9.5 – NOTES Limiting Reactants and Percent Yield - PowerPoint Presentation

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9.5 – NOTESLimiting Reactants and Percent Yield

III. Limiting Reactants

A. Why do reactions stop? Are all reactions 100% efficient?

Reactions might

stop

for a number of reasons including:

consumption

of one or multiple reactants; occurrence of side reactions;

B. Calculating the product when a reactant is limited

Limiting

reactant:

limits the extent to which a reaction can

proceed

, thus determining the amount of

product

formed;

the

portion of all the other reactants that remain after the

reaction

stops =

excess

reactants;

Theoretical yield:

the

maximum

amount of product

that can

be formed from the quantities of reactants given

To determine the limiting reactant: 1. Read the question and write the balanced equation.

2. Label all given amounts (usually 2 reactant masses).

3. Do stoichiometry

twice

, determining the amount of product

(

or of one product if there is more than one), starting with each

given

reactant.

4. Choose the

smaller

answer – that is the

theoretical

yield

.

The

reactant that gives you this answer is the

limiting reactant

.

The

other reactant is the

excess reagent

.

Examples: a. If 5.0 grams of nickel and 5.0 grams of chlorine gas are reacted completely, how many grams of product could be formed, theoretically?

Ni + Cl

2



NiCl

2

Given: 5.0 g Ni ( = 58.7g/

mol

) Find: g NiCl

2

( = 129.7g/

mol

)

5.0 g Cl

2

( =71.0g/

mol

)

5.0g Ni x

1

mol

Ni

x

1

mol

NiCl

2

x

129.7g NiCl

2

58.7g Ni 1

mol

Ni 1

mol

NiCl

2

= 11g NiCl

2

5.0g Cl

2

x

1

mol

Cl

2

x

1

mol

NiCl

2

x

129.7g NiCl

2

71.0 g Cl

2

1

mol

Cl

2

1

mol

NiCl

2

= 9.1g NiCl

2

It is possible to determine the limiting reactant from the mole ratio; however, most problems ask for the mass.

b. If you react 13.5 grams of

HCl

with 14.8 grams of

FeS

, how many grams of H

2

S could be produced?

2HCl +

FeS



H

2

S + FeCl

2

Given: 13.5g

HCl

(= 36.5g/

mol

) Find: g H

2

S (= 33.1g/

mol

)

14.8g

FeS

(=87.9g/

mol

)

13.5g

HCl

x

1

mol

HCl

x

1

mol

H

2

S

x

33.1g H

2

S

36.5g

HCl

2

mol

HCl

1

mol

H

2

S = 6.12g H

2

S

14.8g

FeS

x

1

mol

FeS

x

1

mol

H

2

S

x

33.1g H

2

S

87.9g

FeS

1

mol

FeS

1

mol

H

2

S = 5.57g H

2

S

C. Why use an excess of a reactant?

Helps

force

the reaction to completion and can sometimes speed up a

reaction

often

the least

expensive

reactant is used as the excess

reactant

think

of burner – use excess

air

to create a blue flame which completely burns the fuel rather than fuel as the excess which creates a residue (carbon) on glassware.

To determine the quantity of excess:

1

. Read the question and write the balanced equation.

2. Use stoichiometry, determine the quantity of reactant needed to completely consume the limiting reactant.

3

. Subtract the quantity used from the quantity given.

Example: Determine the amount of

HCl

in excess using the above example.

2HCl +

FeS



FeCl

2

+ H

2

S

14.8g

FeS

x

1

mol

FeS

x

2

mol

HCl

x

36.5g

HCl

= 12.3g

HCl

(used)

87.9g

FeS

1

mol

FeS

1

mol

HCl

13.5g – 12.3 = 1.2g excess

IV. Percent yield How efficient was your process?

Percent yield =

experimental yield

x 100

theoretical yield

Most reactions never succeed in producing the predicted amount of product

Why experimental yields < theoretical yields?

not

every reaction goes

cleanly

or

completely

many

reactions

stop

before all the reactants are used up, so the actual amount of product is less than

expected

liquid

reactants or products may

adhere

to the surface of containers or

evaporate

solid

product is always left behind on

filter

paper or lost in the purification

process

products

other than the intended ones may be formed by

competing

reactions, thus reducing the yield of the desired product.

Examples: 1. You reacted 2.78 grams of magnesium with excess sulfuric acid. You massed the resulting magnesium sulfate and found it had a mass of 3.45 grams. What was your percentage yield?

Mg + H

2

SO

4



MgSO

4

+ H

2

Given: 2.78g Mg (=24.3g/

mol

) Find: g MgSO

4

(=120.4g/

mol

)

3.45g MgSO

4

produced

2.78g Mg x

1

mol

Mg

x

1

mol

MgSO

4

x

120.4g MgSO

4

24.3g Mg 1

mol

Mg 1

mol

MgSO

4

=13.8g MgSO

4

% yield =

3.45g MgSO

4

13.8 g x 100 = 25.0%

2. If you react 7.82 grams of copper with excess silver nitrate solution, how many grams of silver would be produced if the reaction were 85% efficient?

Cu

+ 2AgNO

3



Cu(NO

3

)

2

+

2Ag

Given: 7.82g Cu (63.5g/

mol

) Find: g Ag ( 108g/

mol

)

7.82g Cu x

1

mol

Cu

x

2

mol

Ag

x

108 g Ag

63.5g

Cu 1

mol

Cu 1

mol

Ag = 26.6g Ag

26.6g Ag x .85 = 22.6g Ag