II. Limiting ReactantsII. Limiting Reactants

Stoichiometry – 3.7Stoichiometry – 3.7

A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants

Available IngredientsAvailable Ingredients• 4 slices of bread• 1 jar of peanut butter• 1/2 jar of jelly

Limiting ReactantLimiting Reactant• bread

Excess ReactantsExcess Reactants• peanut butter and jelly

A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants

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

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

A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants

Limiting ReactantLimiting 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 ReactantExcess Reactant• added to ensure that the other reactant is

completely used up• cheaper & easier to recycle

B. Limiting ReactantsB. Limiting ReactantsB. Limiting ReactantsB. Limiting Reactants

1. Write a balanced equation.

2. For each reactant, calculate the

amount of product formed.

3. Smaller answer indicates:

• limiting reactant

• amount of product

B. Limiting ReactantsB. Limiting ReactantsB. Limiting ReactantsB. Limiting Reactants

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 79.1 g ? L0.90 L

2.5M

B. Limiting ReactantsB. Limiting ReactantsB. Limiting ReactantsB. Limiting Reactants

79.1g Zn

1 molZn

65.39g Zn

= 27.1 L H2

1 molH2

1 molZn

22.4 LH2

1 molH2

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

B. Limiting ReactantsB. Limiting ReactantsB. Limiting ReactantsB. Limiting Reactants

22.4L H2

1 molH2

0.90L

2.5 molHCl

1 L= 25 L

H2

1 molH2

2 molHCl

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

B. Limiting ReactantsB. Limiting ReactantsB. Limiting ReactantsB. Limiting Reactants

Zn: 27.1 L H2 HCl: 25 L H2

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: 25 L H2

left over zinc

LIMITING REACTANTLIMITING REACTANTLIMITING REACTANTLIMITING REACTANT

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.

LIMITING REACTANTLIMITING REACTANTLIMITING REACTANTLIMITING REACTANT

Problem #2:

• HF: limiting

• 4.0 mol excess SiO2

C. Percent YieldC. Percent YieldC. Percent YieldC. Percent Yield1. 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

C. Percent YieldC. Percent YieldC. Percent YieldC. Percent Yield

100yield ltheoretica

yield actualyield %

calculated w/stoich.

measured in lab

C. Percent YieldC. Percent YieldC. Percent YieldC. Percent Yield

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

C. Percent YieldC. Percent YieldC. Percent YieldC. Percent Yield

45.8 gK2CO3

1 molK2CO3

138.21 gK2CO3

= 49.4g KCl

2 molKCl

1 molK2CO3

74.55g KCl

1 molKCl

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g

actual: 46.3 g

Theoretical Yield:

C. Percent YieldC. Percent YieldC. Percent YieldC. Percent Yield

Theoretical Yield = 49.4 g KCl

% Yield =46.3 g

49.4 g 100 =93.7%

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g

actual: 46.3 g