1. Dalton’s Law of Partial Pressures and Gas Law Stoichiometry
Honors Chemistry

2. Dalton’s Law of Partial Pressures
The pressure of each gas in a mixture is called the partial pressure of that gas.
Dalton’s Law states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each of the component gases.
Ptotal = P1 + P2 + P

3. Ptotal = P1 + P2 + P3 + . . . Ptotal = 2.4 + 6.0 = 8.4 atm
PH2 = 2.4 atm
PHe = 6.0 atm
Ptotal = =
8.4 atm

4. If the first three containers are all put into the fourth, we can find the pressure in that container by adding up the pressure in the first 3:
2 atm
+ 1 atm
+ 3 atm
= 6 atm 1 2 3 4

5. The partial pressure of each gas is equal to the mole fraction (X) of each gas times the total pressure
Moles gas x Ptotal = Pgas
Total Moles
Mole fraction is like a percent

6. Sample Problem #1
A mixture of He, Ne and Ar gases have a total pressure of 790 mmHg. If there is 15% Ar, 60% He and 25% Ne, what is the partial pressure of each gas?
Ptotal = PHe + PNe + PAr = 790 mmHg
15% Ar = (0.15)(790 ) = mmHg +
60% He = (0.60)(790) = mmHg +
25% Ne = (0.25)(790) = mmHg
P total = mmHg

7. Sample Problem #2
The partial pressure of CO2 in a mixture of gases is 0.8 atm. If the total pressure is 1.05 atm, what is the mole fraction of CO2 in the mixture?
PCO2 = 0.8atm Ptotal = 1.05atm
(Mole Fraction )Ptotal = PCO2
XCO2 (1.05atm) = 0.8 atm
XCO2 = atm = 0.76
1.05 atm

8. Gas Collected by Water Displacement
A gas collected by water displacement is not pure, but always mixed with water vapor. Some molecules at the surface of water always evaporate and exert a pressure called a vapor pressure.
Therefore, you must subtract the water –vapor pressure from the total pressure to get the pressure just from the gas.

9. Sample Problem #3
50mls of CH4 was collected over water at 25oC and 748mmHg. What is the volume of the gas at STP? (The vapor pressure of water is 23.8mmHg)
Ptotal = PCH4 + PH2O
748mmHg = PCH mmHg
PCH4 = mmHg (dry gas)
P1V1 = P2V (50mls)(724.2mmHg) = (760mmHg)V2
T T K K
Solving for V2 = 43.6 mL

10. Gas Law Stoichiometry
What volume of oxygen will form at 25oC and 1.3atm when 50 grams of KClO3 decomposes according to the following equation?
2KClO3  2KCl + 3O2
First you find out how many moles of oxygen gas will form using stoichiometry =
0.61 moles O2
50 g KClO3
1 mole KClO3
3 moles O2
g KClO3
2 moles KClO3
Then you use the Ideal gas law to get the volume. You cannot use the 22.4 L per 1 mole relationship because you are not at STP!!!!
(1.3atm)V = (0.61 moles) ( L.atm/K.mole) (298K)
V = Liters