1
Mass Calculations for Reactions
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummingsedited by bbg
2
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) 2NH3(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
3
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 NH= 85.0 g NH33
4
How many grams of O2 are needed to produce 0.400 mole Fe2O3 in the following reaction?
4Fe(s) + 3O2(g) 2Fe2O3(s)
1) 38.4 g O2
2) 19.2 g O2
3) 1.90 g O2
Learning Check
5
2) 19.2 g O2) 19.2 g O22
0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2= 19.2 g O19.2 g O22
2 mole Fe2O3 1 mole O2
mole factor molar mass
Solution
6
The reaction between H2 and O2 produces 13.1 g water.
How many grams of O2 reacted?
2H2(g) + O2(g) 2H2O(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
7
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 O11.6 g O22
Calculating the Mass of a Reactant
8
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?
2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)
1) 88.6 g C2H2
2) 44.3 g C2H2
3) 22.2 g C2H2
9
3) 22.2 g C3) 22.2 g C22HH22
2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(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 C= 22.2 g C22HH22
Solution
10
Calculating the Mass of Product
When 18.6 g ethane gas C2H6 burns in oxygen, how
many grams of CO2 are produced?
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(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
11
Calculating the Mass of Product
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 molarmolar mole-mole molar
mass Cmass C22HH66 factor mass CO factor mass CO22
= 54.4 g CO54.4 g CO22
12
Learning Check
How many grams H2O are produced when 35.8 g C3H8 react by the following equation?
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
1. 14.6 g H2O
2. 58.4 g H2O
3. 117 g H2O
13
Solution
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(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 H58.4 g H22O (2)O (2)
