Transcript Modern Chemistry Chapter 9 Stoichiometry

Modern Chemistry Chapter 9
Stoichiometry
Modern Chemistry Chapter 9
Stoichiometry

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composition
stoichiometry deals with
the mass relationships of
elements in compounds.
reaction stoichiometry
involves the mass
relationships between
reactants and products in a
chemical reaction.
Types of Stoichiometry Problems
1)
2)
3)
4)
mole to mole (Both the given
and the unknown quantities are
amounts in moles.)
mole to mass (The given
amount is in moles and the
unknown amount is in grams.)
mass to mole (The given
amount is in grams and the
unknown amount is in moles.)
mass to mass (Both the
given and the unknown amount
is in grams.)
Mole Ratio & Molar Mass


mole ratio- A
conversion factor that
relates the amounts in
moles of any two
substances involved in
a chemical reaction.
Found by using the
coefficients in the
balanced formula
equation of the
reaction.
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molar mass- Equal
to the mass in grams
of one mole of an
element or a
compound.
Found by adding the
individual element
atomic masses from
the formula of the
compound.
Section Review Problems

Do section review
problems
#1 through #4 on
page 301 of the
textbook.
Section Review page 301
1- The branch of chemistry that deals with mass
relationships in compounds and in chemical reactions.
2 HgO  2 Hg + O2
2- a)
2 mol HgO
2 mol Hg
&
2 mol HgO
1 mol O2
2 mol Hg
2 mol HgO
&
2 mol Hg
1 mol O2
1 mol O2
2 mol HgO
&
1 mol O2
2 mol Hg
Section Review page 301
3- It is used to convert moles of one
substance into moles of another
substance.
4- The formula equation MUST be
BALANCED so mole ratios can be
determined.
Using Conversion Factors
Amount
of given
substance
in moles
# moles
given
x
x
coefficient
of unknown
coefficient
of known
mole ratio
=
=
amount of
unknown
substance
in moles
#moles of
unknown
Do practice problems #1 & #2 on page 306 of text.
Page 306 #1
3 H2
+ N2

