9.1-9.3 Stoichiometry.ppt

Converting grams to moles.

Determine how many moles there are in 5.17 grams of Fe(C5H5)2.

Goal

= moles Fe(C₅H₅)₂ Given

5.17 g Fe(C₅H₅)₂

Use the molar mass to convert grams to

moles.

Fe(C₅H₅)₂

2 x 5 x 1.001 = 10.01 2 x 5 x 12.011 = 120.11 1 x 55.85 = 55.85

mol

g

185.97

g

185.97

mol

2

Stoichiometry (more working with ratios)

Ratios are found within a chemical equation.

2HCl + Ba(OH)1 ₂  2H₂O + BaCl1 ₂

2 moles of HCl react with 1 mole of Ba(OH)₂ to form 2 moles of H₂O and 1 mole of BaCl₂

When N₂O₅ is heated, it decomposes:

2

N

O

(g)



4

NO

(g) + O

(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

= moles NO

4.3 mol N

O

5 2 2

O

N

mol

2

NO

mol

4

8.6

b. How many moles of O2 can be produced from 4.3 moles of N2O5?

= mole O

4.3 mol N

O

O

N

2mol

O

mol

1

2.2

2

N

O

(g)



4

NO

(g) + O

(g)

4.3 mol

? mol

2

N

O

(g)



4

NO

(g) + O

(g)

4.3 mol ? mol

Mole – Mole Conversions

4

When N

O

is heated, it decomposes:

2

N

O

(g)



4

NO

(g) + O

(g)

a. How many moles of N2O5 were used if 210g of NO2 were produced?

= moles N

O

210 g NO

2 5 2

NO

mol

4

O

N

mol

2

2.28

b. How many grams of N2O5 are needed to produce 75.0 grams of O2?

= grams N

O

75.0

g O

2 5 2

O

1mol

O

N

mol

2

NO

g

0

.

46

NO

mol

O

g

32.0

O

mol

O

N

mol

O

N

g

108

gram ↔ mole and gram ↔ gram conversions

2N₂O₅(g)  4NO₂(g) + O₂(g) 210g

? moles

2N₂O₅(g)  4NO₂(g) + O₂(g) 75.0 g ? grams

Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

First write a balanced equation.

6 Aluminum is an active metal that when placed in hydrochloric acid

produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

Al(s) + HCl(aq) 2 6 2 3 AlCl₃(aq) + H₂(g)

Now let’s get organized. Write the information

below the substances.

3.45 g ? grams

Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

Al(s) + HCl(aq) 2 6 2 3 AlCl₃(aq) + H₂(g)

3.45 g ? grams

Let’s work the problem.

= g AlCl₃

3.45 g Al

Al g 27.0

Al mol

We must always convert to Now use the molar ratio.

Al mol 2 AlCl mol 2 ₃

8 Molarity

Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M)

When working problems, it is a good idea to change M

into its units.

mL

1000

moles

Liter

moles

A solution is prepared by dissolving 3.73 grams of AlCl

in

water to form 200.0 mL solution. A 10.0 mL portion of

the solution is then used to prepare 100.0 mL of solution.

Determine the molarity of the final solution.

1

:

= mol

L

3.73 g

g

133.4

mol

200.0 x 10

L

0.140

2

:

M

V

= M

V

(0.140 M)(10.0 mL) = (? M)(100.0 mL)

0.0140 M = M

molar mass of AlCl3

dilution formula

50.0 mL of 6.0 M H

SO

(battery acid) were spilled and

solid NaHCO

(baking soda) is to be used to neutralize the

acid. How many grams of NaHCO

must be used?

H

SO

(aq) + 2NaHCO



2H

O(l) + Na

SO

(aq) + 2CO

(g)

12

50.0 mL

6.0 M

L

mol

6.0

? g

Look!

A conversion

factor!

50.0 mL of 6.0 M H

SO

(battery acid) were spilled and

solid NaHCO

(baking soda) is to be used to neutralize the

acid. How many grams of NaHCO

must be used?

H

SO

(aq) + 2NaHCO



2H

O(l) + Na

SO

(aq) + 2CO

(g)

Solution Stoichiometry

50.0 mL

6.0 M

L

mol

6.0

? g

50.0 mL of 6.0 M H

SO

(battery acid) were spilled and

solid NaHCO

(baking soda) is to be used to neutralize the

acid. How many grams of NaHCO

must be used?

H

SO

(aq) + 2NaHCO



2H

O(l) + Na

SO

(aq) + 2CO

(g)

Solution Stoichiometry

=

Our Goal

= g

Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H₂SO₄ solution.

First write a balanced Equation.

16 Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H₂SO₄ solution.

Now, let’s get organized. Place numerical Information and accompanying UNITS below each

compound.

____ 2 1 2 1NaOH + ____H₂SO₄  ____H₂O + ____Na₂SO₄ 0.102 M

L mol

? mL

35.0 mL

mL 1000

mol 0.125

L mol 0.125



Since 1 L = 1000 mL, we can use

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H₂SO₄ solution.

Now let’s get to work

____ 2 1 2 1NaOH + ____H₂SO₄  ____H₂O + ____Na₂SO₄ 0.102 M L mol

? mL

35.0 mL mL 1000 mol 0.125 L mol 0.125 

= mL NaOH

H₂SO₄

35.0 mL H_{0.125 mol} 2SO4 1000 mL H₂SO₄

NaOH 2 mol 1 mol H₂SO₄

1000 mL NaOH

0.102 mol NaOH 85.8

Units Match

Solution Stoichiometry:

What volume of 0.40 M HCl solution is needed to

completely neutralize 47.1 mL of 0.75 M Ba(OH)

?

