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Proportional Relationships

Proportional Relationships. Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation. 2 Mg + O 2  2 MgO. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem.

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Proportional Relationships

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  1. Proportional Relationships • Stoichiometry • mass relationships between substances in a chemical reaction • based on the mole ratio • Mole Ratio • indicated by coefficients in a balanced equation 2 Mg + O2 2 MgO Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  2. 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 Mole Mole 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

  3. Visualizing a Chemical Reaction 2 2 Na + Cl2 NaCl 10 5 10 10 ? 10 5 ___ mole Na ___ mole Cl2 ___ mole NaCl

  4. Formation of Ammonia

  5. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. • Mole ratio = moles  moles • Molar mass = moles  grams • Avogadro’s number = particles  moles • Mole ratio = moles  moles Core step in all stoichiometry problems!! 4. Check answer. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  6. Stoichiometry Problems • How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3 2KCl + 3O2 ? mol 9 mol 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  7. Stoichiometry Problems • How many grams of silver will be formed from 12.0 g copper? Cu + 2AgNO3 2Ag + Cu(NO3)2 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 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  8. B2H6 + O2 B2O3 + H2O Rocket Fuel The compound diborane (B2H6) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10 kg of diborane completely? (The products are B2O3 and H2O). B2H6 + O2 B2O3 + H2O Chemical equation Balanced chemical equation 3 3 10 kg x g 1000 g B2H6 1 mol B2H6 3 mol O2 32 g O2 x g O2 = 10 kg B2H6 1 kg B2H6 28 g B2H6 1 mol B2H6 1 mol O2 X = 34,286 g O2

  9. Limiting Reactants • Limiting Reactant • used up in a reaction • determines the amount of product • Excess Reactant • added to ensure that the other reactant is completely used up • cheaper & easier to recycle Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  10. measured in lab calculated on paper Percent Yield actual yield x 100 = % yield theoretical yield

  11. Actual Yield Theoretical Yield % Yield = When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. actual yield 46.3 g K2CO3 +  2KCl + H2O + CO2 2HCl ? g excess 45.8 g theoretical yield Theoretical yield 1 mol K2CO3 2 mol KCl 74.5 g KCl = 49.4 g KCl x g KCl = 45.8 g K2CO3 49.4 g 49.4 g KCl 1 mol K2CO3 1 mol KCl 138 g K2CO3 46.3 g KCl = x 100 % Yield % Yield = 93.7% efficient

  12. Exothermic Reaction Reactants  Products + Thermal Energy (Enthalpy) Energy of reactants Energy of products Reactants Energy -DH Products Reaction Progress

  13. Activation Energy Endothermic Reaction Thermal Energy + Reactants  Products (Enthalpy) Products Energy +DH Reactants Reaction progress

  14. Barbecue An LP gas tank in a home barbecue contains 11.8 X 103g of propane (C3H8). Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank. The heat of reaction is -2044 kJ. __C3H8 + __O2(g)  __CO2(g) + __H2O(g) 1 mol C3H8 -2044 kJ x H = 11.8 X 103 g C3H8 1 mol C3H8 44 g C3H8 X = -5.47 X 105 kJ

  15. C3H8 + 5 O2(g)  3 CO2(g) + 4 H2O(g)

  16. Enthalpy Changes H2(g) + ½ O2(g) DH = +242 kJ Endothermic -242 kJ Exothermic -286 kJ Endothermic DH = -286 kJ Exothermic H2O(g) Energy • 44 kJ • Exothermic +44 kJ Endothermic H2O(l) H2(g) + 1/2O2(g)  H2O(g) + 242 kJ DH = -242 kJ Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211

  17. Enthalpy Changes DH1 = DH2 +DH3 (Hess’s Law) Change in enthalpy is the same, regardless of path of reaction a) H2(g) + 1/2O2(g)  H2O(g) + 242 kJ DH2 = -242 kJ b) H2O(g)  H2O(l) + 44 kJ DH3 = -44 kJ c) H2(g) + 1/2O2(g)  H2O(l) + 286 kJ DH1 = -286 kJ

  18. Enthalpy Changes a) N2(g) + O2(g)  2NO(g) - 180 kJ DH2 = +180 kJ b) 2NO(g) + O2(g)  2NO2(g) + 112 kJ DH3 = -112 kJ c) N2(g) + 2O2(g)  2NO2(g) - 68 kJ DH1 = +68 kJ

  19. Hess’s Law Calculate the enthalpy of formation of carbon dioxide from its elements. C(g) + 2O(g)  CO2(g) Use the following data: 2O(g)  O2(g)DH = - 250 kJ C(s)  C(g) DH = +720 kJ CO2(g)  C(s) + O2(g) DH = +390 kJ 2O(g)  O2(g)DH = - 250 kJ C(g)  C(s) DH = - 720 kJ C(s) + O2(g)  CO2(g)DH = - 390 kJ C(g) + 2O(g)  CO2(g) DH = -1360 kJ Smith, Smoot, Himes, pg 141

  20. Hess’s Law • Calculate DH for the synthesis of diborane from its elements. 2B(s) + 3H2(g)  B2H6DH = ? a) 2B(s) + 3/2O2(g)  B2O3(s) DH = -1273 kJ b) B2O6(g) + 3O2(g)  B2O3(s) + 3H2O(g) DH = -2035 kJ c) H2(g) + 1/2O2(g)  H2O(l) DH = - 286 kJ d) H2O(l)  H2O(g) DH = + 44 kJ

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