2 NH3
6 mol H2 x 2 mol NH3
3 mol H2
= 4mol NH3
Page 306 #2
2 KClO3  2 KCl + 3 O2
15 mol O2 x 2 mol KClO3 = 10 mol KClO3
3 mol O2
Chapter 9 quiz #1- mole to mole calculations
6 NaBr + Mg3(PO4)2  2 Na3PO4 + 3 MgBr2
Use the above balanced formula equation to solve
the following:
1- 7.0 moles of NaBr will produce ? moles Na3PO4
2- 3.0 moles of Mg3(PO4)2 will yield ? moles MgBr2
3- 0.5 moles of NaBr will react with ? moles Mg3(PO4)2
4- 2.5 moles of NaBr will yield ? moles Na3PO4
5- 2.5 moles NaBr will produce ? moles MgBr2
6 NaBr + Mg3(PO4)2  2 Na3PO4 + 3 MgBr2
1-
7 mol NaBr x 2 mol Na3PO4
6 mol NaBr
= 2.3 mol Na3PO4
2-
3 mol Mg3(PO4)2 x 3.0 mol MgBr2 = 9 mol MgBr2
1 mol Mg3(PO4)2
3-
0.5 mol NaBr x 1 mol Mg3(PO4)2 = 0.083 mol
6 mol NaBr
4-
2.5 mol NaBr x 2 mol Na3PO4 = 0.83 mol Na3PO4
6 mol NaBr
5-
2.5 mol NaBr x 3 mol MgBr2 = 1.25 mol MgBr2
6 mol NaBr
Using Conversion Factors
amount of
given
substance
in moles
# moles
given
x moles unknown
moles known
x
mole
ratio
x
x molar
=
mass of
unknown
mass in
grams of
unknown
substance
molar
= mass of unknown
mass
(in grams)
unknown
Do practice problems #1 & #2 on page 308 of the textbook.
Page 308 #1 & 2
2 Mg + O2  2 MgO
2.00 mol Mg x 2 mol MgO x 40.3 g MgO = 80.6 g MgO
2 mol Mg
1 mol MgO
6 CO2 + 6 H2O  C6H12O6 + 6 O2
10 mol CO2 x 1 mol C6H12O6 x 180 g C6H12O6 = 300 g C6H12O6
6 mol CO2
1 mol C6H12O6
Chapter 9 Quiz #2- mole-mass problems
3 MgF2 + Al2O3  3 MgO + 2 AlF3
Use the above balanced formula equation to answer
the following questions.
1- 2.0 mol MgF2 will yield ? grams of MgO
2- 4.0 mol of Al2O3  ? grams of AlF3
3- If 6.0 mol of MgO is produced, ? grams of AlF3
4- 0.6 mol MgF2  ? grams of AlF3
5- 2.75 mol Al2O3  ? grams of MgO
3 MgF2 + Al2O3  3 MgO + 2 AlF3
1- 2.0 mol MgF2 x 3 mol MgO x 40.3 g MgO = 80.6 g
3 mol MgF2
1 mol MgO
MgO
2- 4.0 mol Al2O3 x 2 mol AlF3 x 84.0 g AlF3 = 672.0g
1 mol Al2O3
1 mol AlF3
AlF3
3- 6.0 mol MgO x 2 mol AlF3 x 84.0 g AlF3 = 336 g AlF3
3 mol MgO
1 mol AlF3
4- 0.6 mol MgF2 x 2 mol AlF3 x 84.0 g AlF3 = 33.6 g AlF3
3 mol MgF2 1 mol AlF3
5- 2.75 mol Al2O3 x 3 mol MgO x 40.3 g MgO = 332gMgO
1 mol Al2O3 1 mol MgO
Using Conversion Factors
mass (g) x
of given
substance
grams x
1 mol given
molar mass
of given
x
mol unknown
mol given
= moles of
unknown
substance
1
x mole ratio = moles unknown
molar mass
Do practice problems #1 & #2 on page 309 of the textbook.
Practice problems page 309
2 HgO  2 Hg + O2
125 g O2 x 1 mol O2
32 g O2
x 2 mol HgO = 7.81 mol HgO
1 mol O2
125 g O2 x 1 mol O2 x 2 mol Hg = 7.81 mol Hg
32 g O2
1 mol O2
chapter 9 quiz #3- mass-mole problems
Na2O + CaF2  2 NaF + CaO
Use the above equation to solve the problems.
1- 156.1 grams of CaF2  ? mol CaO
2- 186 g Na2O  ? mol NaF
3- 31 g Na2O  ? mol CaO
4- 31 g Na2O  ? mol NaF
5- A yield of 84 g NaF  ? mol CaO
Na2O + CaF2  2 NaF + CaO
1- 156.1 g CaF2 x 1 mol CaF2 x 1 mol CaO = 2.00 mol CaO
78.1 g CaF2 1 mol CaF2
2-
186.0 g Na2O x 1 mol Na2O x 2 mol NaF = 6.0 mol NaF
62 g Na2O
1 mol Na2O
3-
31.0 g Na2O x 1 mol Na2O x 1 mol CaO = 0.5 mol CaO
62.0 g Na2O
1 mol Na2O
4-
31.0 g Na2O x 1 mol Na2O x 2 mol NaF
62.0 g Na2O
1 mol Na2O
5-
84.0 g NaF x 1 mol NaF x
42.0 g NaF
1 mol CaO
2 mol NaF
= 1.0 mol NaF
= 1.0 mol CaO
Chemistry Chapter 9- Stoichiometry Practice Problems
2 NaF + CaO  Na2O + CaF2
1- 4.5 moles of NaF will produce -?- moles of
Na2O ?
4.5 mol NaF x 1 mol Na2O/2 mol NaF =
2.25 moles Na2O
2- 3.2 moles of CaO will produce -?of CaF2 ?
grams
3.2 mol CaO x 1 mol CaF2/1 mol CaO x
78.1 g CaF2/mol CaF2 = 249.9 g CaF2
2 NaF + CaO  Na2O + CaF2
3-
168.0 grams of NaF will produce -?moles of Na2O ?
168.0 /42.0
4-
x
1 /2 = 2.0 mol Na2O
112.2 grams of CaO will produce -?moles of CaF2 ?
112.2/56.1 x 1/1 = 2.0 mol CaF2