1st write out

a balanced chemical

equation

20

What volume of 0.40 M HCl solution is needed to

completely neutralize 47.1 mL of 0.75 M Ba(OH)

?

2HCl(aq) + Ba(OH)

(aq)



2H

O(l) + BaCl

0.40 M

47.1 mL

0.75 M

? mL

= mL

Ba(OH)

47.1 mL

2 2

Ba(OH) Ba(OH)

mL

1000

0.75mol

1 mol

Ba(OH)

HCl

2 mol

0.40 mol

HCl

HCl

1000 mL

176

Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

First write a balanced chemical reaction.

____2 1 2 1HCl(aq) + ____Ba(OH)₂(aq)  ____H₂O(l) + ____BaCl₂(aq) 23.28 mL

0.135 mol L

24 Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

____2 1 2 1HCl(aq) + ____Ba(OH)2(aq)  ____H2O(l) + ____BaCl2(aq) 23.28 mL 0.135 mol L 25.00 mL ? mol L

= mol Ba(OH)₂ L Ba(OH)2

25.00 x 10-3 L

Ba(OH)₂

Units Already Match on Bottom!

HCl mL 23.28 HCl HCl

mL

1000

mol

0.135

mol

2

mol

l

26

48.0 mL of Ca(OH)

solution was titrated with 19.2

mL of 0.385 M HNO

. Determine the molarity of the

Ca(OH)

solution.

We must first

write a balanced

equation.

48.0 mL of Ca(OH)

solution was titrated with 19.2

mL of 0.385 M HNO

. Determine the molarity of the

Ca(OH)

solution.

Ca(OH)

(aq) + HNO

2

(aq)



2

H

O(l) + Ca(NO

)

(aq)

48.0 mL

19.2 mL

0.385 M

L

mol

0.385



=

mol

L

19.2 mL

HNO

mL

1000

mol

0.385

HNO

2mol

Ca(OH)

1mol

48.0 x 10

L

? M

units match!

0.0770

Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO₂, is used in rebreathing gas masks to generate oxygen.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

First copy down the

the BALANCED

equation!

Now place

numerical the

information below

30 Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

0.15 mol

0.10 mol

? moles

Two starting

amounts?

Where do we

start?

Hide

Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO₂, is used in rebreathing gas masks to generate oxygen.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

0.15 mol

0.10 mol

? moles

Based on:

KO₂ 0.15 mol KO2

= mol O

2 2

KO

4mol

O

mol

3

32

Potassium superoxide, KO₂, is used in rebreathing gas masks to generate oxygen.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

0.15 mol

0.10 mol

? moles

Based on:

KO₂ 0.15 mol KO2

= mol O

2 2

KO

4mol

O

mol

3

0.1125

Hide

Based on:

H₂O

= mol O

0.10 mol H₂O

O

H

2mol

O

mol

3

2 2

Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO₂, is used in rebreathing gas masks to generate oxygen.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant.

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

0.15 mol

0.10 mol

? moles

Based on:

KO₂ 0.15 mol KO2

= mol O

2 2

KO

4mol

O

mol

3

0.1125

H₂O

= mol O

0.10 mol H₂O

O

H

2mol

O

mol

3

0.150

What is the theoretical yield?

Hint: Which is the smallest amount? The is based upon the

It was limited by the amount of KO₂.

34

Theoretical yield

vs.

Actual yield

Suppose the theoretical yield for an

experiment was calculated to be

19.5 grams, and the experiment

was performed, but only 12.3

grams of product were recovered.

Determine the % yield.

Theoretical yield = 19.5 g based on limiting reactant

Actual yield = 12.3 g experimentally recovered

100

x

yield

l

theoretica

yield

actual

yield

%



yield

63.1%

100

x

19.5

12.3

yield

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

If a reaction vessel contains 120.0 g of KO

and 47.0 g of H

O,

how many grams of O

can be produced?

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

120.0 g

Hide one

47.0 g

? g

Based on:

KO

= g O

120.0 g KO₂

g

1

.

71

mol

2 2

KO

4mol

O

mol

3

2 2

O

mol

O

g

0

.

32

36

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

If a reaction vessel contains 120.0 g of KO

and 47.0 g of H

O,

how many grams of O

can be produced?

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

120.0 g

47.0 g

? g

Based on:

KO

= g O

120.0 g KO₂

g

1

.

71

mol

KO

4mol

O

mol

3

O

mol

O

g

0

.

32

H₂O = g O2

Question if only 35.2 g of O2 were recovered, what was the percent yield?

yield

86.9%

100

x

51

.

40

2

.

35

100

x

l

theoretica

actual





47.0 g H₂O

O

H

g

02

.

18

O

H

mol

O

H

mol

2

O

mol

3

O

mol

O

g

0

.

32

If a reaction vessel contains 120.0 g of KO

and 47.0 g of H

O,

how many grams of O

can be produced?

4KO

(s) + 2H

O(l)



4KOH(s) + 3O

(g)

120.0 g

47.0 g

? g

Based on:

KO

= g O

120.0 g KO₂

g

1

.

71

mol

KO

4mol

O

mol

3

O

mol

O

g

0

.

32

H₂O = g O2

47.0 g H₂O

O

H

g

02

.

18

O

H

mol

O

H

mol

2

O

mol

3

O

mol

O

g

0

.

32

Determine how many grams of Water were left over.

The Difference between the above amounts is directly RELATED to the XS H2O.

125.3 - 40.51 = 84.79 g of O₂ that could have been formed from the XS water.

= g XS H₂O

84.79 g O₂

Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO₃)₃ in 455 mL of solution.

L

mol

0.264

L

10

x

455

g

213

mole

g

25.6

3

-



worked the problem, click

here to see setup answer