3 - Limiting Rxn.pptx

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LIMITING REACTANTS,

THEORETICAL YIELD,

PERCENT YIELD

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REACTION STOICHIOMETRY: AN

EXAMPLE

• Suppose there are 45.98 g of sodium. How many grams

of chlorine will be needed for this reaction? How many

grams of NaCl will be produced?

• Can’t convert grams of Na directly to grams of Cl

₂

or NaCl.

• _{Must convert grams of Na to moles of Na (using}

_{molar mass}

₎

and use the

coefficients of the balanced equation

to find moles

of Cl

₂

and NaCl, and then convert those numbers to grams



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REACTION STOICHIOMETRY: AN

EXAMPLE



2 Na

₍

_s

₎



1 Cl

₂₍

_g

₎



2 NaCl

_(s)



Convert g Na to mol Na : 45.98 g Na 1 mol Na

22.989768 g Na  2.00002... mol Na Convert mol Na to mol Cl₂ :

2.00002... mol Na 1 mol Cl2

2 mol Na = 1.00001... mol Cl2 Convert mol Cl₂ to g Cl₂ :

1.00001... mol Cl₂ 70.9054 g Cl₂

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REACTION STOICHIOMETRY: AN

EXAMPLE



2 Na

₍

_s

₎



Cl

₂₍

_g

₎



2 NaCl

_(s)



Or we can put it all together in one calculation : 45.98 g Na 1 mol Na

22.989768 g Na

1 mol Cl₂ 2 mol Na

70.9054 g Cl₂

1 mol Cl₂  70.90611 g Cl2 = 70.91 g Cl2 How many grams of NaCl will be produced?

45.98 g Na 1 mol Na 22.99 g Na

2 mol NaCl 2 mol Na

58.442468 g NaCl

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EXAMPLES: REACTION STOICHIOMETRY

1a. How many moles of CO

₂

can we make from 2.0 moles of

C

₃

H

₈

?

FW_C3H8= 44.09652 g FW_O2 = 31.9988 g FW_CO2= 44.0098 g FW_H2O = 18.01528 g



C

₃

H

_8(g)



5 O

₂

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EXAMPLES: REACTION STOICHIOMETRY

1b. How many moles of H

₂

O can we make from 2.0 moles of

C

₃

H

₈

?



C

₃

H

_8(g)



5 O

₂

(7)

EXAMPLES: REACTION STOICHIOMETRY

1c. How many moles of O

₂

are needed to react with 2.0 moles of

C

₃

H

₈

?



C

₃

H

_8(g)



5 O

₂

(8)

EXAMPLES: REACTION STOICHIOMETRY

1d. How many moles of CO

₂

can be produced from 3.5 mol O

₂

?



C

₃

H

_8(g)



5 O

₂

(9)

EXAMPLES: REACTION STOICHIOMETRY

1e. How many grams of CO

₂

can be produced from 50.0 g of

C

₃

H

₈

?



C

₃

H

_8(g)



5 O

₂

(10)

EXAMPLES: REACTION STOICHIOMETRY

2. In 2012, the world burned 134 billion barrels of gasoline,

roughly equivalent to 5.9x10

14

g of gasoline (C

8

H

18

). How

much CO

₂

is released into the atmosphere from the

combustion of this much gasoline?





2C

₈

H

₁₈₍_l ₎



25 O

₂

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EXAMPLES: REACTION STOICHIOMETRY

3. Aqueous sodium hypochlorite (NaOCl), best known as

household bleach, is prepared by reaction of sodium

hydroxide with chlorine:

How many grams of NaOH are needed to react with 25.0 g of

Cl

₂

?

FWNaOH= 39.997108g FWCl2 = 70.9054g FWNaOCl = 74.441868g FWH2O = 18.01528g FWNaCl= 58.442468g

Ans = 28.2 g NaOH





2 NaOH

₍_aq₎



Cl

₂

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YIELDS OF CHEMICAL REACTIONS



In the previous examples, we have assumed that all of

the reactions “go to completion” – in other words all of

the reactants are used up and completely converted to

products. In real life, some product is almost always

lost due to contamination on the glassware, impurities

in the reactants, incomplete reactions, insufficient

heating, inability to recover all the product, or side

reactions that form other products.



The

theoretical yield

is the amount that would be

obtained if the reaction goes to completion (the

maximum amount

that could be made)



The

actual yield

of a reaction is the amount that is

actually recovered. Actual yield should be <

theoretical yield.

