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.

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

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

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)

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 = g of Al 1 mol Al and g Al g Al 1 mol Al 1.50 g Al x 1 mol Al = mol of Al g Al

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 = g of Al 1 mol Al and g Al g Al 1 mol Al 1.50 g Al x 1 mol Al = mol of Al g Al

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 mol Al x 3 mol Cl 2 = mol of Cl 2 2 mol Al

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 = 300. K V = nRT = ( mol Cl 2 )( L atm/mol K)(300. K) P 1.20 atm = 1.71 L of Cl 2

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 What volume (L) of O 2 at 24 °C and 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

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 = g of NH 3 1 mol NH 3 and g NH g NH 3 Al 1 mol NH g NH 3 x 1 mol NH 3 = 1.64 mol of NH g NH 3 Solution

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 mol NH 3 x 5 mol O 2 = 2.05 mol of O 2 4 mol NH 3 Solution (continued)

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 = 297 K Place the moles of O 2 in the ideal gas law. V = nRT =(2.05 mol)( L atm/mol K)(297 K) P atm = 52.6 L of O 2 Solution (continued)

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

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 L O 2 x 1 mol O 2 = mol of O 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 mol O 2 x 4 mol Fe = mol of Fe 3 mol O 2 Solution

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 = g of Fe 1 mol Fe and g Fe g Fe 1 mol Fe mol Fe x 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 g Fe = 18.3 g of Fe 22.4 L O 2 3 mol O 2 1 mol Fe Solution (continued)