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Chapter 12 - Stoichiometry

Chapter 12 - Stoichiometry. “SUPER DIMENSIONAL ANALYSIS”. Review of moles…. 1 mole = 6.022 x 10 23 particles Molar mass =. ( ). X grams. X = molar mass of substance. 1 mole. Use atomic mass. Calculate molar mass of CaBr 2. Ca:. x 1. = 40.1 g/mol. 40.1. Br:. x 2.

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Chapter 12 - Stoichiometry

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  1. Chapter 12 - Stoichiometry “SUPER DIMENSIONAL ANALYSIS”

  2. Review of moles….. • 1 mole = 6.022 x 1023 particles • Molar mass = ( ) X grams X = molar mass of substance 1 mole Use atomic mass

  3. Calculate molar mass of CaBr2 Ca: x 1 = 40.1 g/mol 40.1 Br: x 2 79.9 = 159.8 g/mol = 199.9 g/mol

  4. What is mass of 0.250 mol CO2? C: x 1 = 12 g/mol 12 x 2 O: 16 = 32 g/mol = 44 g/mol 0.250 mol CO2 44 g CO2 = 11 g CO2 1 mol CO2

  5. Stoichiometry • Defn – study of mass relationships between reactants and products in a chemical reaction • What does this mean? • How much of something (products) can be made? • How much of starting materials were there?

  6. Stoichiometry balanced chemical reaction You must have a _____________________ to do stoichiometry calculations.

  7. To make a basic cheeseburger: 2 buns + 1 meat patty + 1 cheese 1 cheeseburger 1 c-burger 1 c-burger 1 c-burger 2 buns 1 meat patty 1 cheese 1 meat patty 1 cheese 2 buns 1 meat patty 2 buns 1 cheese

  8. Mole Ratio • Defn – ratio btwn # of moles of any two substances in a balanced chem rxn

  9. Ex reaction 2 Al + 3 Br2 2 AlBr3 • give all the mole ratios 2 mol Al 3 mol Br2 2 mol AlBr3 3 mol Br2 2 mol Al 2 mol Al 2 mol Al 3 mol Br2 2 mol AlBr3 2 mol AlBr3 3 mol Br2 2 mol AlBr3

  10. 4 steps to a basic stoichiometry problem • Balance the equation • Identify the given, convert to moles • Identify unknown, do a mole to mole ratio between given and unknown (KEY STEP) • Convert unknown to unit specified in problem

  11. Stoichiometry Flow Chart Molar Mass A Molar Mass B Mole Ratio B A grams A moles A moles B grams B

  12. examples • How many moles of H2 are formed when 65 g HCl is used? • If 3.7 mol KBr reacts with calcium, how many moles of CaBr2 are formed? B A A B

  13. Mole A to Mole B Conversions Molar Mass A Molar Mass B Mole Ratio B A grams A moles A moles B grams B

  14. Mole A to Mole B Conversions __ C3H8 + __ O2 __ CO2 + __ H2O • How many moles of CO2 are produced when 10.0 moles of O2 is used? 1 5 3 4 ? mol 10.0 mol B A 10.0 mol O2 3 mol CO2 = 6 mol CO2 5 mol O2 Mole Ratio B A

  15. Mole A to Mass B Conversion Molar Mass A Molar Mass B Mole Ratio B A grams A moles A moles B grams B

  16. Mole A to Mass B Conversion 2 1 2 ___ Na + ___ Cl2 ___ NaCl • How many grams of sodium chloride is formed when 1.25 moles of sodium react w/ chlorine gas? B A ? g 1.25 mol 58.5 g NaCl 1.25 mol Na 2 mol NaCl = 73.1 g NaCl 2 mol Na 1 mol NaCl Mole Ratio B A Molar Mass B

  17. Mass A to Mole B Conversion Molar Mass A Molar Mass B Mole Ratio B A grams A moles A moles B grams B

  18. Mass A to Mole B Conversion B A 2 1 2 ___ Na + ___ Cl2 ___ NaCl • How many moles of chlorine gas is needed to make 50 g NaCl? ? mol 50 g 1 mol NaCl 1 mol Cl2 50 g NaCl = 0.43 mol Cl2 58.5 g NaCl 2 mol NaCl Molar Mass A Mole Ratio B A

