1. Gas Mixtures Because the gas laws apply to ideal gases, they also apply to gas mixtures. Laws frequently used: Ideal gas law Dalton’s Law for partial pressures (including mole fraction)

2. Collection of Gases By Water Displacement Example:2KClO 3 (g) 2KCl (s) + 3O 2 (g) Example:2KClO 3 (g)  2KCl (s) + 3O 2 (g)

3. Dalton’s Law of Partial Pressures

4. Dalton’s Law (based on Avogadro’s Law) For two gases in a mixture: P total = total gas pressure P A = partial pressure of gas ‘A’ (etc.)   P total = P A + P B +....  The total pressure of a gas mixture is the sum of the partial pressures of the components of the gas.

5. Dalton’s Law  P N2 = 0.78 atm P O2 = 0.21 atm Example: air is 78% N 2, 21% O 2, and 1% other gases. At 1 atm:

6. Example - 1  A mixture of gases in scuba diving tank: (at 25 o C, 1atm) He...... 46 L O 2...... 12 L tank volume = 5.0 L ¿ (a)partial pressure of each gas? (b)total pressure inside tank? Strategy – for (a) (i)to use PV = nRT, we need the number of moles of each. (ii)then we can determine the partial pressure for each

7.  A mixture of gases in scuba diving tank: (at 25 o C, 1atm) He...... 46 L O 2...... 12 L tank volume = 5.0 L ¿ (i) number of moles of each gas? (ii) partial pressure of each gas? (a)partial pressure of each gas?

8.  A mixture of gases in scuba diving tank: (at 25 o C, 1atm) He...... 46 L O 2...... 12 L tank volume = 5.0 L ¿ (b) total pressure inside tank? P total = P A + P B +.... P total = 9.3 atm + 2.4 atm = 11.7 atm

9.  Example - 2 Potassium chlorate (KClO 3 ) was heated in a test tube and decomposed by the following reaction: ¿ Question: a. a.What is the partial pressure of O 2 in the gas collected? b. b.What was the mass of KClO 3 in the original sample? 2KClO 3 (s)  2KCl (s) + 3O 2 (g)  The oxygen is collected by water displacement at 22 o C, at a total pressure of 754 torr, for a total volume of 0.650 L. (P H2O = 21 torr.)

10.  ¿ 2KClO 3 (s)  2KCl (s) + 3O 2 (g) (a) What is the P O2 ? P total = P O2 + P H2O P O2 = 754 torr – 21 torr = 733 torr

11.  ¿ 2KClO 3 (s)  2KCl (s) + 3O 2 (g) (b) What was the mass of KClO 3 in the original sample? iii. Number of moles of O 2 ii. Number of moles of KClO 3 / i. Grams KClO 3

12.  Example - 3 Two bulbs are separated by a valve. ¿ BulbGasP init V init A Ne1.09 atm1.12 L B CO0.773 atm2.18 L When the valves are opened, and the gases are allowed to reach equilibrium, what is the final pressure inside the bulbs? (Assume constant temperature.)

13.  ¿ Strategy (iv) Final pressure (iii) pressure from PV = nRT (T total = V A + V B + V C ) (ii) total number of moles (n total = n A + n B + n C ) (i) mole of each gas

14. ¿ (i) mole of each gas: (ii)total moles: (We don’t know Temp, but RT will cancel when calculating P total )

15. ¿ (iii) P Total :

16. Mole Fraction The ratio of the number of moles of given component in a mixture to the total number of moles in the mixture.  for component gas, A: Combining ideal gas laws for both, and cancelling out R, V and T, and rearrange: The fraction is called the mole fraction of A (= X A ) P A = X a P total