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Chapter 12

Chapter 12. Stoichiometry. 2H 2 + O 2  2H 2 O Pg. 354 What can you tell me about this in terms of moles, particles, and mass? 2 molecules of H 2 + 1 molecule of O 2  2 molecules of H 2 O 2 mol H 2 + 1 mol O 2  2 mol H 2 O 2 mol H 2 x 2.016 g H 2 / 1 mol H 2 = 4.032 g H 2

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Chapter 12

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  1. Chapter 12 Stoichiometry

  2. 2H2 + O2 2H2O Pg. 354 • What can you tell me about this in terms of moles, particles, and mass? • 2 molecules of H2 + 1 molecule of O2  2 molecules of H2O • 2 mol H2 + 1 mol O2  2 mol H2O • 2 mol H2x 2.016 g H2/ 1 mol H2 = 4.032 g H2 • 1 mol O2x 31.998 g O2/ 1 mol O2 = 31.998 g O2 • 2 mol H2O x 18.015 g H2O / 1 mol H2O = 36.03 g H2O

  3. From the previous equation, • the mass of the reactants were: • H2 = 4.032 g and O2 = 31.998 g which totals 36.03 g • The mass of the product was: • H2O = 36.03 g • What law is shown by these calculations? • Law of Conservation of Mass

  4. What is Molar Mass? • Molar mass is the grams of a substance per 1 mol of that substance. • Ex. 65.39 g Zn / 1 mol Zn, 31.998 g O2/ 1 mol O2, 18.015 g H2O / 1 mol H2O

  5. Mole Ratios • All reactions occur in certain ways • There is a relationship between every part of a chemical reaction. • Everything in a chemical reaction occurs in certain ratios. • These are called mole ratios.

  6. Mole Ratios • Pg 356 1C • mol ratios: • 4 mol Zn / 10 mol HNO3, 4 mol Zn / 4 mol Zn(NO3)2, 4 mol Zn / 1 mol N2O, 4 mol Zn / 5 mol H2O, 10 mol HNO3/ 4 mol Zn, 10 mol HNO3/ 1 mol N2O, 10 mol HNO3/ 5 mol H2O …

  7. By using mole ratios and molar mass you can calculate unknown parts of a chemical reaction based on information that you are given. • mol mol • y molxmol ratio = z mol • mol mass • y molxmol ratio x molar mass = z mass • mass  mass • y g x molar mass xmol ratio x molar mass = z g

  8. Examples • 2CH4 + S8 2CS2 + 4H2S • mol  mol • You have 1.3 mol of CH4, calculate the mol for S8 • 1.3 mol CH4x 1 mol S8/ 2 mol CH4 = 0.65 mol S8 • mol  mass • You have 6.1 mol H2S, calculate the mass of S8 • 6.1 mol H2S x 1 mol S8 / 4 mol H2S x 256.56 g S8/ 1 mol S8 = 391.25 g S8

  9. Mass  mass • You have 7.19 g S8, calculate the mass of 2CS2 • 7.19 g S8x 1 mol S8/ 256.56 g S8x 2 mol CS2 / 1 mol S8x 76.15 g CS2/ 1 mol CS2 = 4.27 g CS2 • You do: • C12H22O11 + 3O2  2H3C6H5O7 + 3H2O 2.50 mol x g

  10. Homework • Pg 378 53-55, 62, 63, 65, 67, 72, 74, 75

  11. Limiting and Excess Reactants • Limiting Reactants • Limits the extent of the reaction and determines the amount of product produced. • Excess reactants • Available in more than the required amount

  12. Steps to Finding Limiting and Excess Reactants • Na + Fe2O3 Na2O + Fe 100.0 g 100.0 g • Balance the equation • 6Na + Fe2O3  3Na2O + 2Fe • Now you know that 6 mol Na reacts with 1 mol Fe2O3 (mol ratio)

  13. 2. Calculate the moles of each reactant 100 g Na x 1 mol Na / 22.99 g Na = 4.350 mol Na 100 g Fe2O3x 1 mol Fe2O3/ 159.70 g Fe2O3 = 0.6262 mol Fe2O3

  14. 3. Compare the mol ratio of the reactant in the balanced chemical equation with the experimental mol ratio. • Put the smallest on bottom and scale to 1. • Eq: 6 mol Na / 1 mol Fe2O3 • Given: 4.350 mol Na / 0.6262 mol Fe2O3 • In order to scale to 1 divide each number by the bottom number. • Given: 6.947 mol Na / 1 mol Fe2O3

  15. 4. If the numerator in the given is larger than the numerator from the equation then it is excess if it is smaller then it is limiting. • Do not use the ratios for any other calculation. • From the example problem, because the numerator in the given is higher it is in excess and the other is limiting.

  16. You may be ask to find the grams of a product based on your previous calculations. • 5. Calculate the product • Always use the limiting reactant. • Why? • From the equation, find the mass of iron produced. (mass  mass) • 100 g Fe2O3x 1 mol Fe2O3/ 159.70 x 2 mol/ 1 mol Fe2O3x 55.85 g Fe / 1 mol Fe = 69.94 g Fe (7.0 x 101 g Fe)

  17. 6. Mass of excess remaining • 1st calculate the excess used. • Always use the limiting reactant. • Mass of limiting  mass of excess • Take your calculated value of the mass of excess and subtract that from what you were given. • 100.0 g Na – 86.37 g Na = 13.63 g Na in excess

  18. Example • Pg380 #81 • Determine excess and limiting • Calculate product • Calculate excess remaining

  19. Percent Yield • Actual / Theoretical x 100% Or • Experimental / Calculated x 100%

  20. Example Pg. 372 #27 • Calculate the mass of AlCl3. • Always use the limiting reactant. • Mass  mass • You should get 24.0 g AlCl3 • This is your calculated value. The actual or experimental value you were given was 22.0 g. • 22.0 g AlCl3/ 24.0 g AlCl3x 100% = 91.7% • 91.7% is percent yield

  21. You do • Pg. 368 #21

  22. Homework • Pg. 381 87-89

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