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# II. Limiting Reactants - PowerPoint PPT Presentation

Stoichiometry – 3.7. II. Limiting Reactants. A. Limiting Reactants. Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly. Limiting Reactant bread. Excess Reactants peanut butter and jelly. A. Limiting Reactants.

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Stoichiometry – 3.7

### II. Limiting Reactants

• Available Ingredients

• 1 jar of peanut butter

• 1/2 jar of jelly

• Limiting Reactant

• Excess Reactants

• peanut butter and jelly

• In a laboratory, usually one or more of the reactants are present in excess. There is more than the exact amount required to react

• Once one of the reactants is used up, no more product can form

• Limiting Reactant

• used up in a reaction

• Limits the amount of reactant that can combine and determines amount of product

• determines the amount of product that can form

• Excess Reactant

• added to ensure that the other reactant is completely used up

• cheaper & easier to recycle

1. Write a balanced equation.

2. For each reactant, calculate the amount of product formed.

• limiting reactant

• amount of product

• 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?

Zn + 2HCl  ZnCl2 + H2

? L

79.1 g

0.90 L

2.5M

Zn + 2HCl  ZnCl2 + H2

? L

79.1 g

0.90 L

2.5M

79.1

g Zn

1 mol

Zn

65.39

g Zn

1 mol

H2

1 mol

Zn

22.4 L

H2

1 mol

H2

= 27.1 L

H2

Zn + 2HCl  ZnCl2 + H2

? L

79.1 g

0.90 L

2.5M

0.90

L

2.5 mol

HCl

1 L

1 mol

H2

2 mol

HCl

22.4

L H2

1 mol

H2

= 25 L H2

left over zinc

Zn: 27.1 L H2 HCl: 25 L H2

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: 25 L H2

• Try Example Problem #2

• Method 1: Convert both reactants to product. See which is less.

• Method 2: Convert one reactant to another. See how much is needed.

• Problem #2:

• HF: limiting

• 4.0 mol excess SiO2

1. actual yield: measured amount of product obtained from a reaction; measured in actual lab; less than theoretical yield due to experimental errors

2. theoretical yield: maximum amt. of product that could ideally be obtained from a given amount of reactant

measured in lab

calculated w/stoich.

• When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.

K2CO3 + 2HCl  2KCl + H2O + CO2

45.8 g

? g

actual: 46.3 g

K2CO3 + 2HCl  2KCl + H2O + CO2

Theoretical Yield:

45.8 g

? g

actual: 46.3 g

45.8 g

K2CO3

1 mol

K2CO3

138.21 g

K2CO3

2 mol

KCl

1 mol

K2CO3

74.55

g KCl

1 mol

KCl

= 49.4

g KCl

46.3 g

49.4 g

K2CO3 + 2HCl  2KCl + H2O + CO2

Theoretical Yield = 49.4 g KCl

45.8 g

49.4 g

actual: 46.3 g

 100 =

93.7%

% Yield =