Practice Problem   If 10.0 g if sodium peroxide, Na2O2 , reacts with water to produce sodium hydroxide and oxygen, how many liters of oxygen will be produced at 20. oC and 750. torr? P1V1 = P2V2 T1 T2 PV = nRT 1 atm = 760 torr R = 0.0821 atm*L / mol*K 2 Na2O2 (s) + 2 H2O (l) → 4 NaOH (aq) + O2 (g)

2 Na2O2 (s) + 2 H2O (l) → 4 NaOH (aq) + O2 (g) 10.0 g Na2O2 (1mol) = 0.128 mol Na2O2 (78.0g) 0.128 mol Na2O2 (1 mol O2) = 0.0641 mol O2 (2 mol Na2O2) P1V1 = P2V2 T1 T2 PV = nRT

PV = nRT P = 750 760 V = nRT = 0.0641 mol O2 (0.0821 atm L / mol K) 293 K P 0.987 atm V = 1.56 L

P1V1 = P2V2 T1 T2 0.0641 mol O2 (22.4 L) = 1.44 L O2 @ STP (1 mol) 760 torr 1.44 L = 750 torr V V = 1.56 L 273K 293 K

1 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) Practice Problem 1 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)   What volume of oxygen will react with 15.0 L of propane (C3H8) to form carbon dioxide and water? What volume of carbon dioxide will be formed? What volume of water vapor will be formed? All the components are gases so the mole ratios = volume ratios when the gases are at the same pressure and temperature.

1 C3H8 + 5 O2 → 3 CO2 + 4 H2O 1 C3H8 15.0 L C3H8 X = 75.0 L O2 5 O2 = X L O2   1 C3H8 15.0 L C3H8 X = 45.0 L CO2 3 CO2 = X L CO2 1 C3H8 15.0 L C3H8 X = 60.0 L H2O 4 H2O = X L H2O

Do Problems: 38 & 40 pg 461 42 & 44 pg 463