1
Chapter 10-Gas Mixtures & Reactions
10.1 Mixtures of Gases
 Dalton hypothesized that gas particles behaved independently and
that the pressure exerted by a particular is the same whether it is
by itself or is part a mixture of gases(if temperature is constant)
 see Table 1 p.460 in text
Table 1: Dalton’s Analysis of Gases in the Atmosphere
Chemical name of the component
Dalton
IUPAC
azotic gas
oxygenous gas
aqueous vapour
carbonic acid gas
nitrogen
oxygen
water
carbon dioxide
Pressure
(mm
(kPa)
Hg)
593.3
79.11
157
20.9
11
1.5
0.5
0.07
Composition
%
78.08
20.95
varies
0.04
 this resulted in Dalton’s Law of Partial Pressures
 states “ the total pressure of a mixture of non-reacting gases is
equal to the sum of the partial pressures of the individual gases
i.e. PTOTAL=P1+P2+P3+…….
 Partial pressure is the pressure (P) that a gas in a mixture
would exert if it were the only gas present in the same volume
and at the same temperature
Sample Problem 1
Q: In a compressed air tank for scuba diving to a depth of 30m, a
mixture with an oxygen partial pressure of 28 atm and a nitrogen
partial pressure of 110 atm is used. What is the total pressure in
the tank?
A: PO2=28 atm
PN2=110 atm
2
PTOTAL= ?
PTOTAL= PO2 + PN2
=28 atm + 110 atm
PTOTAL= 138 atm
Finish questions 1-4 on p. 461
Therefore, the total pressure in the tank is 138 atm.
 Dalton’s Law of Partial Pressure can be explained by 2 concepts of
KMT;
1. the pressure of a gas is caused by collisions of molecules
with the walls of the container.
2. gas molecules act independently of each other
Finish questions 5-7 on page 463.
Applications of Partial Pressure
1. Gas exchange between living organisms and the environment
depends on solubility and partial pressure.
2. Collection of gases by displacement of water.
e.g. H2 + O2
 water vapour is a gas and the pressure exerted by a gas above a
liquid is called its vapour pressure
 vapour pressure of water at different temperatures is well known
Fig. 3 p.464
 we can use Dalton`s Law of Partial Pressures and a table of
known vapour pressures of water to determine the pressure
3
of “dry” gas that has been collected
Sample Problem 2
Q: In a lab, oxygen gas was collected by water displacement at an
atmospheric pressure of 96.8kPa and at a temperature of 22ºC.
Calculate the partial pressure of dry oxygen.
A: PTOTAL=96.8kPa
PH2O=2.64kPa
PO2 = ?
PTOTAL= PO2 + PH2O
PO2 = PTOTAL - PH2O
PO2 = 96.8kPa – 2.64kPa
PO2 = 94.2kPa
Therefore the partial pressure of the dry oxygen is 94.2kPa.
Do questions 8-10 on page 465.
10.2 Reactions of Gases
 based on work by Joseph Gay-Lussac ( 1809)
 Law of Combining Volumes – when measured at the same
temperature & pressure, volumes of gaseous reactants and
products of chemical reactions are always in simple ratios of whole
numbers
4
e.g. hydrogen + chlorine----hydrogen chloride
1.0L
1.0L
2.0L
Ratio: 1
:
1
:
2
 when gases are at the same temperature & pressure, we can use
the Law of Combining Volumes to predict volumes of other gases
involved in a chemical reaction
(2 years later-Avogadro proposed Avogadro’s Theory)
e.g. burning butane in a lighter
2C4H10(g) + 13O2(g) -----8CO2(g) + 10H2O(g)
amounts:
volumes:
or
2mol
2L
4L
13mol
13L
26L
8mol
8L
16L
10mol
10L
20L
if we have 120mL of butane, how much O2 is needed?
For every 13mol of O2 we need 2mol of C4H10 ,
therefore the volume of O2 would be greater than 120mL by a factor
of 13/2;
therefore VO2=120mL x 13/2 = 780mL
Sample Problem 1
Q: A catalytic converter in the exhaust system of a car uses oxygen
(from the air) to convert carbon monoxide to carbon dioxide,
which is released through the tailpipe. If we assume the same
temperature and pressure, what volume of oxygen is required to
react with 125L of carbon monoxide produced during a 100km
trip?
A: 2CO(g) + O2(g)-----2CO2(g)
125L
V=?
VO2=125L CO x 1mol/2mol
=62.5L ,therefore the volume of O2 required is 62.5L
Answer questions 1, 3-5 on pages 468 & 469.
5
 to suggest an explanation between volume & coefficient ratios
Avogadro proposed a theory that stated “equal volumes of
gases at the same temperature and pressure would contain
equal numbers of molecules”
Molar Volume of Gases
 for all gases at a special temperature & pressure, there must be a
certain volume that contains one mole of particles…….called the
molar volume.
 it has been determined empirically that the molar volume of a gas
at SATP is 24.8L/mol
 the molar volume of a gas at STP is 22.4L/mol
which gives us n=v/V where n=moles, v=volume and V=molar
volume
Sample problem 2
Q: What volume is occupied by 0.024mol of carbon dioxide gas at
SATP?
A: nCO2=0.024mol
VSATP=24.8L/mol
V= ?
