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Chapter 5 Chemical Reactions

5.9cMass Calculations for Reactions

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Steps in Finding the Moles and Masses in a Chemical Reaction

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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

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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

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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

Learning Check

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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

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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

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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

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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

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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

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Calculating the Mass of Product

When 18.6 g ethane gas C2H6 burns in oxygen, howmany 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

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Calculating the Mass of Product

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)

The setup would be18.6 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

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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 H2O2) 58.4 g H2O3) 117 g H2O

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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