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

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

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 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)

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.

In this reaction, H 2 an O 2 combine to form water. The balanced reaction is: 2H 2 + O 2 2H 2 O H 2 is the limiting reagent

N 2(g) + 3H 2(g) 2NH 3(g) The mole ratios are: 1 mole N 2 2 moles NH 3 1 mole N 2 2 moles NH 3 3 moles H 2 2 moles NH 3 3 moles H 2 1 mole N 2 3 moles H 2

If you have less than 1 mole of nitrogen, or less than 3 moles of hydrogen, you won’t get 2 moles of ammonia N 2(g) + 3H 2(g) 2NH 3(g)

You have 2 moles of N 2 and 2 moles of H 2. – What is the limiting reagent? – How much ammonia will this reaction yield? N 2(g) + 3H 2(g) 2NH 3(g)

You have 2 moles of N 2 and 2 moles of H 2. – What is the limiting reagent? – Choose one reactant arbitrarily and plug it into the equation: H 2 is the limiting reagent because we would need 6 moles of it to react with 2 moles N 2 N 2(g) + 3H 2(g) 2NH 3(g) 2 mol N2 3 mol H2 1 mol N2 6 mol H2

You have 2 moles of N 2 and 2 moles of H 2. – How much ammonia will this reaction yield? – Plug the limiting reagent into the equation with the mole ratio of the product N 2(g) + 3H 2(g) 2NH 3(g) 2 mol H2 2 mol NH3 3 mol H mol NH3 limiting reagent

2Cu + S Cu 2 S You have 80 g CU and 25 g S – What is the limiting reagent? – What is the maximum amount of Cu 2 S yielded?

2Cu + S Cu 2 S 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 S You have 80 g CU and 25 g S – What is the maximum amount of grams of Cu 2 S produced? – Plug the limiting reagent into the equation with the mole ratio of the product 1.26 mol Cu 1 mol Cu2S 2 mol Cu.63 mol Cu2S The molar mass of Cu 2 S is 159 g, so the yield of Cu 2 S is.63 mol x 159 g/mol Cu 2 S = g Cu 2 S limiting reagent

Percent yield Actual yield / theoretical yield X 100% Duh!!