# Mass Calculations for Reactions

## Presentation on theme: "Mass Calculations for Reactions"— Presentation transcript:

Mass Calculations for Reactions
Copyright © by Pearson Education, Inc. Publishing as Benjamin Cummings edited by bbg

Moles to Grams Suppose we want to determine the mass (g) of NH3
that can form from 2.50 moles N2. N2(g) + 3H2(g) NH3(g) The plan needed would be moles N moles NH 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 g NH3 1 mole N mole NH3 given mole-mole factor molar mass = g NH3

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

Solution 2) g O2 0.400 mole Fe2O3 x 3 mole O2 x g O2= g O2 2 mole Fe2O 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? 2H2(g) O2(g) 2H2O(g) ? g g The plan and factors would be g H2O mole H2O mole O 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 g O g H2O moles H2O 1 mole O2 molar mole-mole molar mass H2O factor mass O2 = g O2

Learning Check 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) 1) 88.6 g C2H2
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? 2C2H2(g) O2(g) CO2(g) + 2H2O(g) 1) g C2H2 2) g C2H2 3) g C2H2

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

Calculating the Mass of Product
When 18.6 g ethane gas C2H6 burns in oxygen, how many grams of CO2 are produced? 2C2H6(g) O2(g) CO2(g) + 6H2O(g) 18.6 g ? g The plan and factors would be g C2H mole C2H mole CO g CO2 molar mole-mole molar mass C2H factor mass CO2

Calculating the Mass of Product
The setup would be 18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x g CO2 30.1 g C2H moles C2H mole CO2 molar mole-mole molar mass C2H factor mass CO2 = g CO2

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

Solution C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
35.8 g C3H8x 1 mole C3H8 x 4 mole H2O x g H2O 44.1 g C3H mole C3H mole H2O molar mole-mole molar mass C3H factor mass H2O = g H2O (2)