1. STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE

2. 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. USING EQUATIONS

3. 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. USING EQUATIONS

4. The calculation of quantities in chemical 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. USING EQUATIONS

5. 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 FSW 2 HP 2. The balanced equation for the production of 1 bike is. USING EQUATIONS F +S+2W+H+2P  FSW 2 HP 2

6. 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 FSW 2 HP 2 =2W (balanced eqn) What is unknown? –# of wheels = ? wheels USING EQUATIONS

7. The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation. 640 FSW 2 HP 2 1 FSW 2 HP 2 2 W = 1280 Wheels We can make the same kinds of connections from a chemical rxn eqn. N 2 (g) + 3H 2 (g)  2NH 3 (g) The key is the “coefficient ratio”.

8. –The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chem rxn. 1 mole of N 2 reacts with 3 moles of H 2 to produce 2 moles of NH 3. –N 2 and H 2 will always react to form ammonia in this 1:3:2 ratio of moles. So if you started with 10 moles of N 2 it would take 30 moles of H 2 and would produce 20 moles of NH 3

9. Using the coefficients, from the balanced rxn equation as ratios 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 value, the real world isn’t always perfect.

10. MOLE – MOLE EXAMPLE The following rxn shows the synthesis of aluminum oxide. 3O 2 (g) + 4Al(s)  2Al 2 O 3 (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 Al 2 O 3 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s)

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

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

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

14. 1.How many mols of hydrogen will be produced if 0.44 mol of CaH 2 reacts according to the following equation? CaH 2 + 2H 2 O  Ca(OH) 2 + 2H 2 2.Iron will react with oxygen to produce Iron III oxide. How many moles of Iron III oxide will be produced if 0.18 mol of Iron reacts? Practice for You… (.89 mol H 2 ) (.090 mol Fe 2 O 3 )

15. 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.

16. Acetylene gas (C 2 H 2 ) is produced by adding water to calcium carbide (CaC 2 ). How many grams of C 2 H 2 are produced by adding water to 5.00 g CaC 2 ? CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2 MASS – MASS CALCULAT’NS 1 CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2

17. What do we know? –Given mass = 5.0 g CaC 2 –Mole ratio: 1 mol CaC 2 = 1 mol C 2 H 2 –MM of CaC 2 = 64.0 g CaC 2 –MM of C 2 H 2 = 26.0g C 2 H 2 What are we asked for? –grams of C 2 H 2 produced MASS – MASS CALCULAT’NS 1

18. Step 1: “Get to Moles!” in this case that can be done by using the Molar Mass of your given compound. 5.0 g CaC 2 64.0 g CaC 2 1 mol CaC 2 =.07813 mol CaC 2 Step 2: Now we are ready for the KEY step…converting from mols of our given to mols of unknown using the mole ratio..07813 mol CaC 2 1 mol CaC 2 1 mol C 2 H 2 =.07813 mol C 2 H 2

19. Step 3: Since we are asked for mass of our unknown in this problem, we need to use our molar mass of our unknown and convert our newly calculated mols into grams..07813 mol C 2 H 2 1 mol C 2 H 2 26.0 g C 2 H 2 = 2.03 g C 2 H 2 Summary of 3 Steps: 1.Get to Moles 2.Mole Ratio 3.Get to desired final unit

20. The double replacement reaction between Lead II nitrate and Potassium Iodide produces a bright yellow precipitate that can be used as a color additive in paint. How many grams of Potassium iodide would we need to completely react 25.3 g of Lead II nitrate? MASS – MASS CALCULAT’NS 2 Pb(NO 3 ) 2 + 2 KI  PbI 2 + 2 KNO 3 mass A  mols A  mols B  mass B

21. 25.3 g Pb(NO 3 ) 2 331.2g Pb(NO 3 ) 2 1mol Pb(NO 3 ) 2 MASS – MASS CALCULAT’NS 2 1mol Pb(NO 3 ) 2 2mol KI 1mol KI 166 g KI = 25.4 g KI

22. 1.What mass of Barium chloride is needed to react completely with 46.8 g of Sodium phosphate accor-ding to the following rxn equation? BaCl 2 + Na 3 PO 4  Ba 3 (PO 4 ) 2 + NaCl 2.Use the equation to determine what mass of FeS must react to form 326g of FeCl 2. FeS + HCl  H 2 S + FeCl 2 Practice for You… (89.2 g BaCl 2 ) (226 g FeS)

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 the mole ratio from 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

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

27. What do we know? –Mass of H 2 O = 29.2 g H 2 O –2 mol H 2 O = 1 mol O 2 (from balanced equation) –MM of H 2 O = 18.0 g H 2 O –1 mol O 2 = 6.02x10 23 molecules of O 2 What are we asked for? – molecules of O 2 MORE MOLE EXAMPLES

28. mass A  mols A  mols B  molecules B 29.2 g H 2 O 18.0 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O 2 6.02x10 23 molecules O 2 = 4.88 x 10 23 molecules O 2

29. The last step in the production of nitric acid is the reaction of NO 2 with H 2 O. 3NO 2 +H 2 O  2HNO 3 +NO How many liters of NO 2 must react with water to produce 5.00x10 22 molecules of NO? MORE MOLE EXAMPLES

30. What do we know? –Molecules NO = 5.0x10 22 molecules NO –1 mol NO = 3 mol NO 2 (from balanced equation) –1 mol NO = 6.02x10 23 molecules NO –1 mol NO 2 = 22.4 L NO 2 What are we asked for? – Liters of NO 2 MORE MOLE EXAMPLES

31. molecules A  mols  mols B  volume B 5.0x10 22 mol- ecules NO 6.02x10 23 mol- ecules NO 1 mol NO 3 mol NO 2 = 5.58 L NO 2 1 mol NO 2 22.4 L NO 2

32. 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? C 7 H 6 O 3 + C 4 H 6 O 3  C 9 H 8 O 4 + HC 2 H 3 O 2 Salicylic acid Acetic anhydride aspirinvinegar

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 C 7 H 6 O 3 = 138.10g –MM C 4 H 6 O 3 = 102.06g What are we asked for? – Mass of salicylic acid in kgs + 10% – Mass of acetic anhydride in kgs + 10%

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

35. 22,549.4 mols aspirin 1 mol asp 1 mol C 7 H 6 O 3 136.10g C 7 H 6 O 3 1000 g 1 kg = 3068.97 kg salicylic acid + (306.897 g) Salicylic Acid: = 3380 kg of salicylic acid

36. 22,549.4 mols aspirin 1 mol asp 1 mol C 4 H 6 O 3 102.06g C 4 H 6 O 3 1000 g 1 kg = 2301.39 kg Acetic anhydride + 230.139 kg Acetic Anhydride: = 2530 kg Acetic anhydride