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YIELDS OF CHEMICAL REACTIONS



The

percent yield (%yield)

is the actual yield

expressed as a percentage of the theoretical yield:



The reactant the is used up first, which limits the

yield of the reaction, is the

limiting reactant

(or

limiting reagent

).



The reactant in excess is any reactant that is

present in a larger amount than what is required to

react completely with the limiting reactant.



%

anything



part

whole

x

100% % Yield =

actual yield

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EXAMPLES: PERCENT YIELD

4. Methyl

tert

-butl ether (MTBE, C

₅

H

₁₂

O), a substance

used as an octane booster in gasoline, can be made by

reaction of isobutylene (C

₄

H

₈

), with methanol

(CH

₃

OH). What is the percent yield of the reaction if

32.8 g of MTBE is obtained from reaction of 26.3 g of

isobutylene with sufficient methanol?

Answer: 79.4%





C

₄

H

₈₍_g₎



CH

₃

OH

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LIMITING REACTANTS



When we are given a reaction between two or more

reactants, one may be completely consumed before the

other(s).

The reaction must

stop

at this point,

leaving us

with the remaining reactants

in excess.



The amount of this reactant, then, determines the

maximum amount of the product(s) that can form, and is

known as the

limiting reactant.

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LIMITING REACTANTS



For example, suppose we were making

4-door cars, and we had the following list of

“ingredients.” How many cars could we

make?

4 engines

4 drivers’ seats

4 steering wheels

4 rear-view mirrors

15 doors

8 windshield wipers

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LIMITING REACTANTS AND PIZZA

1 crust + 5 oz. tomato sauce + 2 cups cheese



1 pizza

If we have 4 crusts, 10 cups of cheese, and 15 oz. of

tomato sauce, how many pizzas can we make?

Tomato sauce is the

limiting reagent

, and the

theoretical yield

is

3 pizzas.

http://phet.colorado.edu/en/simulation/reactants-products-and-leftov ers



4 crusts 1 pizza

1 crust  4pizzas 10 cups cheese 1 pizza

2 cups cheese  5pizzas 15 ounces tomato sauce 1 pizza

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LIMITING REACTANTS EXAMPLE



Suppose we mix 1.00 mol of N

₂

and 5.00 mol of H

₂

.

What is the limiting reagent? What is the maximum

amount of NH

₃

that can be produced? How much of

the excess reactant will be left over?

NH

2

H 3 + N

(g) (g)

(g) 2 3

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LIMITING REACTANTS EXAMPLE



Suppose we mix 1.00 mol of N

₂

and 5.00 mol of H

₂

.

What is the limiting reagent? What is the maximum

amount of NH

₃

that can be produced? How much of

the excess reactant will be left over?

1. Set up ICE table (same as BCA table!)



N₂

(g) + 3 H2(g)  2 NH3(g)

(initial) I 1.00 mol 5.00 mol 0 mol (change) C - 1x - 3x + 2x (end) E

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LIMITING REACTANTS EXAMPLE



Suppose we mix 1.00 mol of N

₂

and 5.00 mol of H

₂

.

What is the limiting reagent? What is the maximum

amount of NH

₃

that can be produced? How much of the

excess reactant will be left over?

1.

Set up ICE table.

2.

Determine which reactant is limiting by solving for x for

each reactant. Smallest value of x is limiting reagent



N₂

(g) + 3 H2(g)  2 NH3(g)



x  moles of reactant actually present

moles of reactant needed XN2 =

1.00

1 =1 XH2 

5.00

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LIMITING REACTANTS EXAMPLE



Suppose we mix 1.00 mol of N

₂

and 5.00 mol of H

₂

.

What is the limiting reagent? What is the maximum

amount of NH

₃

that can be produced? How much of

the excess reactant will be left over?

1.

Set up ICE table.

2.

Determine which reactant is limiting by solving for x

for each reactant. Smallest value of x is limiting

reagent.

N

₂

is the limiting reagent.



1.00

1 =1 XH2 

5.00

(24)

LIMITING REACTANTS EXAMPLE



Suppose we mix 1.00 mol of N

₂

and 5.00 mol of H

₂

. What is the

limiting reagent? What is the maximum amount of NH

₃

that can be

produced? How much of the excess reactant will be left over?

3.

