Chemistry 20

PowerPoint presentation byR. Schultz

robert.schultz@ei.educ.ab.ca

Chapter 8

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

8.1 Limiting and Excess Reagents

• What would happen in the following situation?

2 sandwiches 6 sandwiches5 sandwiches

Only 2 sandwiches be made because …..

figure 8.1, page 296

8.1 Limiting and Excess Reagents

• toast is the limiting reagent

• Do Thought Lab 8.1, page 296

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

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 –

looking forgiven

any product!

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

pick a product – it doesn’t matter which, and find out which makes least number of moles of productI’ll use C3H3N and call it n3

n3

1 3

1000 124 24 24

42.09 1gmol

gn mol n mol mol

2 3

600 135.0 35.0 17.6

17.04 2gmol

gn mol n mol mol

limiting reagent is not necessarily the one with smaller mass

limiting

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)

3 Mg(NO3)2(aq) + 2 Na3PO4(aq) Mg3(PO4)2(s) + 6 NaNO3(aq)n1

100.0 mL0.5 mol/L

n2

125.0 mL1.2 mol/L

n3

1

13 3

0.5 0.1000 0.05

0.05 0.017

molL

n c v

L mol

n mol mol

2

13 2

1.2 0.1250 0.15

0.15 0.075

molL

n c V

L mol

n mol mol

limiting

8.1 Limiting and Excess Reagents

• 8. Calculate the mass of Mg3(PO4)2(s)formed

• n = 0.017 mol x 262.87 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)

8.1 Limiting and Excess Reagents

• Worksheet BLM 8.1.3

• Worksheet BLM 8.1.5, questions 1-3 only

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

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)

8.2 Predicted and Experimental Yields

• Example: question 4 page 311

experimental yield% 100%

predicted yieldyield

8.2 Predicted and Experimental Yields

2 NaCl(aq) + 1 Pb(NO3)2(aq) 1 PbCl2(s) + 2 NaNO3(aq)n1

0.58 gn2

3.50 gprecipitate

n3

m=?

1

3

0.580.0099

58.441

0.0099 0.00502

gmol

gn mol

n mol mol

a)

2

3

3.500.0106

331.21

0.0106 0.01061

gmol

gn mol

n mol mol

limiting

0.0050 278.1 1.4gmolm mol g

Worksheet BLM 8.2.1

8.2 Predicted and Experimental Yields

Worksheet BLM 8.2.1

b)

experimental yield% 100%

predicted yield

1.22100% 88%

1.4

yield

gg

8.3 Acid-Base Titration

Titration Set-up: fig 8.5, page 312

Titration talk:

“titration ofwith

sampletitrant”

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

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

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

8.3 Acid-Base TitrationHIn(aq)

HIn(aq)

HIn(aq)In‾(aq

)In‾(aq)

In‾(aq)

green

8.3 Acid-Base Titration

• Indicators used to show endpoint

• Discuss questions 4-6, page 314

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 mLc=?

n1

v=(29.51-1.50) mL

c=0.150 mol/L1 0.150 0.02801 0.00420mol

Ln c v L mol

to be continued …….

8.3 Acid-Base Titration

• Practice Problem 21, page 315, continued

2

10.00420 0.00420

1mol

Ln mol

0.00420( ) 0.210

0.02000mol

Lmol

HCl aqL

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

8.3 Acid-Base Titration

• Investigation 8.C, page 316

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

8.3 Acid-Base Titration

• Discuss questions 8, 9, 10 page 319

• Thought Lab 8.2 page 319 – Plotting a Titration Curve

8.3 Acid-Base Titration

0

2

4

6

8

10

12

14

0 20 40 60 80 100

volume of HNO3(aq)

pH

8.3 Acid-Base Titration

• Chapter Review