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Chapter 3 Lesson 4

Limiting Reactants

Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Limiting Reactant

A limiting reactant in a chemical reaction is the

substance that • Is used up first.• Stops the reaction.• Limits the amount of product that can form.

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Reacting Amounts

In a table setting, there is 1

plate, 1 fork, 1 knife, and

1 spoon.

How many table settings are

possible from 5 plates, 6 forks,

4 spoons, and 7 knives?

What is the limiting item?

Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Reacting Amounts

Four table settings can be made.

Initially Use Left over

plates 5 4 1

forks 6 4 2

spoons 4 4 0

knives 7 4 3

The limiting item is the spoon.

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Limiting Reactants

When 4.00 mol H2 is mixed with 2.00 mol Cl2,how

many moles of HCl can form?

H2(g) + Cl(g) 2HCl (g)

4.00 mol 2.00 mol ??? mol

• Calculate the moles of product from each reactant, H2 and Cl2.

• The limiting reactant is the one that produces the smaller amount of product.

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Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

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Limiting Reactants Using Moles

HCl from H2

4.00 mol H2 x 2 mol HCl = 8.00 mol HCl

1 mol H2 (not possible)

HCl from Cl2

2.00 mol Cl2 x 2 mol HCl = 4.00 mol HCl

1 mol Cl2 (smaller number)

The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl.

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Checking Calculations

Initially H2

4.00 mol

Cl2 2.00 mol

2HCl

0 mol

Reacted/

Formed

-2.00 mol -2.00 mol +4.00 mol

Left after reaction

2.00 mol

Excess

0 mol

Limiting

4.00 mol

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Limiting Reactants Using Mass

If 4.80 mol Ca mixed with 2.00 mol N2, which is thelimiting reactant? 3Ca(s) + N2(g) Ca3N2(s)

Moles of Ca3H2 from Ca

4.80 mol Ca x 1 mol Ca3N2 = 1.60 mol Ca3N2 3 mol Ca (Ca used up)

Moles of Ca3H2 from N2

2.00 mol N2 x 1 mol Ca3N2 = 2.00 mol Ca3N2

1 mol N2 (not possible)

All Ca is used up when 1.60 mol Ca3N2 forms. Thus, Ca is the limiting reactant.

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Limiting Reactants Using Mass

Calculate the mass of water produced when

8.00 g H2 and 24.0 g O2 react?

2H2(g) + O2(g) 2H2O(l)

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Limiting Reactants Using Mass

Calculate the grams of H2 for each reactant.

H2: 8.00 g H2 x 1 mol H2 x 2 mol H2O x 18.02 g H2O

2.016 g H2 2 mol H2 1 mol H2O

= 71.5 g H2O (not possible)

O2:

24.0 g O2 x 1 mol O2 x 2 mol H2O x 18.02 g H2O 32.00 g O2 1 mol O2 1 mol H2O

= 27.0 g H2O (smaller)

O2 is the limiting reactant.