Chapter 5 Chemical Reactions and Quantities 5.8 Mass Calculations for Reactions

Steps in Finding the Moles and Masses in a Chemical Reaction

Moles to Grams Suppose we want to determine the mass (g) of NH3 that can form from 2.50 moles N2. N2(g) + 3 H2(g) 2 NH3(g) The plan needed would be moles N2 moles NH3 grams NH3 The factors needed would be: mole factor NH3/N2 and the molar mass NH3

Moles to Grams The setup for the solution would be: 2.50 mole N2 x 2 moles NH3 x 17.0 g NH3 1 mole N2 1 mole NH3 given mole-mole factor molar mass = 85.0 g NH3

Learning Check How many grams of O2 are needed to produce 0.400 mole Fe2O3 in the following reaction? 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) 1) 38.4 g O2 2) 19.2 g O2 3) 1.90 g O2

Solution 2) 19.2 g O2 0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2= 19.2 g O2 2 mole Fe2O3 1 mole O2 mole factor molar mass

Calculating the Mass of a Reactant The reaction between H2 and O2 produces 13.1 g water. How many grams of O2 reacted? 2 H2(g) + O2(g) 2 H2O(g) ? g 13.1 g The plan and factors would be g H2O mole H2O mole O2 g O2 molar mole-mole molar mass H2O factor mass O2

Calculating the Mass of a Reactant The setup would be: 13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2 18.0 g H2O 2 moles H2O 1 mole O2 molar mole-mole molar mass H2O factor mass O2 = 11.6 g O2

Learning Check Acetylene gas C2H2 burns in the oxyacetylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g CO2? 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(g) 1) 88.6 g C2H2 2) 44.3 g C2H2 3) 22.2 g C2H2

Solution 3) 22.2 g C2H2 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(g) 75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x 26.0 g C2H2 44.0 g CO2 4 moles CO2 1 mole C2H2 molar mole-mole molar mass CO2 factor mass C2H2 = 22.2 g C2H2

Calculating the Mass of Product When 18.6 g ethane gas C2H6 burns in oxygen, how many grams of CO2 are produced? 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g) 18.6 g ? g The plan and factors would be g C2H6 mole C2H6 mole CO2 g CO2 molar mole-mole molar mass C2H6 factor mass CO2

Calculating the Mass of Product 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g) The setup would be 18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2 30.1 g C2H6 2 moles C2H6 1 mole CO2 molar mole-mole molar mass C2H6 factor mass CO2 = 54.4 g CO2

Learning Check How many grams H2O are produced when 35.8 g C3H8 react by the following equation? C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) 1) 14.6 g H2O 2) 58.4 g H2O 3) 117 g H2O

Solution 2) 58.4 g H2O C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) 35.8 g C3H8x 1 mole C3H8 x 4 mole H2O x 18.0 g H2O 44.1 g C3H8 1 mole C3H8 1 mole H2O molar mole-mole molar mass C3H8 factor mass H2O = 58.4 g H2O