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Unit 12 Stoichiometry

Unit 12 Stoichiometry. Stoichiometry. describing the relative quantities of reactants and products in a chemical reaction. http://visual.ly/chocolate-chip-cookies. How Big is a Mole?. One mole of marbles would cover the entire Earth (oceans included) for a depth of two miles.

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Unit 12 Stoichiometry

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  1. Unit 12 Stoichiometry

  2. Stoichiometry describing the relative quantities of reactants and products in a chemical reaction http://visual.ly/chocolate-chip-cookies

  3. How Big is a Mole? One mole of marbles would cover the entire Earth (oceans included) for a depth of two miles. One mole of $1 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times. It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills.

  4. Amedeo Avogadro 1 mole = 602213673600000000000000 or 6.022 x 1023 ? quadrillions thousands trillions billions millions

  5. Reminders STAAR Chart Molar Mass Al2O3 = Conversion 56.0 g Al2O3 = ? moles Al2O3 2(27.0) + 3(16.0) = 102.0 g/mol 56.0 g Al2O3x1 mol Al2O3= 0.549 mol Al2O3 102.0 g Al2O3

  6. Mass, Volume, Mole Relationship

  7. In a balanced chemical equation, the coefficients describe the mole ratio. 2 H2 + O2 2 H2O The mole ratios for this equation are: 2 mole H2: 1 mole O2 2 mole H2: 2 mole H2O 1 mole O2: 2 mole H2O

  8. Stoichiometry Island Diagram Known Unknown Substance A Substance B M Mass Mass Mountain Mass Mole Island Volume Mole Mole Volume V Liter Lagoon Particles Particles P Particle Place Stoichiometry Island Diagram

  9. Stoichiometry Island Diagram Known Unknown Substance A Substance B Mass Mass 1 mole = molar mass (g) 1 mole = molar mass (g) Use coefficients from balanced chemical equation Volume Mole Mole Volume 1 mole = 22.4 L @ STP 1 mole = 22.4 L @ STP (gases) (gases) 1 mole = 6.022 x 1023 particles (atoms or molecules) 1 mole = 6.022 x 1023 particles (atoms or molecules) Particles Particles Stoichiometry Island Diagram

  10. Practice 1: How many molecules of chlorine are required to form 100.g sodium chloride? Na + Cl2 NaCl+ 441 kJ 1st: balance the equation 2nd: write the info from the word problem above the equation 3rd: use mole island diagram to map out your steps 4th: use dimensional analysis and mole ratios to solve

  11. 1st: balance the equation 2 Na + Cl2 2 NaCl + 441 kJ

  12. 2nd: write the info from the word problem above the equation How many molecules of chlorine are required to form 100.g sodium chloride? ?molecules 100.g 2 Na + Cl2 2 NaCl + 441 kJ

  13. 3rd: use mole island diagram to map out your steps Stoichiometry Island Diagram 100. g Known Unknown Substance A Substance B Mass Mass NaCl Cl2 1 mole = molar mass (g) 1 mole = molar mass (g) Use coefficients from balanced chemical equation Volume Mole Mole Volume 1 mole = 22.4 L @ STP 1 mole = 22.4 L @ STP (gases) (gases) ? molecules 1 mole = 6.022 x 1023 particles (atoms or molecules) 1 mole = 6.022 x 1023 particles (atoms or molecules) Particles Particles Stoichiometry Island Diagram

  14. 4th: use dimensional analysis and mole ratios to solve ?molecules 100.g unknown known 2 Na + Cl2 2 NaCl + 441 kJ 100.g NaClx 1 mole NaClx 1 mole Cl2x (6.02 x 1023) molecules Cl2 = 58.5 g NaCl 2 mole NaCl 1 mole Cl2 Here’s how you set that up: (100 x 1 x 1 x 6.02 x 1023)  (58.5 x 2 x 1) =

