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STOICHIOMETRY

STOICHIOMETRY. USING THE REACTION EQUATION LIKE A RECIPE. USING EQUATIONS. Nearly everything we use is manufactured from chemicals. Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes.

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STOICHIOMETRY

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  1. STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE

  2. USING EQUATIONS • Nearly everything we use is manufactured from chemicals. • Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes. • For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them. • Chemical processes carried out in industry must be economical, this is where balanced equations help.

  3. USING EQUATIONS • Equations are a chemist’s recipe. • Eqs tell chemists what amounts of reactants to mix and what amounts of products to expect. • When you know the quantity of one substance in a rxn, you can calculate the quantity of any other substance consumed or created in the rxn. • Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

  4. USING EQUATIONS • The calculation of quantities in chem-ical reactions is called stoichiometry. • Imagine you are in charge of manu-facturing for Rugged Rider Bicycle Company. • The business plan for Rugged Rider requires the production of 128 custom-made bikes each day. • You are responsible for insuring that there are enough parts at the start of each day.

  5. USING EQUATIONS • Assume that the major components of the bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P). • The finished bike has a “formula” of FSW2HP2. • The balanced equation for the production of 1 bike is. F +S+2W+H+2P FSW2HP2

  6. + + + + H P F S 2W + + + + FSW HP 2 2

  7. USING EQUATIONS • Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes? • What do we know? • Number of bikes = 640 bikes • 1 FSW2HP2=2W (balanced eqn) • What is unknown? • # of wheels = ? wheels

  8. 1280 wheels = • The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation. 2 W 640 FSW2HP2 1 FSW2HP2 • We can make the same kinds of connections from a chemical rxn eqn. N2(g) + 3H2(g)  2NH3(g) • The key is the “coefficient ratio”.

  9. The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical rxn. • 1 mole of N2 reacts with 3 moles of H2 to produce 2moles of NH3. • N2 and H2 will always react to form ammonia in this 1:3:2 ratio of moles. • So if you started with 10 moles of N2 it would take 30 moles of H2 and would produce 20 moles of NH3

  10. Using the coefficients, from the balan-ced rxn equation to make connections between reactants and products, is the most important information that a rxn equation provides. • Using this information, you can calculate the amounts of the reactants involved and the amount of product you might expect. • Any calculation done with the next process is a theoretical number, the real world isn’t always perfect.

  11. Using the coefficients of balanced rxn equations and our knowledge of mole conversions we can perform powerful calculations. A.K.A. stoichiometry. • A balanced rxn equation is essential for all calculations involving amounts of reactants and products. • If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn equation.

  12. MOLE – MOLE EXAMPLE • The following rxn shows the synthesis of aluminum oxide. 3O2(g) + 4Al(s)  2Al2O3(s) 3O2(g) + 4Al(s)  2Al2O3(s) • If you only had 1.8 mols of Al how much product could you make? Given: 1.8 moles of Al Uknown: ____ moles of Al2O3

  13. Mole Ratio MOLE – MOLE EXAMPLE 3O2(g) + 4Al(s)  2Al2O3(s) • Solve for the unknown: 2 mol Al2O3 1.8 mol Al = 0.90mol Al2O3 4 mol Al

  14. MOLE – MOLE EXAMPLE 2 • The following rxn shows the synthesis of aluminum oxide. 3O2(g) + 4Al(s)  2Al2O3(s) • If you wanted to produce 24 mols of product how many mols of each reactant would you need? Given: 24 moles of Al2O3 Uknown: ____ moles of Al ____ moles of O2

  15. MOLE – MOLE EXAMPLE 2 3O2(g) + 4Al(s)  2Al2O3(s) • Solve for the unknowns: 4 mol Al 24 mol Al2O3 = 48 mol Al 2 mol Al2O3 3 mol O2 24 mol Al2O3 = 36 mol O2 2 mol Al2O3

  16. MASS – MASS CALCULAT’NS • No lab balance measures moles directly, generally mass is the unit of choice. • From the mass of 1 reactant or prod-uct, the mass of any other reactant or product in a given chemical equation can be calculated, provided you have a balanced rxn equation. • As in mole-mole calcs, the unknown can be either a reactant or a product.

  17. MASS – MASS CALCULAT’NS 1 Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2). CaC2 + 2H2O  C2H2 + Ca(OH)2 How many grams of C2H2 are produced by adding water to 5.00 g CaC2?

  18. MASS – MASS CALCULAT’NS 1 • What do we know? • Given mass = 5.0 g CaC2 • Mole ratio: 1 mol CaC2 = 1 mol C2H2 • MM of CaC2 = 64.0 g CaC2 • MM of C2H2 = 26.0g C2H2 • What are we asked for? • grams of C2H2 produced

  19. MASS – MASS CALCULAT’NS 1 mass A  moles A  moles B  mass B 5.0 g CaC2 1 mol CaC2 1 mol C2H2 64.0 g CaC2 1mol CaC2 26.0 g C2H2 1mol C2H2 = 2.03 g C2H2

  20. MASS – MASS CALCULAT’NS 2 You’ve recently learned that Copper will replace silver ions out of solution. You’re eyes light up with this money making opportunity. However, you decide it might be best if you did some preliminary calculations to determine to the feasibility of this get rich scheme. Copper is not very hard to find, however the largest size of Silver nitrate found in the Flinn Catalog is the 500 g size and it costs $305.91. Currently Silver sells for $9.00/ounce on the stock market. How much money could you sell your manufactured Silver for?

