1. 1 Chapter 11 Gases 11.9 Gas Laws and Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2. 2 Gases in Equations The volume or amount of a gas in a chemical reaction can be calculated from STP conditions or the ideal gas law. Mole factors from the balanced equation.

3. 3 STP and Gas Equations What volume (L) of O 2 gas is needed to completely react with 15.0 g of aluminum at STP? 4Al(s) + 3O 2 (g) 2Al 2 O 3 (s) Plan: g Al mol Al mol O 2 L O 2 (STP) 15.0 g Al x 1 mol Al x 3 mol O 2 x 22.4 L (STP) 26.98 g Al 4 mol Al 1 mol O 2 = 9.34 L O 2 at STP

4. 4 Ideal Gas Equation and Reactions What volume (L) of Cl 2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s)

5. 5 Ideal Gas Equation and Reactions 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 1.5 g ? L 1.2 atm, 300K 1. Calculate the moles of Cl 2 needed. 1.5 g Al x 1 mol Al x 3 mol Cl 2 = 0.083 mol Cl 2 26.98 g Al 2 mol Al 2. Place the moles Cl 2 in the ideal gas equation. V = nRT =(0.083 mol Cl 2 )(0.0821 L atm/mol K)(300 K) P 1.2 atm = 1.7 L Cl 2

6. 6 What volume (L) of O 2 at 24°C and 0.950 atm is needed to react with 28.0 g NH 3 ? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Learning Check

7. 7 1. Calculate the moles of O 2 needed. 28.0 g NH 3 x 1 mol NH 3 x 5 mol O 2 17.03 g NH 3 4 mol NH 3 = 2.06 mol O 2 2. Place the moles of O 2 in the ideal gas equation. V = nRT =(2.06 mol)(0.0821 L atm/mol K)(297 K) P 0.950 atm = 52.9 L O 2 Solution