  15. example 2KClO3  2KCl + 3O2 How many grams of KClO3 are required to produce 9.00 L of O2 at STP? ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L O2 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3 32.8 g 2 mol KClO3 122.55 g KClO3 1 mol O2 x g KClO3 = 9.00 L O2 = 32.8 g KClO3 3 mol O2 1 mol KClO3 22.4 L O2 O2 KClO3 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  16. example Cu + 2 AgNO3 2 Ag + Cu(NO3)2 How many grams of silver will be formed from 12.0 g copper? 12.0 g ? g 12.0 g Cu 1 mol Cu 63.55 g Cu 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag = 40.7 g Ag 40.7 g 2 mol Ag 107.87 g Ag 1 mol Cu x g Ag = 12.0 g Cu = 40.7 g Ag 1 mol Cu 1 mol Ag 63.55 g Cu Cu Ag Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  17. Practice 2: How many grams of Na will be needed to react with 10.L of Cl2 gas at STP? 2Na + Cl2 2NaCl + 441 kJ

  18. Practice 3: Carbon reacts with sulfur dioxide to form carbon disulfide and carbon monoxide in an exothermic reaction that releases 160 kJ. • How many moles of CS2 form when 6.0 moles of C reacts with excess SO2? b. How many grams of carbon are needed to form 56.0 grams of CS2?

  19. Limiting Reactants • Used up first in a reaction • Limit how much product is made • Which reactant limits the number of burgers? • The leftovers are called excess reactants. http://www.shs.d211.org/science/faculty/hlg/AP%20web%20pages/LR%20gus%20danica/limiting%20reactants.html

  20. Practice 4 (limiting):Which is the limiting reactant? http://www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/cr/m2/s3/

  21. Practice 5(limiting):Which is the limiting reactant? Oxygen Hydrogen http://wps.prenhall.com/wps/media/objects/165/169519/blb9ch0307.html

  22. Determining limiting reactant (without a diagram) 1st: balance equation 2nd: pick a product (it is wise to choose the simplest product) 3rd: using the starting amounts of each reactant, use stoichiometry to calculate the amount (in grams) of chosen product 4th: analyze your results; the limiting reactant is the reactant which gives you less product

  23. Practice 6: If there are 100.0 grams of each reactant available, determine which one is the limiting reactant. Ca + AlBr3 CaBr2 + Al 1st: Balance equation

  24. Practice 6: If there are 100.0 grams of each reactant available, determine which is the limiting reactant. Ca + AlBr3 CaBr2 + Al • 2nd: pick a product …..Al I chose Aluminum because it is simpler than CaBr2.

  25. Practice 6: If there are 100.0 grams of each reactant available, determine which is the limiting reactant. 100. g 100. g ? 3 Ca + 2 AlBr3 3 CaBr2 + 2 Al 3rd: using the starting amounts of each reactant, use stoichiometry to calculate the amount (in grams) of chosen product 100. g Cax 1 mole Ca x 2 mole Al x 27.0 g Al = 44.8 g Al 40.1g Ca 3 mole Ca 1 mole Al 100. g AlBr3 x 1 mole AlBr3 x 2 mole Al__ x 27.0 g Al = 10.1 g Al 266.7g AlBr3 2 mole AlBr3 1 mole Al

  26. Practice 6: If there are 100.0 grams of each reactant available, determine which is the limiting reactant. 100. g 100. g ? 3 Ca + 2 AlBr3 3 CaBr2 + 2 Al 4th: analyze your results; the limiting reactant is the reactant which gives you less product • 100. g Ca produced 44.8 g Al • 100. g AlBr3 produced only 10.1 g Al • So… AlBr3 is your limiting reactant. Which means that after 10.1 g Al is made, you are out of AlBr3 and the reaction stops.

  27. Practice 7: 39.5 grams of zinc are reacted with 42.6 grams of hydrochloric acid. Which is the limiting reactant? Zn + HCl ZnCl2 + H2

  28. Actual and Theoretical Yields • Theoretical Yield: calculated amount of products • Actual Yield: amount of product formed in laboratory experiment C C + + A A B B   In a lab: 5g A + 5g B  8g C 5g A + 5g B  10g C

  29. Calculating Percent Yield • Percent Yield: indicates how much of expected product was obtained in a reaction C • Theoretical yield: 10 g Actual yield: 8g • Percent yield = 8g (100) = 80% yield • 10 g ) (

  30. Percent Yield Practice

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