1. 1 Chapter 9 Mole Factors Calculations with Equations Limiting Reactions Percent Yield

2. 2 Mole-Mole Factor Shows the mole-to-mole ratio between two of the substances in a balanced equation Derived from the coefficients of any two substances in the equation

3. 3 Writing Mole Factors 4 Fe + 3 O 2 2 Fe 2 O 3 Fe and O 2 4 mol Feand3 mol O 2 3 mol O 2 4 mol Fe Fe and Fe 2 O 3 4 mol Feand2 mol Fe 2 O 3 2 mol Fe 2 O 3 4 mol Fe

4. 4 O 2 and Fe 2 O 3 3 mol O 2 and2 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O 2

5. 5 Learning Check S1 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mole factor for H 2 and N 2 is 1) 3 mol N 2 2) 1 mol N 2 3) 1 mol N 2 1 mol H 2 3 mol H 2 2 mol H 2 B. A mole factor for NH 3 and H 2 is 1) 1 mol H 2 2) 2 mol NH 3 3) 3 mol N 2 2 mol NH 3 3 mol H 2 2 mol NH 3

6. 6 Solution S1 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mole factor for H 2 and N 2 is 2) 1 mol N 2 3 mol H 2 B. A mole factor for NH 3 and H 2 is 2) 2 mol NH 3 3 mol H 2

7. 7 Chemical Calculations 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react? 6.0 mol O 2 x mol Fe 2 O 3 = 4.0 mol Fe 2 O 3 mol O 2

8. 8 Learning Check 2 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 mol of O 2 ? 1) 3.00 mol Fe 2) 9.00 mol Fe 3) 16.0 mol Fe

9. 9 Solution S2 4 Fe + 3 O 2 2 Fe 2 O 3 12.0 mol O 2 x mol Fe = 16.0 mol Fe mol O 2 4 3

10. 10 Learning Check S3 4 Fe + 3 O 2 2 Fe 2 O 3 How many grams of O 2 are needed to produce 0.400 mol of Fe 2 O 3 ? 1) 38.4 g O 2 2) 19.2 g O 2 3) 1.90 g O 2

11. 11 Solution S 3 0.400 mol Fe 2 O 3 x 3 mol O 2 x 32.0 g O 2 2 mol Fe 2 O 3 1 mol O 2 = 19.2 g O 2

12. 12 Calculating Mass of A Substance Balance equation Convert starting amount to moles Use coefficients to write a mol-mol factor Convert moles of desired to grams

13. 13 Calculation The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? Write the equation H 2 (g) + O 2 (g)H 2 O (g) Balance the equation 2 H 2 (g) + O 2 (g)2 H 2 O (g)

14. 14 Organize data mol bridge 2 H 2 (g) + O 2 (g)2 H 2 O (g) ? g 13.1 g Plan g H 2 O mol H 2 O mol O 2 O 2 Setup 13.1 g H 2 O x 1 mol H 2 O x 1 mol O 2 x 32.0 g O 2 8.0 g H 2 O2 mol H 2 O 1 mol O 2 = 11.6 g O 2

15. 15 Points to Remember 1. Read an equation in moles 2. Convert given amount to moles 3. Use mole factor to give desired moles 4. Convert moles to grams grams (givengrams (desired) moles (given)moles (desired)

16. 16 Learning Check S 4 How many O 2 molecules will react with 505 grams of Na to form Na 2 O? 4 Na + O 2 2 Na 2 O Complete the set up: 505 g Na x 1 mol Na x ________ x _______ 23.0 g Na

17. 17 Solution S 4 4 Na + O 2 2 Na 2 O 505 g Na x 1 mol Na x 1 mol O 2 x 6.02 x 10 23 23.0 g Na 4 mol Na 1 mol O 2 = 3.30 x 10 24 molelcules

18. 18 Learning Check S5 Acetylene gas C 2 H 2 burns in the oxyactylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g of CO 2? 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O 75.0 g CO 2 x _______ x _______ x _______

19. 19 Solution S5 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O 75.0 g CO 2 x 1 mol CO 2 x 2 mol C 2 H 2 x 26.0 g C 2 H 2 44.0 g CO 2 4 mol CO 2 1 mol C 2 H 2 = 22.2 g C 2 H 2

20. 20 Pathways for Problems Using Equations Given (A) Find (B) grams (A)grams (B) moles (A) moles (B) particles (A)particles (B)

21. 21 Pathways for Problems Using Equations Given (A) Find (B) grams (A)grams (B) molar molar mass (A) mass (B) coefficients moles (A) moles (B) Avogadro's Avogradro’s number number particles (A)particles (B)

22. 22 Limiting Reactants If the amounts of two reactants are given, the reactant used up first determines the amount of product formed.

