lesson 16: theoretical yield actual yield percent yield · o are produced, what is the percent...

Lesson 16:

Theoretical Yield

Actual Yield

Percent Yield

Do Now (5pts) April 3, 2019• Copy down info from CJ board.

• Take out calculators to be checked.

• Answer questions in Box 1 of Lesson 16 note packet.

• Progress reports will be handed out next class. Quiz correction due date will be announced then.

• Who needs to make a make-up quiz? Please come see me at the podium.

2 1 4 1

2 flags + 1 pole + 4 wheels + 1 car body 1 car

If you had 62 flags, 37 poles, 102 wheels, and 30 car bodies, how many complete

toy cars could you make? What reagent is limiting?

62 flags / 2 per car = 3137 poles / 1 per car = 37102 wheels / 4 per car = 25.530 car bodies / 1 per car = 30

Has the smallest equivalent, the wheels are the limiting reagentYou can only make 25 complete cars

SiO2 (s) + 4HF (g) SiF4 (g) + 2H2O (l)If you started with 0.910 mol of silicon dioxide and 3.51 mol of HF, what is the limiting reactant?What is the theoretical yield of water (in moles)?

HW review

Review: Limiting and Excess Reagents

One reactant almost always limits the amount of product produced in a reaction.

Once one of the reactant is used up, ____________________________________________.

The substance that is used up first is the ________________________________.

Identifying the limiting reagent: calculate the ___________________________________.

The limiting reagent has the lesser equivalent (you will run out of that reactant

first).

then no more product can be formedlimiting reagent

molar equivalents

1SiO2 + 4HF → 2 SiF4 + 2 H2O

4.5 mol 6.0 mol

What is the limiting reagent?

Amount that you

start with

Co-efficient of the

reactant

4.5

1vs.

6.0

4

4.5 1.5These

numbers are called “ molar equivalents”

This number is SMALLER, therefore it is the limiting

reagent.

HF

HF

SiO2

Amount that you

start with

Co-efficient of the

reactant

Finding the “molar equivalent” of each reactant:

This has to be in compounds, molecules, or moles.

N2H4 (l) + 2 H2O2 (l) → N2 (g) + 4 H2O (g)

0.750 mol 5.2 mol What is the limiting reagent?

Amount that you

start with

Co-efficient of the

reactant

0.750

1vs.

5.2

2

0.750 2.6These

numbers are called “ molar equivalents”

This number is SMALLER, therefore it is the limiting

reagent.

H2O2

N2H4

N2H4

Let’s look at #3 on p.3

If 500 mol CO and 750 mol H2 are present, which is the limiting reagent?

500 mol CO1

750 mol H2

2

500 375 H2 is the limiting reagent

Methanol, CH3OH is the simplest of alcohols. It is synthesized by the

reaction of hydrogen and carbon monoxide.

CO (g) + 2H2 (g) CH3OH

Methanol, CH3OH is the simplest of alcohols. It is synthesized by the

reaction of hydrogen and carbon monoxide.

CO (g) + 2H2 (g) CH3OH

How many moles of the excess reagent remain unchanged?

750 mol H2

mol H2

mol COmol CO×

1=

2

1375 will be used up

Amount we started with – amount consumed in reaction

500 mol CO – 375 mol CO = 125 mol CO

Start with the limiting reagent:

Methanol, CH3OH is the simplest of alcohols. It is synthesized by the

reaction of hydrogen and carbon monoxide.

CO (g) + 2H2 (g) CH3OH

How many moles of CH3OH are formed?

750 mol H2

mol H2

mol CH3OH×

1=

2

Start with the limiting reagent:

1 mol CH3OH375 will be produced

Let’s look at the second problem on p.3

The black oxide of iron, Fe3O4, occurs in nature as the mineral magnetite. The substance can also be used in the laboratory by the reaction between red-hot iron and steam according to the following reaction:

3Fe(s) + 4H2O (g) Fe3O4(s) + 4H2 (g)

When 36 g of H2O are mixed with 67.0 g Fe, which is the limiting reagent?

36 g H2O

18.016 g H2O

1 mol H2O

67.0 g Fe

55.85 g Fe

1 mol Fe

×

×

1

1

=

=

2 mol H2O

1.19964 mol Fe

2 mol H2O

4

1.19964 mol Fe

3

=

=

0.5

0.39988

LR

FIRST – find how many moles of each reactant you have.THEN – calculate the mole equivalents.

molar mass

molar mass

mole equivalent

mole equivalent

What mass in grams of black iron oxide is produced?

3Fe(s) + 4H2O (g) Fe3O4(s) + 4H2 (g)LR

36 g

This is a grams grams conversion.

