1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 11 Gases 11.9 Gas Laws and Chemical Reactions Under water, the pressure on a diver is greater than the atmospheric pressure.

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 Gases in Equations The volume or amount of a gas in a chemical reaction can be calculated from the ideal gas law mole-mole factors from the balanced equation molar mass

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Reactions Involving Gases 3

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Example of Using the Ideal Gas Law with an Equation What volume, in L, of Cl 2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s)

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 Example of Using the Ideal Gas Law with an Equation (continued) STEP 1 Calculate the moles of given using molar mass or ideal gas law. 1 mol of Al = 26.98 g of Al 1 mol Al and 26.98 g Al 26.98 g Al 1 mol Al 1.50 g Al x 1 mol Al = 0.0556 mol of Al 26.98 g Al

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 Example of Using the Ideal Gas Law with an Equation (continued) STEP 1 Calculate the moles of given using molar mass or ideal gas law. 1 mol of Al = 26.98 g of Al 1 mol Al and 26.98 g Al 26.98 g Al 1 mol Al 1.50 g Al x 1 mol Al = 0.0556 mol of Al 26.98 g Al

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 Example of Using the Ideal Gas Law with an Equation (continued) STEP 2 Determine the moles of needed using a mole-mole factor. 0.0556 mol Al x 3 mol Cl 2 = 0.0834 mol of Cl 2 2 mol Al

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 Example of Using the Ideal Gas Law with an Equation (continued) STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law. To determine liters of gas, use the ideal gas law arranged to solve for V. T = 27 °C + 273 = 300. K V = nRT = (0.0834 mol Cl 2 )(0.0821 L atm/mol K)(300. K) P 1.20 atm = 1.71 L of Cl 2

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 What volume (L) of O 2 at 24 °C and 0.950 atm is needed to react with 28.0 g of NH 3 ? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Learning Check

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 STEP 1 Calculate the moles of given using molar mass or ideal gas law. 1 mol of NH 3 = 17.03 g of NH 3 1 mol NH 3 and 17.03 g NH 3 17.03 g NH 3 Al 1 mol NH 3 28.0 g NH 3 x 1 mol NH 3 = 1.64 mol of NH 3 17.03 g NH 3 Solution

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 STEP 2 Determine the moles of needed using a mole-mole factor. 5 mol of O 2 = 4 mol of NH 3 4 mol NH 3 and 5 mol O 2 5 mol O 2 4 mol NH 3 1.64 mol NH 3 x 5 mol O 2 = 2.05 mol of O 2 4 mol NH 3 Solution (continued)

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law. To determine liters of gas, use the ideal gas law arranged to solve for V. T = 24 °C + 273 = 297 K Place the moles of O 2 in the ideal gas law. V = nRT =(2.05 mol)(0.0821 L atm/mol K)(297 K) P 0.950 atm = 52.6 L of O 2 Solution (continued)

13. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 What mass of Fe will react with 5.50 L of O 2 at STP? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 1) 13.7 g of Fe 2) 18.3 g of Fe 3) 419 g of Fe Learning Check

14. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 STEP 1 Calculate the moles of given using molar mass or ideal gas law. Use molar volume at STP to calculate moles of O 2. 5.50 L O 2 x 1 mol O 2 = 0.246 mol of O 2 22.4 L O 2 STEP 2 Determine the moles of needed using a mole- mole factor. 4 mol of Fe = 3 mol of O 2 4 mol Fe and 3 mol O 2 3 mol O 2 4 mol Fe 0.246 mol O 2 x 4 mol Fe = 0.328 mol of Fe 3 mol O 2 Solution

15. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law. 1 mol of Fe = 55.85 g of Fe 1 mol Fe and 55.85 g Fe 55.85 g Fe 1 mol Fe 0.328 mol Fe x 55.85 g Fe = 18.3 g of Fe 1 mol Fe Placing all three steps in one setup gives (STEP 1) (STEP 2) (STEP 3) 5.50 L O 2 x 1 mol O 2 x 4 mol Fe x 55.85 g Fe = 18.3 g of Fe 22.4 L O 2 3 mol O 2 1 mol Fe Solution (continued)