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Chemistry 20 PowerPoint presentation by R. Schultz Chapter 8.

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Presentation on theme: "Chemistry 20 PowerPoint presentation by R. Schultz Chapter 8."— Presentation transcript:

1 Chemistry 20 PowerPoint presentation by R. Schultz Chapter 8

2 8.1 Limiting and Excess Reagents Recall the illustration from Chapter 7: 3 slices toast + 2 slices turkey + 4 strips bacon 1 sandwich 6 slices toast + 4 slices turkey + 8 strips bacon 2 sandwiches

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

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 C 3 H 6 (g) + 2 NH 3 (g) + 2 O 2 (g) C 3 H 3 N(g) + HCN(g) + 4 H 2 O(g) n kg n g pick a product – it doesn’t matter which, and find out which makes least number of moles of product I’ll use C 3 H 3 N and call it n 3 n3n3 limiting reagent is not necessarily the one with smaller mass limiting

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 identify the limiting reagent – find which makes least moles of Mg 3 (PO 4 )(s) 3 Mg(NO 3 ) 2 (aq) + 2 Na 3 PO 4 (aq) Mg 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) n mL 0.5 mol/L n mL 1.2 mol/L n3n3 limiting

9 8.1 Limiting and Excess Reagents 8. Calculate the mass of Mg 3 (PO 4 ) 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 Mg 3 (PO 4 ) 2 (s)

10 8.1 Limiting and Excess Reagents Worksheet BLM 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(NO 3 ) 2 (aq) 1 PbCl 2 (s) + 2 NaNO 3 (aq) n g n g precipitate n 3 m=? a) limiting Worksheet BLM 8.2.1

15 8.2 Predicted and Experimental Yields Worksheet BLM b)

16 8.3 Acid-Base Titration Titration Set-up: 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 Point where indicator changes colour: endpoint if indicator is properly chosen, endpoint occurs at equivalence point

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 CO 2 (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) → H 2 O(l) + NaCl(aq) n 2 v=20.00 mL c=? n 1 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 HNO 3 (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 Chapter Review


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