1 / 16

# Homework 8: due Today Homework 9 posted and due Monday 4/2

Homework 8: due Today Homework 9 posted and due Monday 4/2. Cut and try approach example #3: mixed  grams. C 12 H 22 O 11 + 12 O 2 --------  12 CO 2 + 11H 2 O MW 342 g/mol 32 g/mol 44 g/mol 18 g/mol

## Homework 8: due Today Homework 9 posted and due Monday 4/2

E N D

### Presentation Transcript

1. Homework 8: due Today Homework 9 posted and due Monday 4/2

2. Cut and try approach example #3: mixed  grams C12H22O11 + 12O2 -------- 12CO2 + 11H2O MW 342 g/mol 32 g/mol 44 g/mol 18 g/mol 1*1023 molecules 72.73 g ?? g How many grams of CO2 will be produced if 1023 molecules of sucrose, C12H22O11, are combined with 72.73 grams of O2 according to the balanced equation above? 88 g CO2 (sucrose limits)

3. 2C8H18 + 25O2 -------- 16CO2 + 18H2O MW 114 g/mol 32 g/mol 44 g/mol 18 g/mol 1.7051022 molecules 1.136 g ?? g Given 1.705*1022 molecules of C8H18 and 1.136 g O2, how many grams of CO2form ? • 10.0 g • 1.0 g • 0.11 g • 1.56 g

4. Need more practice ???

5. % Yield: a non-chemical analogy Billy comes home with his math test. He got 40 out of 75 questions right. Did he `pass’ the test ? Test Score = Test Yield = 100* 40 =53.3% 75

6. Examples of chemical % yield 5O2 + C3H83CO2 + 4H2O Given mol excess 0.250.7(obs) Exp. Product mol 0.7 Theory product mol 4 x 0.25 =1 (theory=max)1 % yield =exp. mol x100 theory mol =0.7x 1001 = 70%

7. Examples of chemical % yield (continued) 5O2 + C3H83CO2 + 4H2O MW 32 44 44 18 Given excess 1.1 g 0.5 g Exp. Mol CO2 = 0.5= 0.01136 44 Mol C3H8 = 1.1=0.025 44 Product CO2 moles theory = 3 x 0.025 =0.075 1 % yield = 100 * Exp Mol Theory Mol =0.01136 x 100% =15.1% 0.075

8. Bob reports making 4.0 g H2O from 2.2 g C3H8. Is Bob crazy or okay ? Examples of chemical % yield (continued) 5O2 + C3H83CO2 + 4H2O MW 32 44 44 18 Bob Reports 4.0 g H2O Given excess 2.2 g Mol C3H8 =2.2 =0.050 44 Product H2O moles theory = 4 x 0.05 =0.2 mol 1 Maximum g H2O=18 g/mol*0.2 mol=3.6 g Bob’s an idiot

9. MW 16 32 18 CH4 + 2O2 CO2 + 2H2OA 1 g sample of methane (CH4) is burned in excess O2 according to the equation above. You collect 1.125 g of H2O. What is your % yield ? • 50% • 89 % • 25 % • 100% • I have no idea

10. Need more practice ????

11. Where we are on the mole road trip…. We is done with moles !!!

12. And now, after many weeks of &^%!! Moles ….for something completely different…. CLASSICAL REACTIONs

13. Today’s Student Learning Objective (SLO): Students will be able to: write, balance, identify and predict common reaction classes (metatheses, acid-base,redox). SLO #6 Translation: What happens when I add this to that ?

14. CLASSICAL REACTIONs (continued) What to read… chapter 4 ( + a bit of Ch. 14) • Metatheses 131-140 • Acid-base 140-147; 549-552 • Oxidation-reduction 147-159

15. Classical Reactions 1) Metathesis (double replacement) Complete Molecular reaction Pb(NO3)2 (aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) Complete Ionic Reaction Pb2+ +2NO3-+2K+ +2I- PbI2(s) + 2K+ +2NO3- Net Ionic Reaction Pb2+ + 2I-PbI2(s) METATHESIS REACTIONS (Two labs ago)

More Related