Plug in the value of x for the limiting reagent into the ice

table and solve for the end part of the table.



N₂

(g) + 3 H2(g)  2 NH3(g)

(initial) I 1.00 mol 5.00 mol 0 mol

(change) C - 1(1.00) - 3(1.00) + 2(1.00) (end) E



N₂

(g) + 3 H2(g)  2 NH3(g)

(initial) I 1.00 mol 5.00 mol 0 mol (change) C - 1.00 mol - 3.00 mol + 2.00 mol (end) E 0 2.00 mol 2.00 mol

(25)

LIMITING REACTANTS EXAMPLE



Now suppose we mix 2.15 mol of N

₂

and 6.15 mol of

H

₂

. What is the theoretical yield of NH

₃

?

NH

2

H 3 + N

(g) (g)

(g) 2 3

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LIMITING REACTANTS EXAMPLE



Now suppose we mix 2.15 mol of N

₂

and 6.15 mol of

H

₂

. What is the theoretical yield of NH

₃

?

NH 2 H 3 + N (g) (g)

(g) 2 3

2 



N₂

(g) + 3 H2(g)  2 NH3(g)

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LIMITING REACTANTS EXAMPLE



Now suppose we mix 2.15 mol of N

₂

and 6.15 mol of

H

₂

. What is the theoretical yield of NH

₃

?

NH 2 H 3 + N (g) (g)

(g) 2 3

2 



N₂

(g) + 3 H2(g)  2 NH3(g)



2.15

1 = 2.15 XH2 

6.15

3  2.05 Therefore H₂ is limiting and X = 2.05

(28)

LIMITING REACTANTS EXAMPLE



Now suppose we mix 2.15 mol of N

₂

and 6.15 mol of

H

₂

. What is the theoretical yield of NH

₃

?

NH 2 H 3 + N (g) (g)

(g) 2 3

2 



N₂

(g) + 3 H2(g)  2 NH3(g)

(initial) I 2.15 mol 6.15 mol 0 mol

(change) C - 1(2.05) - 3(2.05) + 2(2.05) (end) E



2.15

1 = 2.15 XH2 

6.15

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LIMITING REACTANTS EXAMPLE



Now suppose we mix 2.15 mol of N

₂

and 6.15 mol of

H

₂

. What is the theoretical yield of NH

₃

?

NH 2 H 3 + N (g) (g)

(g) 2 3

2 



N₂

(g) + 3 H2(g)  2 NH3(g)

(initial) I 2.15 mol 6.15 mol 0 mol (change) C - 2.05 mol - 6.15 mol + 4.10 mol (end) E 0.10 mol 0 mol 4.10 mol



2.15

1 = 2.15 XH2 

6.15

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EXAMPLES: LIMITING REACTANTS

5. Butane, C

₄

H

₁₀

, undergoes combustion with oxygen,

O

₂

, to form carbon dioxide and water:

If 100. g of C

₄

H

₁₀

and 100. g of O

₂

are mixed,

a. Which of the two reactants is the limiting reagent?

g/mol 18.02 g/mol 44.01 g/mol 32.00 g/mol 58.12 (g) 2 ) ( 2 ) ( 2 ) ( 10

4

H

+

13 O

8 CO

+

10 H

O

C

(31)

EXAMPLES: LIMITING REACTANTS

7. Butane, C

₄

H

₁₀

, undergoes combustion with oxygen,

O

₂

, to form carbon dioxide and water:

If 100. g of C

₄

H

₁₀

and 100. g of O

₂

are mixed,

b. How many grams of carbon dioxide will be formed? Grams of H₂O formed?

c. How many grams of excess reactant will be left over?

g/mol 18.02 g/mol 44.01 g/mol 32.00 g/mol 58.12 (g) 2 ) ( 2 ) ( 2 ) ( 10

4

H

+

13 O

8 CO

+

10 H

O

C

(32)

EXAMPLES: LIMITING REACTANTS

7. Butane, C

₄

H

₁₀

, undergoes combustion with oxygen,

O

₂

, to form carbon dioxide and water:

If 100. g of C

₄

H

₁₀

and 100. g of O

₂

are mixed,

d. If the actual yield of CO₂ had been 75.0 g, what would be the percent yield of the reaction?



2 C

₄

H

₁₀₍_g₎

+ 13 O

₂₍_g₎



8 CO

₂₍_g₎

+ 10 H

₂

O

_(g)