Stoichiometry cont...
(12.3)
Stoichiometry in the
Real World
Limiting Reactants
b
Available Ingredients
• 4 slices of bread
• 1 jar of peanut butter
• 1/2 jar of jelly
b
Limiting Reactant
• bread
b
Excess Reactants
• peanut butter and jelly
A. Limiting Reactants
b
Available Ingredients
• 24 graham cracker squares
• 1 bag of marshmallows
• 12 pieces of chocolate
b Limiting Reactant
• chocolate
b Excess Reactants
• Marshmallows and
graham crackers
A. Limiting Reactants
b
Limiting Reactant
• one that is used up in a reaction
• determines the amount of
product that can be produced
b
Excess Reactant
• added to ensure that the other
reactant is completely used up
• cheaper & easier to recycle
A. Limiting Reactants
1. Write the balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
• limiting reactant
• amount of product actually
possible
A. Limiting Reactants
b
79.1 g of zinc react with 68.1 g HCl.
Identify the limiting and excess
reactants. How many grams of
hydrogen can be formed?
Zn + 2HCl →
79.1 g 68.1 g
ZnCl2 + H2
?g
A. Limiting Reactants
Zn + 2HCl →
79.1 g 68.1 g
ZnCl2 + H2
?g
79.1 1 mol 1 mol 2.02 g
g Zn Zn
H2
H2
= 2.44 g
65.39 1 mol 1 mol
H2
g Zn
Zn
H2
A. Limiting Reactants
Zn + 2HCl →
79.1 g 68.1 g
68.1 1 mol
g HCl HCl
ZnCl2 + H2
?g
1 mol 2.02 g
H2
H2
= 1.89
36.46 2 mol 1 mol g H2
g HCl HCl
H2
A. Limiting Reactants
Zn: 2.44 g H2
HCl: 1.89 g H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 1.89 g H2
left over zinc
A. Limiting Reactants #2
b
5.42 g of magnesium ribbon react
with 4.00 g of oxygen gas. Identify
the limiting and excess reactants.
How many grams of magnesium
oxide are formed?
2Mg + O2 →
5.42 g 4.00 g
2MgO
?g
A. Limiting Reactants #2
2Mg + O2
5.42 g
4.00 g
→
2MgO
?g
5.42 1 mol 2 mol 40.31 g
g Mg Mg MgO MgO
= 8.99 g
24.31 2 mol 1 mol
MgO
g Mg Mg
MgO
A. Limiting Reactants #2
2Mg + O2
5.42 g
4.00
g O2
1 mol
O2
32.00
g O2
4.00 g
→
2MgO
?g
2 mol 40.31
MgO g MgO
= 10.1 g
1 mol 1 mol MgO
O2
MgO
A. Limiting Reactants #2
Mg: 8.99 g MgO
O2: 10.1 g MgO
Limiting reactant: Mg
Excess reactant: O2
Product Formed: 8.99 g MgO
Excess oxygen
A. Limiting Reactants
b
What other information could you
find in these problems?
• How much of each reactant is
used – in grams, liters, moles
• How much of excess reactant is
left over – in grams, liters,
moles
B. Percent Yield
measured in lab
calculated on paper
B. Percent Yield
b
When 45.8 g of K2CO3 react
with excess HCl, 46.3 g of KCl
are formed. Calculate the
theoretical and % yields of KCl.
Ch. 13 - States of
Matter
Liquids & Solids
Physical
Properties
A. Liquids vs. Solids
IMF Strength
Fluid
Density
Compressible
Diffusion
LIQUIDS
SOLIDS
Stronger than
in gases
Very strong
Y
N
high
high
N
N
slower than in
gases
extremely slow
B. Liquid Properties
 Surface
Tension
• attractive force between particles in a
liquid that minimizes surface area
C. Types of Solids
 Crystalline
- repeating geometric
pattern
• covalent network
• metallic
• ionic
• covalent molecular
 Amorphous
decreasing
m.p.
- no geometric pattern
C. Types of Solids
Ionic
(NaCl)
Metallic
C. Types of Solids
Covalent
Molecular
Covalent
Network
(H2O)
(SiO2 - quartz)
Amorphous
(SiO2 - glass)
Liquids & Solids
III. Changes of State
A. Phase Changes
A. Phase Changes
 Evaporation
• molecules at the surface gain enough
energy to overcome IMF
 Volatility
• measure of evaporation rate
• depends on temp & IMF
A. Phase Changes
 Equilibrium
• trapped molecules reach a balance
between evaporation & condensation
A. Phase Changes
Pressure
• pressure of vapor above
a liquid at
equilibrium
• depends on temp & IMF
• directly related to volatility
v.p.
 Vapor
temp
v.p.
IMF
temp
v.p.
A. Phase Changes
 Boiling
Point
• temp at which v.p. of liquid
equals external pressure
• depends on Patm & IMF
• Normal B.P. - b.p. at 1 atm
Patm
b.p.
IMF
b.p.