5-Calculate the molar mass of each of the reactants &
products of the above balanced formula equation. Use
the molar masses in the following problems.
a- AlN =
(1 x 27.0)
b- Na2O =
(2 x 23.0)
c- Al2O3 =
(2 x 27.0)
d- Na3N =
(3 x 23.0)
+ (1 x 14.0) = 41.0 g/mol
+ (1 x 16.0) = 62.0 g/mol
+ (3 x 16.0) = 102.0 g/mol
+ (1 x 14.0) = 83.0 g/mol
2 AlN + 3 Na2O  Al2O3 + 2 Na3N
6-
82.0 grams of AlN will produce -?moles of Al2O3 ?
82.0/41.0 x 1/2 = 1.0 mol Al2O3
7-
164.0 grams AlN will produce -?- moles
of Na3N ?
164.0/41.0 x 2/2 = 4.0 mol Na3N
2 AlN + 3 Na2O  Al2O3 + 2 Na3N
8-
2.5 moles of Na2O will produce -?grams of Na3N ?
2.5 x 2/3 x 83.0 = 138.3 g Na3N
9-
0.75 moles of Na2O will produce -?moles of Al2O3 ?
0.75 x 1/3 = 0.25 mol Al2O3
10- 11.0 moles of Na2O will produce -?grams of Na3N ?
11.0 x 2/3 x 83.0 = 608.7 g Na3N
2 H2 + O2  2 H2O
11- 8.0 grams of H2 will react with -?- moles
of O2 ?
8.0/2.0 x 1/2 = 2.0 mol O2
12- 64.0 grams of O2 will produce -?- moles
of H2O ?
64.0/32.0 x 2/1 = 4.0 mol H2O
13- 0.25 moles of H2 will produce -?- moles of
H2O ?
0.25 x 2/2 = 0.25 mol H2O
2 H2 + O2  2 H2O
14- 1.5 moles of H2 will produce -?- grams
of H2O ?
1.5 x 2/2 x 18.0 = 27.0 g H2O
15- 14 moles of O2 will produce -?- moles of
H2O ?
14 x 2/1 = 28.0 mol H2O
Using Conversion Factors
mass (g) x 1 mol given x mol unknown x molar mass unknown = mass
of given
molar mass
mol given
1 mol unknown
of unsubstance
of given
known
grams x
1
x mole ratio x molar mass = mass of unknown
given molar mass
Do practice problems #1, #2, & #3 on page 311 of textbook.
Do section review problems #1 - #5 on page 311 of textbook.
Practice problems page 311
NH4NO3  N2O + 2 H2O
32 g N2O x 1 mol N2O x 1 mol NH4NO3 x 80 g NH4NO3 = 60 g NH4NO3
44 g N2O
1 mol N2O
1 mol NH4NO3
Chapter 9 quiz #4- mass-mass problems
Na2O + CaF2  2 NaF + CaO
1-
124 g Na2O  ? grams NaF
2-
124 g Na2O  ? grams CaO
3-
234.3 g CaF2  ? g NaF
4-
234.3 g CaF2  ? g CaO
5-
84.0 g NaF  ? g CaO
Na2O + CaF2  2 NaF + CaO
1-
124 g Na2O x 1 mol Na2O x 2 mol NaF x 42.0 g NaF = 168 g
62 g Na2O
1 mol Na2O 1 mol NaF
2- 124 g Na2O x 1 mol Na2O x 1 mol CaO x 56.1 g CaO = 112.2 g
62 g Na2O
1 mol Na2O 1 mol CaO
3- 234.3 g CaF2 x 1 mol CaF2 x 2 mol NaF x 42.0 g NaF = 252 g
78.1 gCaF2 1 mol CaF2
1 mol NaF
4- 234.3 g CaF2 x 1 mol CaF2 x 1 mol CaO x 56.1 g CaO = 168.3 g
78.1 gCaF2 1 mol CaF2
1 mol CaO
5- 84.0 g NaF x 1 mol NaF x 1 mol CaO x 56.1 g CaO = 56.1 g
42.0 g NaF
2 mol NaF
1 mol CaO
The “MOLE HILL”
x mole ratio
# moles known
# moles unknown
÷ molar mass
of known
x molar mass
of unknown
mass of known
mass of unknown
Stoichiometry Practice Problems
2 H2 + O2  2 H2O
1)
2.5 mol H2 x 1 mol O2 = 1.25 mol O2
2 mol H2
2)
2.5 mol H2 x 2 mol H2O = 2.5 mol H2O
2 mol H2
3)
2.5 mol H2 x 1 mol O2 x 32 g O2 = 40 g O2
2 mol H2
1 mol O2
4)
2.5 mol H2 x 2 mol H2O x 18 g H2O = 45 g H2O
2 mol H2
1 mol H2O
Stoichiometry Practice Problems
2 H2 + O2  2 H2O
5)
16 g H2 x 1 mol H2 x 1 mol O2 = 4.0 mol O2
2 g H2
2 mol H2
6)
16 g H2 x 1 mol H2 x 2 mol H2O = 8.0 mol H2O
2.0 g H2 2 mol H2
7) 16 g H2 x 1 mol H2 x 1 mol O2 x 32 g O2 = 128 g O2
2.0 g H2 2 mol H2 1 mol O2
8) 16 g H2 x 1 mol H2 x 2 mol H2O x 18 g H2O = 144 g H2O
2.0 g H2
2 mol H2
1 mol H2O
Limiting Reactant & Percentage Yield
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
limiting reactant is the
reactant that limits the
amount of the other
reactant that can
combine and the amount
of product that can be
formed in a chemical
reaction.
excess reactant is the
substance that is NOT
completely used up in a
chemical reaction.
Sample & Practice Problems
See sample problem F on page 313 of textbook.
Do practice problems #1 on page 313.
See sample problem G on pages 314-315.
Do practice problems #1 & #2 on page 315.
QuickLAB

