Chapter 9 stoichiometry
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Chapter 9 Stoichiometry. 9.3 Limiting reagent and percent yield. Things you will learn. You will be able to determine what the limiting reactant in a chemical reaction is and what the excess reactant is.

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Chapter 9 Stoichiometry

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Chapter 9 stoichiometry

Chapter 9Stoichiometry

9.3

Limiting reagent and percent yield


Things you will learn

Things you will learn

  • You will be able to determine what the limiting reactant in a chemical reaction is and what the excess reactant is.

  • You will be able to calculate percent yield from a theoretical yield and an actual yield.


Chapter 9 stoichiometry

  • Chocolate Chip Cookie Ingredients

  • • 3/4 cup sugar• 3/4 cup packed brown sugar• 1 cup butter, softened• 2 large eggs,beaten• 1 teaspoon vanilla extract• 2 1/4 cups all-purpose flour• 1 teaspoon baking soda• 3/4 teaspoon salt• 2 cups semisweet chocolate chips• if desired, 1 cup chopped pecans, or chopped walnuts


This recipe makes 16 cookies

This recipe makes 16 cookies


Chapter 9 stoichiometry

  • Chocolate Chip Cookie Ingredients

  • • 3/4 cup sugar• 3/4 cup packed brown sugar• 1 cup butter, softened• 2 1 large egg, beaten• 1 teaspoon vanilla extract• 2 1/4 cups all-purpose flour• 1 teaspoon baking soda• 3/4 teaspoon salt• 2 cups semisweet chocolate chips• if desired, 1 cup chopped pecans, or chopped walnuts


Limiting reagent

Limiting reagent

  • In this case, the whole recipe must be scaled back by 50%, and the yield will only be 8 cookies

  • Eggs are the limiting reagent (reactant)


Chapter 9 stoichiometry

A balanced chemical equation is a recipe.

If there is a part of the ingredients which is less than what the recipe calls for, there will be less product and excess reactants which can’t combine with other things.


Chapter 9 stoichiometry

In this reaction, H2 an O2 combine to form water. The balanced reaction is:

2H2 + O2 2H2O

H2 is the limiting reagent


Chapter 9 stoichiometry

N2(g) + 3H2(g) 2NH3(g)

The mole ratios are:

1 mole N2

2 moles NH3

3 moles H2

2 moles NH3

1 mole N2

3 moles H2

2 moles NH3

1 mole N2

2 moles NH3

3 moles H2

3 moles H2

1 mole N2


Chapter 9 stoichiometry

N2(g) + 3H2(g) 2NH3(g)

  • If you have less than 1 mole of nitrogen, or less than 3 moles of hydrogen, you won’t get 2 moles of ammonia


Chapter 9 stoichiometry

N2(g) + 3H2(g) 2NH3(g)

  • You have 2 moles of N2 and 2 moles of H2.

    • What is the limiting reagent?

    • How much ammonia will this reaction yield?


Chapter 9 stoichiometry

N2(g) + 3H2(g) 2NH3(g)

  • You have 2 moles of N2 and 2 moles of H2.

    • What is the limiting reagent?

    • Choose one reactant arbitrarily and plug it into the equation:

      H2 is the limiting reagent because we would need 6 moles of it to react with 2 moles N2

2 mol N2

3 mol H2

1 mol N2

6 mol H2


Chapter 9 stoichiometry

N2(g) + 3H2(g) 2NH3(g)

  • You have 2 moles of N2 and 2 moles of H2.

    • How much ammonia will this reaction yield?

    • Plug the limiting reagent into the equation with the mole ratio of the product

limiting reagent

2 mol H2

2 mol NH3

3 mol H2

1.33 mol NH3


2cu s cu 2 s

2Cu + S Cu2S

  • You have 80 g CU and 25 g S

    • What is the limiting reagent?

    • What is the maximum amount of Cu2S yielded?


2cu s cu 2 s1

2Cu + S Cu2S

  • You have 80 g CU and 25 g S

    • What is the limiting reagent?

    • 80 g Cu is 1.26 mole Cu

    • 25 g S is .78 mole S

    • Choose one reactant arbitrarily and plug it into the equation:

1.26 mol Cu

1 mol S

2 mol Cu

.63 mol S

We have more than .63 mole S, so Cu is the limiting reagent


2cu s cu 2 s2

2Cu + S Cu2S

  • You have 80 g CU and 25 g S

    • What is the maximum amount of grams of Cu2S produced?

    • Plug the limiting reagent into the equation with the mole ratio of the product

limiting reagent

1.26 mol Cu

1 mol Cu2S

2 mol Cu

.63 mol Cu2S

The molar mass of Cu2S is 159 g, so the yield of Cu2S is .63 mol x 159 g/mol Cu2S = 100.2 g Cu2S


Percent yield

Percent yield

  • Actual yield / theoretical yield X 100%

  • Duh!!


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