem #1: if we have 2 mol of N2 reacting with sufficient H2, how many moles of NH3 will be produced? • ratio to set up the proportion: • That means the ratio from the equation is: NH3 N2 2 1 Initial amount Final Amount 2 mole N2x 2 mole NH3 = 4mole NH3 1 mole N2

  16. N2 + 3 H2 ---> 2 NH3 Problem #2: Suppose 6 mol of H2 reacted with sufficient nitrogen. How many moles of ammonia would be produced? 6 mole H2x 2 mole NH3 = 4mole NH3 3 mole H2

  17. N2 + 3 H2 ---> 2 NH3 Problem #2:We want to produce 2.75 mol of NH3. How many moles of nitrogen would be required? 2.75 mole NH3x 1 mole N2 = 1.38 mole NH3 2 mole NH3

  18. Mass-Mass calculation 2 AuCl3 ---> 2 Au + 3 Cl2 Problem #1: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl3? The amount of moles of initial chemical The molar ratio of chemical 64.0 gram AuCl3 x 1 mole AuCl3 = .211 mole AuCl3 303.35 grams AuCl3 The mole ratio of chemicals .211 mole AuCl3 x 3 mole Cl2= .3165 mole Cl2 2 mole AuCl3 .3165 mole Clx 70.9 gram Cl2= 22.4 gram Cl2 1 mole Cl2

  19. 3 AgNO3 + AlCl3 --> 3 AgCl + Al(NO3)3 Problem #2:Calculate the mass of AgCl that can be prepared from 200. g of AlCl3 and sufficient AgNO3? 200. gram AlCl3 x 1 mole AlCl3 = 1.50 mole AlCl3 133.33 grams AlCl3 1.50 mole AlCl3 x 3 mole AgCl= 4.5 mole AgCl 1 mole AlCl3 4.5 mole AgCl x 143.35 gram AgCl= 645. gram Cl2 1 mole AgCl

  20. 2 Au + 3 Cl2 ---> 2 AuCl3 Problem #4: How many grams of AuCl3 can be made from 100.0 grams of chlorine? 100. gram Cl2x 1 mole Cl2 = 1.41mole Cl2 70.9 grams Cl2 1.41 mole Cl2 x 2 mole AuCl3 = .94mole AuCl3 3 mole Cl2 .94 mole AuCl3 x 303.35 grams AuCl3 = 285 gram AuCl3 1 mole AuCl3

  21. Particle-Particle calculation 2 AuCl3 ---> 2 Au + 3 Cl2 Problem #1: How many particles of chlorine can be liberated from the decomposition of 6.02x1023 particles of AuCl3? The amount of moles of initial chemical The partial ratio of chemical 6.02x1023 particles AuCl3 x 1 mole AuCl3 = 1 mole AuCl3 6.02x1023 particles AuCl3 The mole ratio of chemicals 1 mole AuCl3 x 3 mole Cl2= 1.5mole Cl2 2 mole AuCl3 1.5 mole Cl2x 6.02x1023 particles Cl2= 9.03x1023 particles Cl2 1 mole Cl2

  22. 2 AuCl3 ---> 2 Au + 3 Cl2 Problem #2:How many particles of chlorine can be liberated from the decomposition of 150. particles of AuCl3? 150. particles AuCl3 x 1 mole AuCl3 = 2.492x10-22 mole AuCl3 6.02x1023 particles AuCl3 2.492x10-22 mole AuCl3 x 3 mole Cl2= 3.738x10-22 mole Cl2 2 mole AuCl3 3.738x10-22 mole Cl2x 6.02x1023 particles Cl2= 225 particles Cl2 1 mole Cl2

  23. 3 AgNO3 + AlCl3 --> 3 AgCl + Al(NO3)3 Problem #3:Calculate the particles of AgCl that can be prepared from 132particals of AlCl3 and sufficient AgNO3? 132. article AlCl3 x 1 mole AlCl3 = 2.19x1024 mole AlCl3 6.02x1023 particle AlCl3 2.19x1024 mole AlCl3 x 3 mole AgCl= 6.57x1024 mole AgCl 1 mole AlCl3 6.57x1024mole AgClx 6.02x1023 particle AgCl=396. particle Cl2 1 mole AgCl

  24. Volume-Volume calculation 2 AuCl3 ---> 2 Au + 3 Cl2 Problem #1: How many particles of chlorine can be liberated from the decomposition of 22.4 liters of AuCl3? The amount of moles of initial chemical The volume ratio 22.4 liters AuCl3 x 1 mole AuCl3 = 1 mole AuCl3 22.4 liters AuCl3 The mole ratio of chemicals 1 mole AuCl3 x 3 mole Cl2= 1.5mole Cl2 2 mole AuCl3 1.5 mole Cl2x 22.4 liters Cl2= 33.6 liters Cl2 1 mole Cl2

  25. N2 + 3 H2 ---> 2 NH3 Problem #2: Suppose 5.6 liters of H2 reacted with sufficient nitrogen. How many liters of ammonia would be produced? 5.6 liters H2x 1 mole H2 = .25 moles H2 22.4 liters H2 .25 mole H2 x 2 mole NH3= .167mole NH3 3 mole H2 .167 mole NH3x 22.4 liters NH3= 3.7 liters NH3 1 mole NH3

  26. 2 Na + Cl2 ---> 2 NaCl Problem #3:How many liters of Na are required to react completely with 75.0 liters of chlorine? 75. liters Cl2x 1 mole Cl2 = 3.35moles Cl2 22.4 liters Cl2 3.35 mole Cl2 x 2 mole Na = 6.7 mole Na 1 mole Cl2 6.7 mole Na x 22.4 liters Na = 150. liters Na 1 mole Na

  27. Percent Composition Formulas Mass of element in sample of compound X 100= % Element in Compound Mass of sample of compound OR Mass of element in 1 mol of compound X 100= % Element in Compound Molar mass of compound

  28. Percent Composition Calculation Given: Formula Cu2S Problem #1: Percent Composition of Sulfur 2 Mol Cu X 63.55 Cu = 127.1 g Cu mol Cu 1 mol S X 32.07g S = 32.07 g S mol S Molar mass of Cu2S = 159.2 g 32.07 g S x 100 = 20.15% S 159.2 g Cu2S

  29. Problem #2: Calculate the percent composition of carbon in the following: C6H12O6 6 Mol C x 12.01 g/mole = 72.06 g C 12 Mol H x 1.008 g/mole=12.096 g H 6 Mol O x 16.00 g/mole =96.00 g O 180.2 g C6H12O6 72.06 g C x 100 = 39.99% C 180.2 g C6H12O6

  30. Problem #3: Calculate the percent composition of carbon in the following: CO2 1 Mol C x 12.01 g/mole = 12.01 g C 2 Mol O x 16.00 g/mole =32.00 g O 44.01 g CO2 12.01 g C x 100 = 27.29% C 44.01 g CO26

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