  19. Mass A to Mass B Conversion Molar Mass A Molar Mass B Mole Ratio B A grams A moles A moles B grams B

  20. Mass A to Mass B Conversion A B 1 1 2 ___NH4NO3 ___ N2O + ___ H2O • Determine the mass of water formed from decomposition of 25.0 g NH4NO3 25 g ? g 25 g NH4NO3 1 mol NH4NO3 2 mol H2O 18 g H2O 80 g NH4NO3 1 mol NH4NO3 1 mol H2O = 11.25 g H2O

  21. In basic stoichiometry problems, you are provided with one given quantity. In LR problems, you are given both reactants. Before you can solve the problem, you have to determine which of the two given quantities to use as your given. Limiting Reactant

  22. Limiting Reactant You need to choose the one that will run out first, known as the __________________. It controls how much product you can make. The other reactant is known as the _________________ because there will be some left over. limiting reactant excess reactant

  23. Given: 10 pieces of cheese, 50 bread slices 1 cheese + 2 bread slices  1 cheese sandwich How many cheese sandwiches can you make? 10 cheese What is the limiting reactant? What is the excess reactant? bread How many pieces of excess reactant used? 20 How many pieces of excess reactant left over? 30

  24. How to find limiting reactant • Convert both amounts of reactants to moles • Divide the mole amount of each reactant by its coefficient in the balanced equation • Compare two numbers. The one that is smaller is the limiting reactant. Other one is excess reactant.

  25. 2 3 3 2 ____ Al + ___ CuCl2 ___ Cu + ___ AlCl3 Find the limiting reactant if 6.9 g Al and 0.35 mol CuCl2 are available. 1) Convert both amounts of reactants to moles. Reactant #1: Reactant #2: 1 mol Al 6.9 g Al 27 g Al = 0.256 mol Al = 0.35 mol CuCl2

  26. 2 3 3 2 ____ Al + ___ CuCl2 ___ Cu + ___ AlCl3 2) Divide each mole amount by its coefficient in the balanced equation Reactant #1: Reactant #2: 0.256 mol Al 0.35 mol CuCl2 2 3 = 0.117 = 0.128 CuCl2 is the limiting reactant

  27. 2 3 3 2 ____ Al + ___ CuCl2 ___ Cu + ___ AlCl3 Find the limiting reactant if 6.2 g Al and 48.5 g CuCl2 are available. 1) Convert both amounts of reactants to moles. Reactant #1: Reactant #2: 1 mol Al 1 mol CuCl2 48.5 g CuCl2 6.2 g Al 134.5 g CuCl2 27 g Al = 0.230 mol Al = 0.360 mol CuCl2

  28. 2 3 3 2 ____ Al + ___ CuCl2 ___ Cu + ___ AlCl3 2) Divide each mole amount by its coefficient in the balanced equation Reactant #1: Reactant #2: 0.230 mol Al 0.360 mol CuCl2 2 3 = 0.120 = 0.115 Al is the LR CuCl2 is excess reactant

  29. Based on the LR in #4, how many grams of copper will be produced? 2 3 3 2 ____ Al + ___ CuCl2 ___ Cu + ___ AlCl3 6.2 g Al 1 mol Al 3 mol Cu 63.5 g Cu 27 g Al 2 mol Al 1 mol Cu = 21.9 g Cu

  30. Percent Yield • You buy a 500 g ketchup bottle, do you ever use all 500 g? No. Some is still left on the sides you cannot retrieve. Not all 100% is used.

  31. Percent Yield • Theoretical Yield – max amount of product that can be produced (what you expect to get) • Actual Yield – amt of product you actually produced (always less than theoretical) • Percent Yield actual = x 100 theoretical

  32. Ex problem #1 • Joe does experiment to form carbon dioxide. The maximum he can obtain is 34.5 grams. He performs the experiment and only obtains 18.6 grams. What is his percent yield? 18.6 g 34.5 g A T = 53.9% = X 100

  33. Ex problem #2 __K2CrO4 + __ AgNO3 __KNO3 + __Ag2CrO4 1 2 2 1 • What is the theoretical yield of Ag2CrO4 if 0.500 g AgNO3 reacts with excess K2CrO4? ? g 0.500 g 0.5 g K2CrO4 1 mol K2CrO4 1 mol Ag2CrO4 331.7 g 170 g K2CrO4 1 mol K2CrO4 1 mol Ag2CrO4 = 0.49 g Ag2CrO4

  34. Ex problem #2 • If 0.455 g of Ag2CrO4 is produced, what is the percent yield? 0.455 g 0.49 g A T = 93.2% = X 100

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