1. STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE

2. Nearly everything we use is manufactured from chemicals –Soaps, shampoos, conditioners, 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 –Equations 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 reaction, you can calculate the quantity of any other substance consumed or created in the reaction –Quantity meaning the amount of a substance in grams, liters, molecules, or moles USING EQUATIONS

4. Greek for “measuring elements” The calculation of quantities in chemical reactions is called stoichiometry Imagine you are in charge of manufacturing 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 STOICHIOMETRY

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 work week, 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 (from balanced equation) What is unknown? –# of wheels for 640 bikes= ? 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 equation N 2 (g) + 3H 2 (g)  2NH 3 (g) The key is the “coefficient ratio”

9. Use the coefficients from the balanced reaction equation to make mole ratios –Ratios of balanced coefficients = mole ratios –Makes connections between reactants and products 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 reaction 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 reaction 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 CALCULATIONS No lab balance measures moles directly –Generally, mass is the unit of choice From the mass of 1 reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated –You must have a balanced reaction equation! As in mole-mole calculations, the unknown can be either a reactant or a product

16. Basic Steps to Solve Mass-Mass Stoichiometry Problems 1.Balance the equation 2.Convert mass to moles 3.Set up mole ratios 4.Use mole ratios to calculate moles of desired compound 5.Convert moles to grams, if necessary

17. 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 CALCULATIONS 1 CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2

18. 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 CALCULATIONS 1

19. 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 - Convert from moles of our given to moles 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

20. 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 moles 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

21. 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 CALCULATIONS 2 Pb(NO 3 ) 2 + 2 KI  PbI 2 + 2 KNO 3 mass A  mols A  mols B  mass B

22. 25.3 g Pb(NO 3 ) 2 331.2g Pb(NO 3 ) 2 1mol Pb(NO 3 ) 2 MASS – MASS CALCULATIONS 2 1mol Pb(NO 3 ) 2 2mol KI 1mol KI 166 g KI = 25.4 g KI

23. 1.What mass of barium chloride is needed to react completely with 46.8 g of sodium phosphate according to the following reaction equation? BaCl 2 + Na 3 PO 4  Ba 3 (PO 4 ) 2 + NaCl 2.Use the following 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)

24. Recall, 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

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

27. 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

28. 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

29. 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

30. 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

31. 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

32. Molecules A  Mols A  Mols B  Volume B 5.0x10 22 molecules NO 6.02x10 23 molecules NO 1 mol NO 3 mol NO 2 = 5.58 L NO 2 1 mol NO 2 22.4 L NO 2