1 Chapter 5 Chemical Reactions 5.9c Mass Calculations for Reactions

2 Steps in Finding the Moles and Masses in a Chemical Reaction

3 Moles to Grams Suppose we want to determine the mass (g) of NH 3 that can form from 2.50 moles N 2. N 2 (g) + 3 H 2 (g) 2 NH 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

4 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 NH 3

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

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

7 The reaction between H 2 and O 2 produces 13.1 g 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 O 2 molar mole-mole molar mass H 2 O factor mass O 2 Calculating the Mass of a Reactant

8 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 O 2

9 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 CO 2 ? 2 C 2 H 2 (g) + 5 O 2 (g) 4 CO 2 (g) + 2 H 2 O(g) 1) 88.6 g C 2 H 2 2) 44.3 g C 2 H 2 3) 22.2 g C 2 H 2

10 Solution 3) 22.2 g C 2 H 2 2 C 2 H 2 (g) + 5 O 2 (g) 4 CO 2 (g) + 2 H 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 C 2 H 2

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

12 Calculating the Mass of Product 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 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 CO 2

13 Learning Check How many grams H 2 O are produced when 35.8 g C 3 H 8 react by the following equation? C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) 1) 14.6 g H 2 O 2) 58.4 g H 2 O 3) 117 g H 2 O

14 Solution 2) 58.4 g H 2 O C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 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 44.1 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 = 58.4 g H 2 O