1. Chemistry 20 Chapter 8 PowerPoint presentation by R. Schultz

2. 8.1 Limiting and Excess Reagents
Recall the illustration from Chapter 7:
3 slices toast + 2 slices turkey + 4 strips bacon sandwich
6 slices toast + 4 slices turkey + 8 strips bacon sandwiches

3. 8.1 Limiting and Excess Reagents
What would happen in the following situation?
figure 8.1, page 296
2 sandwiches
6 sandwiches
5 sandwiches
Only 2 sandwiches be made because …..

4. 8.1 Limiting and Excess Reagents
toast is the limiting reagent
Do Thought Lab 8.1, page 296

5. 8.1 Limiting and Excess Reagents
limiting reagent is completely consumed in in a particular chemical reaction
excess reagent is partially consumed in a particular chemical reaction
even the identity of products of a chemical reaction are sometimes determined by whether a given reactant is limiting or excess

6. 8.1 Limiting and Excess Reagents
how to identify limiting reagent:
easiest way – find moles of each reactant, use to find which produces the least number of moles of product –
any product!

7. 8.1 Limiting and Excess Reagents
Example: Practice Problem 6, page 299
C3H6(g) + 2 NH3(g) + 2 O2(g) C3H3N(g) + HCN(g) + 4 H2O(g)
n1 1.0 kg
n2 600 g n3
pick a product – it doesn’t matter which, and find out which makes least number of moles of product I’ll use C3H3N and call it n3
limiting reagent is not necessarily the one with smaller mass

8. 8.1 Limiting and Excess Reagents
Once you’ve identified the limiting reagent you can do stoichiometry to calculate expected yields
Example: Practice Problem 7&8, page 303
7. identify the limiting reagent – find which makes least moles of Mg3(PO4)(s)
limiting
3 Mg(NO3)2(aq) + 2 Na3PO4(aq) Mg3(PO4)2(s) + 6 NaNO3(aq)
n mL 0.5 mol/L
n mL 1.2 mol/L n3

9. 8.1 Limiting and Excess Reagents
8. Calculate the mass of Mg3(PO4)2(s) formed
n = mol x g/mol = 4 g
What would happen if you used the wrong substance as limiting reagent?
You would calculate a larger mass of Mg3(PO4)2(s)

10. 8.1 Limiting and Excess Reagents
Worksheet BLM 8.1.3
Worksheet BLM 8.1.5, questions 1-3 only

11. 8.2 Predicted and Experimental Yields
Predicted or theoretical yield – determined by stoichiometry
Experimental or actual yield – what you end up getting
Lab 8A, page 300

12. 8.2 Predicted and Experimental Yields
Factors limiting experimental yield:
• competing reactions
• incomplete reaction (because it’s slow)
• incomplete reaction (because it reaches equilibrium) • reactant purity
• mechanical losses (details page 306)

13. 8.2 Predicted and Experimental Yields
Example: question 4 page 311

14. 8.2 Predicted and Experimental Yields
2 NaCl(aq) + 1 Pb(NO3)2(aq) PbCl2(s) + 2 NaNO3(aq)
n g
n g
precipitate n3 m=?
limiting
Worksheet BLM 8.2.1

15. 8.2 Predicted and Experimental Yieldsb)
Worksheet BLM 8.2.1

16. 8.3 Acid-Base Titration Titration Set-up: Titration talk:
fig 8.5, page 312
Titration talk:
“titration of with
sample
titrant”
Point where erlenmeyer flask contains stoichiometrically equivalent moles of acid and base:
equivalence point
if indicator is properly chosen, endpoint occurs at equivalence point
Point where indicator changes colour:
endpoint

17. 8.3 Acid-Base Titration
standardizing: doing a titration to find the concentration of a titrant solution to be used in further analyses
HCl(aq) needs to be standardized since pure HCl is a gas and escapes from solution
NaOH(aq) needs to be standardized since its solutions absorb CO2(g) from the air causing its pH to drop
popular titrants

18. 8.3 Acid-Base Titration endpoints observed using acid-base indicators
indicators are weak acid/base pairs where the 2 members have different colours
chart page 10 of Data Booklet shows indicator acid/base pairs
HIn(aq)  H+(aq) + In‾(aq)
colour 1
colour 2

19. 8.3 Acid-Base Titration
HIn(aq)
In‾(aq)
green

20. 8.3 Acid-Base Titration Indicators used to show endpoint
Discuss questions 4-6, page 314

21. 8.3 Acid-Base Titration
Titration calculations – solution stoichiometry
Example: Practice Problem 21, page 315
Questions states that “a student titrates HCl(aq) with NaOH(aq)” Which is the titrant?
NaOH(aq)
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
n2 v=20.00 mL c=?
n1 v=( ) mL c=0.150 mol/L
to be continued …….

22. 8.3 Acid-Base Titration Practice Problem 21, page 315, continued
Practice Problem 22, page 315 states that “a student uses NaOH(aq) to titrate HNO3(aq)”
Which is the titrant?
NaOH(aq)
Note that the base isn’t always the titrant
Worksheet BLM 8.3.3, omit 1a, b

23. 8.3 Acid-Base Titration
Investigation 8.C, page 316

24. 8.3 Acid-Base Titration Titration curves:
Titration of a strong acid with a strong base:
Titration of a strong base with a strong acid:
figures 8.8, 8.9, page 318

25. 8.3 Acid-Base Titration Discuss questions 8, 9, 10 page 319
Thought Lab 8.2 page 319 – Plotting a Titration Curve

26. 8.3 Acid-Base Titration

27. 8.3 Acid-Base Titration
Chapter Review