A. Phase Changes
 Melting
Point
• equal to freezing point
IMF
 Which
m.p.
has a higher m.p.?
• polar or nonpolar?
polar
• covalent or ionic?
ionic
A. Phase Changes
 Sublimation
• solid → gas
• v.p. of solid equals
external pressure
 EX:
dry ice, mothballs,
solid air fresheners
Ch. 14 - Gases
III. Physical
Properties
A. Kinetic Molecular Theory
b
Particles in an ideal gas…
• have no volume
• have elastic collisions
• are in constant, random, straightline motion
• don’t attract or repel each other
B. Real Gases
b
Particles in a REAL gas…
• have their own volume
• attract each other
b
Gas behavior is most ideal…
• at low pressures
• at high temperatures
• in nonpolar atoms/molecules
C. Characteristics of Gases
b
Gases expand to fill any container
• random motion, no attraction
b
Gases are fluids (like liquids)
• no attraction
C. Characteristics of Gases
b
Gases can be compressed
• no volume = lots of empty space
b
Gases undergo diffusion & effusion
• random motion
State Changes
D. Describing Gases
b
Gases can be described by their:
• Temperature
•K
• Pressure
• atm
• Volume
•L
• Number of molecules/moles • #
E. Temperature
b
Always use absolute temperature
(Kelvin) when working with gases!
ºF
-459
ºC
-273
K
0
32
212
0
100
273
373
K = ºC + 273
F. Pressure
Which shoes create the most pressure?
F. Pressure
b
Barometer
• measures atmospheric pressure
b
b
exact height of the
Hg depends on
atmospheric
pressure
usually measured
in mm Hg
F. Pressure
b
Manometer
• measures contained gas pressure
b
b
Difference in
height in two arms
of U-tube is
measure of
pressure of gas
sample
measured in
various different
units
F. Pressure
b
KEY EQUIVALENT UNITS
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi
G. STP
STP
Standard Temperature & Pressure
0°C
1 atm
-OR-
273 K
101.325 kPa
H. Pressure Problem 1
b
The average pressure in Denver,
Colorado, is 0.830 atm. Express
this in (a) mm Hg and (b) kPa.
(a) 0.830 atm 760 mm Hg = 631
1 atm
mm Hg
(b) 0.830 atm 101.325 kPa= 84.1
1 atm
kPa
H. Pressure Problem 2
b
Convert a pressure of 1.75 atm to
kPa and mm Hg.
(a) 1.75 atm
101.325 kPa
= 177
1 atm
kPa
(b) 1.75 atm
760 mm Hg
1 atm
= 1330
mm Hg
H. Pressure Problem 3
b
Convert a pressure of 570. torr to
atmospheres and kPa.
(a) 570 torr
1 atm
760 torr
(b) 570 torr
= .750
atm
101.325 kPa
= 76.0
760 torr
kPa
Ch. 14 – The Behavior of Gases
V
P
T
I. The Gas
Laws
A. Boyle’s Law
b
The pressure and volume
of a gas are inversely
related
• at constant mass & temp
PV = k
P
V
A. Boyle’s Law
E. Gas Law Problems
b
A gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN: P↑ V↓ WORK:
V1 = 100. mL
P1V1 = P2V2
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
B. Charles’ Law
b
V
T
The volume and absolute
temperature (K) of a gas
are directly related
• at constant mass &
pressure
E. Gas Law Problems
b
A gas occupies 473 cm3 at 36°C.
Find its volume at 94°C.
CHARLES’ LAW
GIVEN: T↑ V↑
V1 = 473 cm3
WORK:
V1T2 = V2T1
T1 = 36°C = 309K
(473 cm3)(367 K)=V2(309 K)
V2 = ?
T2 = 94°C = 367K
V2 = 562 cm3
C. Gay-Lussac’s Law
b
P
T
The pressure and absolute
temperature (K) of a gas
are directly related
• at constant mass &
volume
E. Gas Law Problems
A gas’ pressure is 765 torr at 23°C.
At what temperature will the
pressure be 560. torr?
GAY-LUSSAC’S LAW
GIVEN: P↓ T↓ WORK:
b
P1 = 765 torr
P1T2 = P2T1
T1 = 23°C = 296K (765 torr)T2 = (560. torr)(296K)
P2 = 560. torr
T2 = 217 K = -56°C
T2 = ?
Combined Gas Law
• P 1V 1 = P 2V 2
T1
T2
• o r : P1V1T2 = P2V2T1
54
E. Gas Law Problems
b
A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
COMBINED GAS LAW
GIVEN: P↑ T↓ V↓ WORK:
V1 = 7.84 cm3
P1V1T2 = P2V2T1
P1 = 71.8 kPa
(71.8 kPa)(7.84 cm3)(273 K)
T1 = 25°C = 298 K
=(101.325 kPa) V2 (298 K)
V2 = ?
3
V
=
5.09
cm
2
P2 = 101.325 kPa
T2 = 273 K
F. The First 4 Gas Laws
b
The Gas Laws Table