Stoichiometry

Calculations based on a balanced chemical equation Chapter 9 (12)

Why do I need to know about stoichiometry?

Example using stoichiometry: Sample recipe for 1 loaf of bread: 8 cups flour 1 cup sugar 2 cups milk 1.5 cups eggs 1/2 cup of butter 1/8 cup yeast How many loaves of bread can be made from: • 3 cups eggs 2 loaves • 1 cup milk 1/2 loaf • 6 cup flour 3/4 loaf

Sample recipe for 1 loaf of bread: 8 cups flour 1 cup sugar 2 cups milk 1.5 cups eggs 1/2 cup of butter 1/8 cup yeast “Chemical Equation” for bread recipe: 8 Fl + 1 S + 2 M + 1.5

E + 1/2 B + 1/8 Y => 1 Lf

Review of Balancing Equations

2 KClO 3 2 2

What do the coefficients mean?

a) Molecules

“2 molecules of O 2 ” KClO 3 produces 3 molecules

Review of Balancing Equations

2 KClO 3 2 2

What do the coefficients mean?

a) Molecules

“2 molecules molecules of KCl are formed when 3 of O 2 are formed”

Review of Balancing Equations

2 KClO 3 2 2

What do the coefficients mean?

b) Moles

“2 moles moles of KCl are formed when 3 of O 2 are formed”

Review of Balancing Equations

2 KClO 3 2 2

What do the coefficients mean?

b) Moles

“2 moles moles of KCl are formed when 2 of KClO 3 are decomposed”

In the following reaction how many moles of PbCl 2 are formed if 5.000 moles of NaCl react?

2 NaCl + Pb(NO 3 ) 2  PbCl 2 + 2 NaNO 3 5.000 moles NaCl 1 moles PbCl 2 2 moles NaCl = 2.500 moles PbCl 2

In the following reaction how many moles of NH 3 are formed if 4.0 moles of H 2 react?

N 2 + 3 H 2 => 2 NH 3 4.0 moles H 2 2 moles NH 3 3 moles H 2 = 2.7 moles NH 3 Complete Problems 1-5 on the practice page.

In the following reaction how many grams of NH 3 are formed if 4.00 moles of H 2 react?

N 2 + 3 H 2 => 2 NH 3 4.00 moles H 2 2 moles NH 3 3 moles H 2 17 g NH 3 1 moles NH 3 = 45.3 g NH 3 1 Moles A mw Grams A coefficients Moles B 1 mw Grams B

In the following reaction how many moles of NH 3 are formed if 10.0 grams of H 2 react?

N 2 + 3 H 2 => 2 NH 3 10.0 grams H 2 1 moles H 2.016 g H 2 2 2 mole NH 3 3 moles H 2 = 3.31 mol NH 3 1 Moles A mw Grams A coefficients Moles B 1 mw Grams B Complete Problems 6-10 on the practice page.

In the following reaction how many grams of NH 3 are formed if 25.0 grams of N 2 react?

N 2 + 3 H 2 => 2 NH 3 25.0 g N 2 1 moles N 2 28.02 g N 2 2 mole NH 3 1 moles N 2 17 g NH 3 = 1 mole NH 3 30.3 g NH 3 Moles A 1 mw Grams A coefficients Moles B 1 mw Grams B Complete Problems 10-15 on the practice page.

How many grams of NH 3 are formed if 25.0 grams of N 2 react with 10.0 g of H 2 ?

N 2 + 3 H 2 => 2 NH 3 (Solve the problem separately with each number) 25.0 g N 2 1 moles N 2 28.02 g N 2 2 mole NH 3 1 moles N 2 17 g NH 3 1 mole NH 3 = 30.3 g NH 3 10.0 grams H 2 1 moles H 2 2 mole NH 3 2 g H 2 3 moles H 2 17 g NH 3 1 mole NH 3 (The smaller answer is the only correct one) = 56.7 g NH 3

Complete problems 16-20.

How many grams of NH 3 are formed if 25.0 grams of N 2 react with 10.0 g of H 2 ?

N 2 + 3 H 2 => 2 NH 3 25.0 g N 2 1 moles N 2 28.02 g N 2 2 mole NH 3 1 moles N 2 17 g NH 3 1 mole NH 3 = 30.3 g NH 3 10.0 grams H 2 1 moles H 2 2 mole NH 3 2 g H 2 3 moles H 2 17 g NH 3 1 mole NH 3 = 56.7 g NH 3 How much of the excess reagent is left over?

How many grams of H 2 (the excess reagent) are required to react with 25.0 g of N 2 (the limiting reagent) ?

N 2 + 3 H 2 => 2 NH 3

25.0 g N 2 1 moles N 2 28 g N 2 3 mole H 2 1 moles N 2 2 g H 2 1 mole H 2 = 5.36 g H 2

REQUIRED Left over = Given amount – Required amount = 10.0 g H 2 - 5.36 g H 2 = 4.64 g H 2

How many grams of NH 3 are formed if 10.0 grams of N 2 react with 15.0 g of H 2 ?

How much of the excess reagent is left over?

N 2 + 3 H 2 => 2 NH 3

Terms: Percent Yield Calculations Theoretical Yield = the

CALCULATED

amount of product expected Actual Yield = the

EXPERIMENTAL

amount that was actually obtained % Yield = Actual Theoretical X 100

What is the percent yield in a reaction where 1.50 mol of NH 3 10.0 g of H 2 was obtained after reacting with excess nitrogen?

N 2 + 3 H 2 => 2 NH 3

10.0 grams H 2 1 moles H 2 2.016 g H 2 2 mole NH 3 3 moles H 2 = 3.31 mol NH 3 Theoretical Yield % yield = 1.50

3.31

X 100 = 45.3%