23. 23 Analogy Suppose you are preparing cheese sandwiches. Each sandwich requires 2 pieces of bread and 1 slice of cheese. If you have 4 slices of cheese and 10 pieces of bread, how many cheese sandwiches can you make?

24. 24 Cheese Sandwich Products Sandwich 1 + + = Sandwich 2 + + =

25. 25 Learning Check S 6 How many sandwiches can you make? ____ slices of bread + ____ slices of cheese = ____ sandwiches What is left over? ________________ What is the limiting reactant?

26. 26 Solution S 6 How many sandwiches can you make? __10__ slices of bread + __4__ slices of cheese = __4__ sandwiches What is left over? _2 slices of bread What is the limiting reactant? cheese

27. 27 Hints for LR Problems 1. For each reactant amount given, calculate the moles (or grams) of a product it could produce. 2.The reactant that produces the smaller amount of product is the limiting reactant. 3. The number of moles of product produced by the limiting reactant is ALL the product possible. There is no more limiting reactant left.

28. LIMITING REACTANT a) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? 4 Na (s) + O 2(g)  2 Na 2 O (s) 1.Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 6.74 g of Na 2 O 5.00g Na ( 1 mole Na ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 6.74 g of Na 2 O 23 g Na 4 mole Na 1 mol Na 2 O 19.38 g of Na 2 O 5.00g O 2 ( 1 mole O 2 ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 19.38 g of Na 2 O 32 g O 2 1 mole O 2 1 mol Na 2 O Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide. Wrong answer

29. LIMITING REACTANT b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4 Na (s) + O 2(g)  2 Na 2 O (s) There are two methods used to answer this question. The Law of Conservation of mass and Stoichiometry. The amount of O 2 used to make 6.74 g of Na 2 O is calculated by: 1.74 g of O 2 was used 5.00g Na ( 1 mole Na ) ( 1 mole O 2 )( 32 g O 2 ) = 1.74 g of O 2 was used 23 g Na 4 mole Na 1 mol O 2 Or use the Law of Conservation of mass: Mass of product (6.74 g) – mass of limiting reactant (5.00 g) = mass of other reactant, in this case oxygen (1.74 g).

30. 30 The amount of oxygen (O 2 ) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction. LIMITING REACTANT

31. PRACTICE PROBLEMS 1. How much AgCl product will be produced if 100.00 g of BaCl 2 reacted with excess AgNO 3 ? 2. How many moles of carbon dioxide could be produced from 220.0 g of C 2 H 2 and 545.0 g of O 2 ? 3. How many grams of CO 2 can be produced by the reaction of 35.5 grams of C 2 H 2 and 45.9 grams of O 2 ? 4. In the reaction between CH 4 and O 2, if 25.0 g of CO 2 are produced, what is the minimum amount of each reactant needed? 5. Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag. When 10.0 g of copper was reacted with 60.0 g of silver nitrate solution, 30.0 g of silver was obtained. What is the percent yield of silver obtained? 137.57 g 50.5 g 13.63 mol 9.09 g of CH 4 & 36.4 g of O 2 88.3%

32. 32 Percent Yield You prepared cookie dough to make 5 dozen cookies. The phone rings while a sheet of 12 cookies is baking. You talk too long and the cookies burn. You throw them out (or give them to your dog.) The rest of the cookies are okay. How many cookies could you have made (theoretical yield)? How many cookies did you actually make to eat? (Actual yield)

33. 33 Vocabulary Actual yield is the amount of product actually recovered from an experiment Theoretical (possible) yield is the maximum amount of product that could be produced from the reactant. Percent Yield is the actual yield compared to the maximum (theoretical yield) possible.

34. 34 Percent Yield Calculation What is the percent yield of cookies? Percent Yield = Actual Yield (g) recovered X 100 Possible Yield (g) % cookie yield = 48 cookies x 100 = 80% yield 60 cookies