1. Agenda 1/28/2013 Turn in work that is due or had been missing Standard check – 3f and 3c (3b, 3d) Limiting reactants and percent yield (mandatory listening/write 2 definitions) “A” Student Assignment The Mole is a number…(mandatory notes) Computer lab /Project peer review, spelling, formatting check (MLA) Homework – complete “World’s …est” Article. Due 11:30 pm Weds turnitin.com

2. wood + oxygen in air →water vapor + carbon dioxide What is the limiting reactant? Limiting Reactants (or why do reactions stop?)

3. Why use an excess of a reactant? Why not use exact mole ratios given in balanced equation? Some reactions stop before reach completion (probability of collisions) inefficient/wasteful

4. Why use an excess of a reactant? Cont. Over time scientists have found often more efficient to use an excess of one reactant (least expensive) “drives” reaction until all limiting reactant used up Can also speed up a reaction

5. Limited air (oxygen) yellow – glowing bits of unburned fuel (sooty) Correct fuel/oxygen ratio – complete combustion Why use an excess of a reactant?

6. Stoichiometry Section 12.3 Limiting Reactants Use your text to define each term. limiting reactant excess reactant 164 New Vocabulary New Vocabulary This is the slide that must be copied.

7. Stoichiometry Section 12.3 Limiting Reactants Use your text to define each term. Limits the extent of the chemical reaction and thereby determines the amount of product limiting reactant excess reactant 164 New Vocabulary New Vocabulary This is the slide that must be copied.

8. Stoichiometry Section 12.3 Limiting Reactants Use your text to define each term. Limits the extent of the chemical reaction and thereby determines the amount of product “leftover” or unused reactants in a chemical reaction limiting reactant excess reactant 164 New Vocabulary New Vocabulary This is the slide that must be copied.

9. Percent Yield Fe 2 O 3 + 3CO →2Fe + 3CO 2 1 mole (160g) of iron (III) oxide will produce 2 moles of pure iron (2 x 56)g or 112g 112g is the “theoretical yield” (maximum amount of product possible In practice what happens? Get less than expected, say 110g “actual yield”

10. Percent Yield Fe 2 O 3 + 3CO →2Fe + 3CO 2 Percent yield = actual yield (from experiment) x 100 Theoretical yield (from stoichiometric calculation) 112g is the “theoretical yield” for Fe 110g “actual yield” of Fe Percent yield Fe = 110g x 100 = 98% 112g

11. “A” Student Assignment – Deadline Friday, February 8th Read section 12.3 Limiting Reactants Complete notes page 165, 1 st 2 sub-headings only (stop when you get to Determining the Limiting Reactant). Read section 12.4 Percent Yield Complete notes pages 167, 168, and 169 On lined paper complete Section 12.3 Assessment Questions 22 through 26 and Section 12.4 Assessment Questions 30 through 34. Create suitable titles. To turn this work in the three Science notebook pages and the separate papers all need to be stapled together before class. Name need only be written on first page.

12. The Mole is the SI base unit for the amount of a substance. The quantity one mole is set by defining one mole of carbon-12 atoms to have a mass of exactly 12 grams. (CST 3b) The Mole is the SI base unit for the amount of a substance. The quantity one mole is set by defining one mole of carbon-12 atoms to have a mass of exactly 12 grams. (CST 3b) Carbon-12 isotope is stable and readily available all over the world

13. The Mole as a Number of Particles The mole is the SI unit to measure the amount of a substance. 12g of carbon-12 is set to define one mole and contains one mole of carbon-12 atoms. One mole equals 6.02 x 10 23 particles of the substance (atoms, molecules, ions). Standard 3c This we already should know “representative particles”

14. The Mole is a Unit song http://www.youtube.com/watch?v=1R7NiIum2TI&NR=1 Think about how we assign a name to numbers all the time: A dozen eggs/roses/bread rolls 2 dozen 5 dozen million = 1 000 000 =1.0 x 10 6

15. The Mole is 6.02 x 10 23 particles (Avogadro’s constant) Fe 2 O 3 + 3CO →2Fe + 3CO 2 1 mole 3 moles 2 moles3 moles 6.02 x 10 23 (formula units) Fe 2 O 3 3 x (6.02 x 10 23 ) = 18.0 x 10 23 = 1.80 x 10 24 molecules of CO

16. The Mole is 6.02 x 10 23 particles Fe 2 O 3 + 3CO →2Fe + 3CO 2 1 mole 3 moles 2 moles3 moles 2 x (6.02 x 10 23 ) atoms Fe 12.0 x 10 23 = 1.2 x 10 24 3 x (6.02 x 10 23 ) = 18.0 x 10 23 = 1.80 x 10 24 molecules of CO 2 Scientific Notation – Appendix B p 889-891 http://www.nyu.edu/pages/mathmol/textbook/scinot.html

17. CST Question – When methane gas is burned in the presence of oxygen, the following chemical reaction occurs Combustion & Exothermic reaction CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g) If 1 mole of methane reacts with 2 moles of oxygen, then (and the answers are in …. Molecules of CO 2 and H 2 O. What do you do?

18. CST Questions – convert from number of moles to number of molecules by multiplying the number of moles by N A (Avogadro’s constant) CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (g) 1 mole2 moles 1 mole 2 moles 6.02x10 23 2x(6.02x10 23 ) 6.02x10 23 1.20x10 24 Molecules CO 2 molecules H 2 O

19. 1 mole6.02 x 10 23 particles 2 moles 2x(6.02x10 23 ) = 1.20x10 24 3 moles 3(6.02 x 10 23 ) = 1.8 x 10 24 10 moles 10(6.02 x 10 23 ) = 6.02 x 10 24 0.5 moles 0.5(6.02 x 10 23 ) = 3.01 x 10 23 0.25 moles 0.25(6.02 x 10 23 ) = 1.5 x 10 23 0.1 moles 0.1(6.02 x 10 23 ) =6.02 x 10 22 Get used to it/comfortable with it

20. Avogadro’s Principle: equal volumes of gases at the same temperature and pressure contain equal numbers of particles. HeH2H2 Cl 2 28.2 cm 22.4L Molar Volume: A mole (6.02 x 10 23 particles) of any gas occupies 22.4 L at STP (0°C which is 273 K and 1 atm).

21. Planning MonTue s WedsThursFri Jan2224 (3e) Mass/equ ations 28 mole/no. of particles 3c,d Computer lab time (peer review) 30 3d and ionic names Project Feb4 ionic names & review 6Test – Ch 6, 7, all mole, equations, naming cmpds 8 Lab 5wk Grades hol1214 hol2022 lab