Chemistry 20
Chapter 8
PowerPoint presentation by
R. Schultz
robert.schultz@ei.educ.ab.ca
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?
figure 8.1, page 296
2 sandwiches
6 sandwiches
5 sandwiches
Only 2 sandwiches be made because …..
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,
looking for
use
to find which produces
given
the least number of moles of product –
any product!
8.1 Limiting and Excess
Reagents
limiting
• Example: Practice Problem 6, page 299
C3H6(g) + 2 NH3(g) + 2 O2(g)
n2
n1
600 g
1.0 kg
C3H3N(g) + HCN(g) + 4 H2O(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
n1 
1000 g
 24 mol
g
42.09 mol
600 g
n2 
 35.0 mol
g
17.04 mol


1
n3  24 mol   24 mol
1
n3  35.0 mol 
1
 17.6 mol
2
limiting reagent is not necessarily the one with smaller mass
8.1 Limiting and Excess
Reagents
limiting
• Once you’ve identified the limiting reagent
you can do stoichiometry to calculate
expected yields
• Example: Practice Problem 7&8, page 303
3 Mg(NO3)2(aq) + 2 Na3PO4(aq)
n2
n1
125.0 mL
100.0 mL
1.2 mol/L
0.5 mol/L
Mg3(PO4)2(s) + 6 NaNO3(aq)
n3
• 7. identify the limiting reagent – find which
makes least moles of Mg3(PO4)(s)
n1  c  v
 0.5 mol L  0.1000L  0.05 mol
n2  c V
 1.2 mol L  0.1250 L  0.15 mol
n3  0.05 mol  31  0.017 mol
n3  0.15 mol  21  0.075 mol
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
experimental yield
% yield 
100%
predicted yield
• Example: question 4 page 311
a)
limiting
8.2 Predicted and Experimental
Yields
2 NaCl(aq) + 1 Pb(NO3)2(aq)
n2
n1
3.50 g
0.58 g
0.58 g
n1 
 0.0099 mol
g
58.44 mol
1
n3  0.0099 mol   0.0050 mol
2
3.50 g
n2 
 0.0106 mol
g
331.2 mol
1
n3  0.0106 mol   0.0106 mol
1
1 PbCl2(s) + 2 NaNO3(aq)
precipitate
n3
m=?
m  0.0050 mol  278.1g mol  1.4 g
Worksheet BLM 8.2.1
b)
8.2 Predicted and Experimental
Yields
% yield 
experimental yield
100%
predicted yield
1.22 g

100%  88%
1.4 g
Worksheet BLM 8.2.1
8.3 Acid-Base Titration
Titration Set-up:
fig 8.5, page 312
Titration talk:
“titration of sample
with 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
8.3 Acid-Base Titration
• standardizing: doing a titration to find the
concentration of a titrant solution to be
used in further analyses
popular titrants
• 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
8.3 Acid-Base Titration
• endpoints observed using acid-base
indicators
• indicators are weak acid/base pairs where
the 2 members have different colours
HIn(aq)  H+(aq) + In‾(aq)
colour 1
colour 2
• chart page 10 of Data Booklet shows
indicator acid/base pairs
8.3 Acid-Base Titration
HIn(aq)
In‾(aq)
HIn(aq)
HIn(aq)
In‾(aq)
green
In‾(aq)
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)
n1
n2
v=20.00 mL v=(29.51-1.50) mL
c=0.150 mol/L
c=?
n1  c  v  0.150 mol L  0.02801L  0.00420 mol
to be continued …….
8.3 Acid-Base Titration
• Practice Problem 21, page 315, continued
1
n2  0.00420 mol L   0.00420 mol
1
0.00420 mol
 0.210 mol L
HCl(aq) 
0.02000 L
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
14
12
pH
10
8
6
4
2
0
0
20
40
60
volume of HNO3(aq)
80
100
8.3 Acid-Base Titration
• Chapter Review