Chapter 5
Gases - 2
Stoichiometry
Dr. Sapna Gupta
Stoichiometry in Gases
Amounts of gaseous reactants and products can be calculated by
utilizing
− The ideal gas law to relate moles to T, P and V.
− Moles can be related to mass by the molar mass
− The coefficients in the balanced equation to relate moles of
reactants and products
Dr. Sapna Gupta/Gases-Stoichiometry
2
Solved Problem:
When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required to
neutralize the spill. What volume of CO2 was released by the neutralization at 735
mmHg and 20.°C?
First, write the balanced chemical equation:
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
Second, calculate the moles of CO2 produced:
Molar mass of CaCO3 = 100.09 g/mol
1.2  103 g CaCO3
1mol CaCO3
1mol CO2
100.09 g CaCO3 1mol CaCO3
= 11.99 mol
n = 11.99 mol
nRT
P = 735 mmHg
V 
P
= 0.967 atm
T = 20°C = 293 K
11.99 mol 0.08206L  atm (293 K)
mol  K 
= 2.98 × 102 L

V
(3 significant figures)
(0.967 atm)
Dr. Sapna Gupta/Gases-Stoichiometry
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Dalton’s Law of Partial Pressure
• Dalton found that in a mixture of unreactive gases, each gas acts as if
it were the only gas in the mixture as far as pressure is concerned.
• The sum of the partial pressures of all the different gases in a mixture
is equal to the total pressure of the mixture:
P = PA + PB + PC + . . .
Partial Pressure
The pressure exerted by
a particular gas in a mixture
PN2
Dr. Sapna Gupta/Gases-Stoichiometry
Ptotal = PN2 + O2
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Solved Problem:
A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain
0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is
the partial pressure of each component and the total pressure of the sample?

1 mol N2
 0.0830 g N2
28.01 g N2


L  atm 
 0.08206
308 K 
mol

K





100.0 mL  13L 
 10 mL 
PN2 
PO2 
 0.749 atm

1 mol O 2
 0.0194 g O 2
32.00 g O 2


L  atm 
 0.08206
308 K 
mol  K 



100.0 mL  13L 
 10 mL 
 0.153 atm
PCO2

1 mol CO 2 
L  atm 
 0.00640 g CO 2
 0.08206
308 K 
44.01 g CO 2 
mol  K 




100.0 mL  13L 
 10 mL 
PH2O

1 mol H2 O 
L  atm 
 0.00441 g H2 O
 0.08206
308 K 
18.01 g H2 O 
mol  K 




100.0 mL  13L 
 10 mL 
P  PN2  PO2  PCO2  PH2O
 0.0368 atm
 0.0619 atm
= 1.00 atm
Dr. Sapna Gupta/Gases-Stoichiometry
5
Solved Problem:
The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows:
nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water
vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?
PN2  570.0 mmHg
PO2  103.0 mmHg
Mole fraction of N2
PCO2  40.0 mmHg
PH2O  47.0 mmHg
P  PN2  PO2  PCO2  PH2O
570.0 mmHg
103.0 mmHg
40.0 mmHg
47.0 mmHg
P = 760.0 mmHg
Mole fraction of CO2
Mole fraction of O2
Mole fraction of H2O
Dr. Sapna Gupta/Gases-Stoichiometry

570.0 mmHg
760.0 mmHg
= 0.7500

40.0 mmHg
760.0 mmHg
= 0.1355

103.0 mmHg
760.0 mmHg
= 0.0526

47.0 mmHg
760.0 mmHg
= 0.0618
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Collecting Gas Over Water
• Gases are often collected over water. The result is a mixture of the gas
and water vapor.
• The total pressure is equal to the sum of the gas pressure and the
vapor pressure of water.
• The partial pressure of water depends only on temperature and is
known (Table 5.6).
• The pressure of the gas can then be found using Dalton’s law of
partial pressures.
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P  PH2  PH2O
PH2  P  PH2O
PH2  769 mmHg  16.5 mmHg
PH2  752.5 mmHg
PH2  753 mmHg
(no decimal places)
Dr. Sapna Gupta/Gases-Stoichiometry
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Solved Problem
You prepare nitrogen gas by heating ammonium nitrite:
NH4NO2(s)  N2(g) + 2H2O(l)
If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of
gas would you obtain from 5.68 g NH4NO2?
P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg
T = 23°C = 296 K
5.68 g NH4NO2
Molar mass NH4NO2 = 64.06 g/mol
1mol NH4NO2
1mol N2
64.04 g NH4NO2 1mol NH4NO2
= 0.08887 mol N2
n = 0.0887 mol
V
nRT
P
0.0887 mol  0.08206 L  atm (296 K)
V 
mol  K 