  21. MASS – MASS CALCULAT’NS 2 Cu + 2AgNO3 2Ag + Cu(NO3)2 Cu + 2AgNO3 2Ag + Cu(NO3)2 • What do we know? • Given mass = 500 g of AgNO3 • Mole ratio: 2 mol AgNO3 = 2 mol Ag • MM of AgNO3: 169.84g = 1mol • MM of Ag: 107.87 g = 1mol • Price of Silver: $9.00 = 1 ounce • Conversion g to oz: 28.23g = 1 oz

  22. MASS – MASS CALCULAT’NS 2 500 g AgNO3 1mol AgNO3 2 mol Ag 169.8gAgNO3 2 mol AgNO3 107.87g Ag 1 oz $9.00 1mol Ag 28.23 g 1 oz = $101.24

  23. A balanced reaction equation indicates the relative numbers of moles of reactants and products. • We can expand our stoichiometric calculations to include any unit of measure that is related to the mole. • The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP. • The problems can include mass-volume, volume-volume, and particle-mass calculations.

  24. In any of these problems, the given quantity is first converted to moles. • Then themole ratiofrom the balanced eqn is used to convert from the moles of given to the number of moles of the unknown • Then the moles of the unknown are converted to the units that the problem requests. • The next slide summarizes these steps for all typical stoichiometric problems

  25. STOICH MAP 23 6.02X10 PARTICLES PARTICLES 23 6.02X10 A B MASS MOLE MOLE MASS MOLE MOLAR MASS MOLAR MASS RATIO A A B B VOLUME VOLUME 22.4 L A B 22.4 L

  26. MORE MOLE EXAMPLES How many molecules of O2 are produced when a sample of 29.2 g of H2O is decomposed by electrolysis according to this balanced equation: 2H2O  2H2 + O2

  27. MORE MOLE EXAMPLES • What do we know? • Mass of H2O = 29.2 g H2O • 2 mol H2O = 1 mol O2 (from balanced equation) • MM of H2O = 18.0 g H2O • 1 mol O2 = 6.02x1023 molecules of O2 • What are we asked for? • molecules of O2

  28. 23 6.02X10 PARTICLES PARTICLES 23 6.02X10 A B MASS MOLE MOLE MASS MOLE MOLAR MASS MOLAR MASS RATIO A A B B VOLUME VOLUME 22.4 L A B 22.4 L 29.2 g H2O 1 mol H2O 1 mol O2 mass A  mols A  mols B  molecules B 18.0 g H2O 2 mol H2O 6.02x1023 molecules O2 1 mol O2 = 4.88 x 1023 molecules O2

  29. MORE MOLE EXAMPLES The last step in the production of nitric acid is the reaction of NO2 with H2O. 3NO2+H2O2HNO3+NO How many liters of NO2 must react with water to produce 5.00x1022 molecules of NO?

  30. MORE MOLE EXAMPLES • What do we know? • Molecules NO = 5.0x1022 molecules NO • 1 mol NO = 3 mol NO2 (from balanced equation) • 1 mol NO = 6.02x1023 molecules NO • 1 mol NO2 = 22.4 L NO2 • What are we asked for? • Liters of NO2

  31. 23 6.02X10 PARTICLES PARTICLES 23 6.02X10 A B MASS MOLE MOLE MASS MOLE MOLAR MASS MOLAR MASS RATIO A A B B VOLUME VOLUME 22.4 L A B 22.4 L 1 mol NO 3 mol NO2 5.0x1022 mol-ecules NO molecules A mols mols B volume B 1 mol NO 6.02x1023 mol-ecules NO 22.4 L NO2 1 mol NO2 = 5.58 L NO2

  32. C7H6O3 + C4H6O3 C9H8O4 + HC2H3O2 Salicylic acid Acetic anhydride aspirin vinegar Aspirin can be made from a chemical rxn between the reactants salicylic acid and acetic anhydride. The products of the rxn are acetyl-salicylic acid (aspirin) and acetic acid (vinegar). Our factory makes 125,000 100-count bottles of Bayer Aspirin/day. Each bottle contains 100 tablets, and each tablet contains 325mg of aspirin. How much in kgs + 10% for production problems, of each reactant must we have in order to meet production?

  33. What do we know? • Make 125,000 aspirin bottles/day • 100 aspirin/bottle • 325 mg aspirin/tablet • Mole ratio of aspirin to salicylic acid (1:1) and acetic anhydride (1:1) • MM aspirin = 180.11g • MM C7H6O3 = 138.10g • MM C4H6O3 = 102.06g • What are we asked for? • Mass of salicylic acid in kgs + 10% • Mass of acetic anhydride in kgs + 10%

  34. 100 tablets 325mg asp. 125,000 bottles 1 bottle 1 tablet 1 g 1mol asp. 1000 mg 180.16g = 22,549.4 mols aspirin

  35. Salicylic Acid: 1 mol C7H6O3 136.10g C7H6O3 22,549.4 mols aspirin 1 mol C7H6O3 1 mol asp 1 kg = 3068.97 kg salicylic acid + (306.897 g) 1000 g = 3380 kg of salicylic acid

  36. Acetic Anhydride: 1 mol C4H6O3 102.06g C4H6O3 22,549.4 mols aspirin 1 mol C4H6O3 1 mol asp 1 kg = 2301.39 kg Acetic anhydride + 230.139 kg 1000 g = 2530 kg Acetic anhydride

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