1 Chapter 5 Chemical Quantities and Reactions 5.8 Mass Calculations for Reactions Copyright © 2009 by Pearson Education, Inc.

2 Moles to Grams Suppose we want to determine the mass (g) of NH 3 that can form from 2.50 moles of N 2. N 2 (g) + 3H 2 (g) 2NH 3 (g) The plan needed would be moles N 2 moles NH 3 grams NH 3 The factors needed would be: mole factor NH 3 /N 2 and the molar mass NH 3

3 Moles to Grams The setup for the solution would be: 2.50 mole N 2 x 2 moles NH 3 x 17.0 g NH 3 1 mole N 2 1 mole NH 3 given mole-mole factor molar mass = 85.0 g of NH 3

4 How many grams of O 2 are needed to produce mole of Fe 2 O 3 in the following reaction? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 1) 38.4 g of O 2 2) 19.2 g of O 2 3) 1.90 g of O 2 Learning Check

5 Solution 2) 19.2 g of O mole Fe 2 O 3 x 3 mole O 2 x 32.0 g O 2 = 19.2 g of O 2 2 mole Fe 2 O 3 1 mole O 2 mole factor molar mass

6 The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? 2 H 2 (g) + O 2 (g) 2 H 2 O(g) ? g 13.1 g The plan and factors would be g H 2 O mole H 2 O mole O 2 g of O 2 molar mole-mole molar mass H 2 O factor mass O 2 Calculating the Mass of a Reactant

7 The setup would be: 13.1 g H 2 O x 1 mole H 2 O x 1 mole O 2 x 32.0 g O g H 2 O 2 moles H 2 O 1 mole O 2 molar mole-mole molar mass H 2 O factor mass O 2 = 11.6 g of O 2

8 Learning Check Acetylene gas, C 2 H 2, burns in the oxyacetylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g of CO 2 ? 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) 1) 88.6 g of C 2 H 2 2) 44.3 g of C 2 H 2 3) 22.2 g of C 2 H 2

9 Solution 3) 22.2 g of C 2 H 2 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) 75.0 g CO 2 x 1 mole CO 2 x 2 moles C 2 H 2 x 26.0 g C 2 H g CO 2 4 moles CO 2 1 mole C 2 H 2 molar mole-mole molar mass CO 2 factor mass C 2 H 2 = 22.2 g of C 2 H 2

10 Calculating the Mass of Product When 18.6 g of ethane gas, C 2 H 6, burns in oxygen, how many grams of CO 2 are produced? 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) 18.6 g ? g The plan and factors would be g C 2 H 6 mole C 2 H 6 mole CO 2 g of CO 2 molar mole-mole molar mass C 2 H 6 factor mass CO 2

11 Calculating the Mass of Product 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) The setup would be: 18.6 g C 2 H 6 x 1 mole C 2 H 6 x 4 moles CO 2 x 44.0 g CO g C 2 H 6 2 moles C 2 H 6 1 mole CO 2 molar mole-mole molar mass C 2 H 6 factor mass CO 2 = 54.4 g of CO 2

12 Study Tip: Check Units Be sure to check that all units cancel to give the needed unit. Note each cancelled unit in the following setup: needed unit g C 2 H 6 x mole C 2 H 6 x moles CO 2 x g CO 2 g C 2 H 6 moles C 2 H 6 mole CO 2 molar mole-mole molar mass C 2 H 6 factor mass CO 2

13 Learning Check How many grams of H 2 O are produced when 35.8 g of C 3 H 8 react by the following equation? C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) 1) 14.6 g of H 2 O 2) 58.6 g of H 2 O 3) 117 g of H 2 O

14 Solution 2) 58.6 g of H 2 O C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) 35.8 g C 3 H 8 x 1 mole C 3 H 8 x 4 mole H 2 O x 18.0 g H 2 O 44.0 g C 3 H 8 1 mole C 3 H 8 1 mole H 2 O molar mole-mole molar mass C 3 H 8 factor mass H 2 O = 58.6 g of H 2 O