N=v/V, v=n x V
V=0.024mol x 24.8L/mol
=0.60L
therefore the volume of the carbon dioxide gas is 0.60L
6
Sample Problem 3
Q: What amount of oxygen in moles is available for a combustion
reaction in a volume of 5.6L at STP?
A: vO2=5.6L
VSTP=22.4L/mol
nO2= ?
n= v --- nO2=5.6L
V
22.4L/mol
nO2=0.25mol
therefore the amount of oxygen available is 0.25mol.
 gases can be & are often liquefied for transport & storage (e.g. O2
& N2)
 liquefied & compressed gases are sold by mass
 we can use molar volume & molar mass to find the volume of gas
available
Q: What volume does 3.5g of helium gas occupy at SATP?
A: nHe=3.50g/ 4.00g/mol
=0.875mol
now convert to volume at SATP
n=v/V so v= n x V
vHe= 0.875mol x 24.8L/mol
= 21.7L
therefore 3.5g of helium gas occupies 21.7L at SATP.
or
7
One Step: vHe=3.50g x 1mol x 24.8L
4.00g 1mol
at SATP.
vHe=21.7L,
therefore 3.5g of helium gas occupies 21.7L
Sample Problem 4
Q: A propane tank for a BBQ contains liquefied propane. If the tank
mass drops by 94kg after a months use, what volume of propane
gas at SATP was used for cooking?
A: mC3H8=9.1kg
MC3H8=44.11g/mol
VSATP=24.8L/mol
nC3H8= 9.1kg/44.11g/mol
= 0.21mol
vC3H8=0.21mol x 24.8L/mol
vC3H8= 5.2kL
or
vC3H8=9.1kg x 1 mol x 24.8L
44.11g
1 mol
vC3H8=5.2kL
therefore the volume of propane used was 5.2kL
Answer questions 7-11 on page 471.
Read & make notes on 10.3-“The Ozone Layer” (p.475-479)
& answer questions 1-7 on page 479
8
10.4 Gas Stoichiometry
Summary of Gravimetric, Gas & Solution Stoichiometry
STEP 1: Write a balanced chemical equation and list the
measurements and conversion factors for the given substance and
the one to be calculated.
STEP 2: Convert the measurement to an amount in moles using the
appropriate conversion factor.
STEP 3: Calculate the amount of the other substance by using the
mole ratio from the balanced equation.
STEP 4: Convert the calculated amount to the final quantity
requested by using the appropriate conversion factor.
Example:
Q: If 275g of propane burns in a gas BBQ, what volume of oxygen
measured at STP is required for the reaction?
A: Step 1:
C3H8(g)
+
5O2(g)=3CO2(g)+4H2O(g)
m=275g
v= ?
M=44.11g/mol V=22.4L/mol
Step 2:
nC3H8=m/M, nC3H8=275g/44.11g/mol
=6.23mol
Step 3:
Use the mole ratio to calculate the amount of oxygen required
i.e. 5 moles of oxygen are needed for every mole of propane
nO2=6.23mol C3H8 x 5mol O2
1mol C3H8
nO2=31.2mol
9
Step 4:
Convert the required amount of O2 (in moles) to the required
quantity (i.e. volume)
i.e. n=v/V, vO2=n x V
=31.2mol x 22.4L/mol
=698L
therefore the volume of oxygen required is 698L.
Sample Problem 1
Q: Hydrogen gas is produced when sodium metal is added to water.
What mass of sodium is necessary to produce 20.0L of hydrogen
at SATP?
A: step 1:
2Na(s) +
2H2O(l) ----- H2(g) + 2NaOH(aq)
m= ?
v=20.0L
M=22.99g/mol
V=24.8L/mol
step 2:
nH2=20.0L/24.8L/mol
=0.806mol
step 3:
nNa=0.806molH2 x 2mol Na
1mol H2
=1.61mol
step 4:
mNa=n x M
=1.61mol x 22.9g/mol
=37.1g
therefore the mass of sodium necessary is 37.1g.
10
Sample Problem 2
Q: What volume of ammonia at 450kPa pressure and 80ºC can be
contained from the complete reaction of 7.5kg of hydrogen?
A: N2(g) + 3H2(g) ---------2NH3(g)
m=7.5kg
P=450kPa
M=2.02g/mol
T=80ºC=353K
R=8.31 kPa*L
Mol*K
nH2=7.5kg
2.02g/mol
nH2=3.7kmol
nNH3=3.7kmol H2 x 2mol NH3
3mol H2
nNH3=2.5kmol
PV=nRT
V=nRT
P
VNH3= 2.5kmol x 8.31 kPa*L/mol*K x 353K
450kPa
VNH3=16kL
therefore the value of ammonia obtained is 16kL.
Do questions 1-3 on page 483.
11
10.5 Applications Of Gases
meteorology
 high and low pressures
 atmosphere is a mixture of gases
 weather balloons
medicine
 artificial ventilation
 use of anaesthetics
other applications
 pressure in tires
 aerosol cans in pressurized aircraft
 deep-sea diving ( air forced into your lungs)-if you come to
surface too quickly N2 cannot leave your body fast enough –
bubbles form in your blood vessels causing “the bends” –
a very painful and serious condition.
Answer Making Connections Questions 4-6 on page 489
& Section 10.5 Questions 1-3 on bottom of same page.