67.0 g Fe1 ×

g Fe

mol Fe

mol Fe

mol Fe3O4

mol Fe3O4

g Fe3O4

______ g Fe3O4

× × =

molar mass molar massMolar ratio of reactant to product in

balanced reaction

55.851 1

3 1231.55

92.6

92.592

67.0 g

ER

Start with the limiting reagent:

What mass in grams of excess reagent remains when the reaction is completed?

3Fe(s) + 4H2O (g) Fe3O4(s) + 4H2 (g)

LR

36 g

Start with the limiting reagent:

67 g Fe1 g Fe

mol Fe

mol Fe

mol H2O× × ×

mol H2O

g H2O=55.85

134

118.016 29 g H2O

consumed during the reaction

Amount we started with – amount consumed in reaction

36g water to start – 29 g used up = 7 g water left over

67.0 g

ER

Lesson 22:

Theoretical Yield

Actual Yield

Percent Yield

Theoretical yield is the _________________________________________________________

when the limiting reagent is completely consumed.

Theoretical yield _____________________________________________________________ and is

calculated using stoichiometry ________________________________________________.

amount of the limiting reagent

from the limiting reagent.

depends solely on the

beginning

THEORETICAL YIELD

maximum amount of product that can be formed

Just as the __________________allows us to calculate limiting and excess reagents and

masses, we can also ____________________ the mass of products in any given reaction.

The theoretical yield represents _____________________________________________________

THEORETICAL YIELD

predict

mole ratio

the maximum quantity of product possible if the reaction proceeds to completion.

This is what you’ve been calculating all along

The ______________________________________________obtained from a

reaction is called the actual yield (or experimental yield) of that

product.

measured amount of a product

ACTUAL YIELD

PERCENT YIELD

Comparing the theoretical and actual yield helps chemists determine the

reaction’s _________________________. efficiency

actual yieldtheoretical yield × 100Percentage yield =

• The percent yield represents the ____________of the actual yield to the theoretical yield.

ratio

Let’s look at the #1 on p.5

CH4 + 2O2 CO2 + 2H2O

How many grams of H2O are expected when 8.00 grams of CH4 reacts

with an excess of O2?

8.00 g CH4

g CH4

mol CH4

mol CH4

mol H2Omol H2O

g H2O

g H2O

Theoretical Yield

× × × =1 16.042

1

1

2

1

18.016

18.0

Start with the limiting reagent:

If only 17.0 grams of H2O are produced, what is the percent yield of this

reaction?

CH4 + 2O2 CO2 + 2H2O

g H2O Theoretical Yield18.0

g H2O ACTUAL Yield17.0

actual yieldtheoretical yield

× 100Percentage yield =

17.0 / 18.0 = 0.9444× 100

94.4 % yield

Let’s look at the #2 on p.5

C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)52.6 g 60.2 gStarting amounts:

Which is the limiting reagent?

WHAT DO I DO?First, convert from grams to moles

using molar mass.

Then, calculate mole equivalents.

C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)52.6 g 60.2 gStarting amounts:

Which is the limiting reagent?

52.6 g C6H6

78.108 g C6H6

1 mol C6H6 0.673 mol C6H6

60.2 g Cl2

70.9 g Cl2

1 mol Cl20.849 mol Cl2

×

×

1

1

=

=

0.673 /1 = 0.673

0.849 /1 = 0.849

LR ER

C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)52.6 g 60.2 gStarting amounts:

LR ERDetermine the amount of excess reagent reacted and amount remaining.

Start with the limiting reagent:

52.6 g C6H6

78.108 g C6H6

1 mol C6H6

1 mol C6H6

1 mol Cl2

1 mol Cl2

70.9 g Cl2× × ×

g chlorineconsumed in reaction

1

47.760.2 g to start – 47.7 g consumed = 12.5 g of Cl2 in excess

Calculate the theoretical yield of product chlorobenzene.

C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)52.6 g 60.2 gStarting amounts:

LR ER

52.6 g C6H6

78.108 g C6H6

1 mol C6H6

1 mol C6H6

1 mol C6H5Cl× × ×

1 1 mol C6H5Cl

112.55 g C6H5Cl

= 80.98 g C6H5Cl

Start with the limiting reagent:

Theoretically made if the reaction was 100% efficientBUT REACTIONS ARE NOT ALWAYS 100% EFFICIENT!

C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)

If 73.9 g of chlorobenzene (C6H5Cl) was actually produced, what was the percent yield?

Actual yield = 73.9 g C6H5Cl

Theoretical yield = 80.98 g C6H5Cl

73.9 / 80.98 = 0.91257× 100

91.3 % yield

actual yieldtheoretical yield

× 100Percentage yield =