1 atm 
 706 mmHg

760 mmHg 


Dr. Sapna Gupta/Gases-Stoichiometry
= 2.32 L of N2
(3 significant figures)
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Solved Problem
Oxygen was produced and collected over water at 22ºC and a pressure of 754 torr.
2 KClO3(s)  2 KCl(s) + 3 O2(g)
325 mL of gas were collected and the vapor pressure of water at 22ºC is 21 torr.
Calculate the number of moles of O2 and the mass of KClO3 decomposed.
Ptotal = PO + PH
2
2O
= PO + 21 torr = 754 torr
2
PO = 754 torr – 21 torr = 733 torr = 733 / 760 atm
2
V = 325 mL = 0.325 L
T = 22ºC + 273 = 295 K
nO

2
 733 760 atm   0.325 L 
 0.08206 L  atm  (295 K)


mol  K 

 1.29 102 mol O 2
2 KClO3(s)  2 KCl(s) + 3 O2(g)
 2 mol KClO3  122.6 g KClO3 
 

1.29 10 mol O 2 
 3 mol O 2  1 mol KClO3 
 1.06 g KClO3
2
Dr. Sapna Gupta/Gases-Stoichiometry
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Applications of the Gas Laws
• Gases are compressible because the gas molecules
are separated by large distances.
• The magnitude of P depends on how often and with
what force the molecules strike the container walls.
• At constant T, as V increases, each particle strikes the
walls less frequently and P decreases.
• To maintain constant P, as V increases T must
increase; fewer collisions require harder collisions.
• To maintain constant P and T, as V increases n must
increase.
• Gas molecules do not attract or repel one another, so
one gas is unaffected by the other and the total
pressure is a simple sum.
Dr. Sapna Gupta/Gases-Stoichiometry
(Boyle’s Law)
(Charles’ Law)
(Avogadros’ Law)
(Dalton’s Law)
11
Speed of Gas
• Root mean square (rms) speed (urms)
• For two gases (1 and 2)
• Effect of Temperature on Molecular Speed (1st graph)
• Effect of Molar Mass on Molecular Speed (2nd graph)
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Diffusion and Effusion
Diffusion
Effusion
The process whereby a gas spreads out
through another gas to occupy the space
uniformly.
Below NH3 diffuses through air. The
indicator paper tracks its progress.
The process by which a gas flows
through a small hole in a container. A
pinprick in a balloon is one example of
effusion.
Dr. Sapna Gupta/Gases-Stoichiometry
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Real Gases
At high pressure the relationship between pressure and volume does not
follow Boyle’s law. This is illustrated on the graph below.
At high pressure, some assumptions of the kinetic theory no longer hold
true. At high pressure:
1. the volume of the gas molecule is not negligible.
2. the intermolecular forces are not negligible.
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Van der Waal’s Equation
An equation that is similar to the ideal gas law, but which includes two
constants, a and b, to account for deviations from ideal behavior.
The term V becomes (V – nb) to account for the space between
molecules.
The term P becomes (P + n2a/V2) to account for attraction/repulsion
between molecules.
Values for a and b are different for different gases and can be found in
Table 5.7
• The ideal gas law
becomes van der Waal’s equation
a and b have specific values for each gas
Dr. Sapna Gupta/Gases-Stoichiometry
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Key Points
• Properties of gases
• Gas pressure
• Units
• Calculation
• Measurement
•
• The gas laws
• Boyle’s law
• Charles’ law
• Avogadro’s law
•
The kinetic molecular theory
• Assumptions
• Application to the gas laws
• Molecular speed
• Diffusion and effusion
Deviation from ideal behavior
• Factors causing deviation
• Van der Waal’s equation
• The ideal gas law
• Reactions with gaseous reactants and products
• Gas mixtures
• Dalton’s law
• Mole fractions
• Partial pressures
Dr. Sapna Gupta/Gases-Stoichiometry
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