1. Ch Gases
II. Ideal Gas Law
Ideal Gas Law and Gas Stoichiometry

2. Part 1
Ideal Gas Law

3. Standard Temperature & Pressure
A. Avogadro’s Principle
Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature & Pressure
0°C and 1 atm

4. A. Avogadro’s Principle
Equal volumes of gases contain equal numbers of moles
at constant temp & pressure
true for any gas
n = number of moles V n

5. UNIVERSAL GAS CONSTANT
B. Ideal Gas Law
Merge the Combined Gas Law with Avogadro’s Principle: V n PV nT PV T
= k
= R
UNIVERSAL GAS CONSTANT
R= Latm/molK
R=8.315 dm3kPa/molK

6. UNIVERSAL GAS CONSTANT
B. Ideal Gas Law
PV=nRT
UNIVERSAL GAS CONSTANT
R= Latm/molK
R=8.315 dm3kPa/molK

7. C. Ideal Gas Law Problems
Calculate the pressure in atmospheres of mol of He at 16°C & occupying L.
GIVEN:
P = ? atm
n = mol
T = 16°C = 289 K
V = 3.25 L
R = Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289)
L mol Latm/molK K
P = 3.01 atm

8. C. Ideal Gas Law Problems
Find the volume of 85 g of O2 at 25°C and kPa.
GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = kPa
R = dm3kPa/molK
WORK:
85 g 1 mol O2 = 2.7 mol
32.00 g O2
= 2.7 mol
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol dm3kPa/molK K
V = 64 dm3

9. D. Applications of Ideal Gas Law
Can be used to calculate the molar mass of a gas from the density
Substitute this into ideal gas law
And m/V = d in g/L, so

10. D. Applications of Ideal Gas Law
The density of a gas was measured at atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas.
GIVEN:
P = 1.50 atm
T = 27°C = 300. K
d = 1.95 g/L
R = Latm/molK
MM = ?
WORK:
MM = dRT/P
MM=(1.95)( )(300.)/1.50
g/L Latm/molK K atm
MM = 32.0 g/mol

11. D. Applications of Ideal Gas Law
Calculate the density of carbon dioxide gas at 25°C and 750. torr.
GIVEN:
d = ? g/L CO2
T = 25°C = 298 K
P = 750. torr
R = Latm/molK
MM = g/mol
WORK:
750 torr 1 atm = .987 atm torr
MM = dRT/P →d = MM P/RT
d=(44.01 g/mol)(.987 atm)
( Latm/molK )(298K)
d = 1.78 g/L CO2
= .987 atm

12. Part 2
Gas Stoichiometry

13. * Stoichiometry Steps Review *
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
Mole ratio - moles  moles
Molar mass - moles  grams
Molarity - moles  liters soln
Molar volume - moles  liters gas
Mole ratio - moles  moles
Core step in all stoichiometry problems!!
4. Check answer.

14. Standard Temperature & Pressure
A. Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature & Pressure
0°C and 1 atm

15. A. Molar Volume at STP LITERS OF GAS AT STP Molar Volume MASS IN GRAMS
(22.4 L/mol)
MASS IN
GRAMS
MOLES
NUMBER OF
PARTICLES
Molar Mass
(g/mol)
6.02  1023
particles/mol

16. B. Gas Stoichiometry Liters of one Gas  Liters of another Gas:
Avogadro’s Principle
Coefficients give mole ratios and volume ratios
Moles (or grams) of A  Liters of B:
STP – use 22.4 L/mol
Non-STP – use ideal gas law & stoich
Non-STP
Given liters of gas?
start with ideal gas law
Looking for liters of gas?
start with stoichiometry conv

17. C. Gas Stoichiometry - Volume
What volume of oxygen is needed for the complete combustion of 4.00 L of propane (C3H8)?
C3H O2  CO2+ H2O
4.00 L ?L
5 L O2
1 L C3H8
4.00 L C3H8
= 20.0 L O2

18. C. Gas Stoichiometry Problem – STP
How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP?
2KClO3  2KCl + 3O2
? g
9.00 L
9.00 L O2
1 mol O2
22.4 L
2 mol
KClO3
3 mol O2
122.55
g KClO3
1 mol
KClO3
= 32.8 g
KClO3

19. D. Gas Stoichiometry Problem – Non-STP
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?
CaCO3  CaO CO2
5.25 g
? L non-STP
Looking for liters: Start with stoich and calculate moles of CO2.
P = 103 kPa
V = ?
n = ?
R = dm3kPa/molK
T = 298K
5.25 g
CaCO3
1 mol
CaCO3
100.09g
1 mol
CO2
CaCO3
NEXT 
= mol CO2
Plug this into the Ideal Gas Law for n to find liters

20. D. Gas Stoichiometry Problem – Non-STP
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPa
V = ?
n = mol
T = 25°C = 298 K
R = dm3kPa/molK
WORK:
PV = nRT
(103 kPa)V =(0.0525mol)(8.315dm3kPa/molK) (298K)
V = 1.26 dm3 CO2

21. B. Gas Stoichiometry Problem
How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0 L
non-STP
? g
GIVEN:
P = 97.3 kPa
V = 15.0 L
n = ?
T = 21°C = 294 K
R = dm3kPa/molK
WORK:
PV = nRT
(97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K)
n = mol O2
Given liters: Start with Ideal Gas Law and calculate moles of O2.
NEXT 

22. B. Gas Stoichiometry Problem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0L
non-STP
? g
Use stoich to convert moles of O2 to grams Al2O3
0.597
mol O2
2 mol Al2O3
3 mol O2
g Al2O3
1 mol
Al2O3
= 40.6 g Al2O3