Do the QuickLAB
titled “Limiting
Reactants in a
Recipe” on page
316 of the
textbook.
Yes, cooking IS
chemistry!
Percentage Yield




theoretical yield is the maximum
amount of product that can be produced
from a given amount of reactant
actual yield of a product is the
measured amount of a product obtained
from a reaction
percentage yield is the ratio of the
actual yield to the theoretical yield
multiplied by 100
percentage yield = actual yield
x 100
theoretical yield
Problems

see sample problem H on pages 317318 of textbook

Do practice problems #1 & #2 on page
318.

Do section review problems
#4 on page 318.

Do critical thinking problems #37,
#38, #39, & #40 on pages 322 & 323.
#1 -
Chapter 9 test review

20 multiple choice questions
• definitions of composition & reaction
stoichiometry
• mole ratios: their definition & use
• SI units of molar mass
• identify mole ratio from balanced
formula equation
• 5 mole-mole problems
• Definitions & practical applications of
excess reactant & limiting reactant
• definitions & practical applications of
theoretical yield, actual yield, & % yield
Honors Chemistry Chapter 9 Test Review



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
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
25 multiple choice
Know the definitions of reaction & composition
stoichiometry, mole ratio, and units of molar mass.
Know what mole ratio means and how it is used in
stoichiometry.
Determine mole ratio using balanced formula equation.
(2)
Perform mole to mole stoichiometric calculations (4).
Perform mole to mass stoichiometric calculations (1).
Perform mass to mole stoichiometric calculations (1).
Perform mass to mass stoichiometric calculations (1).
Know definitions and applications of limiting and excess
reactants.
Know the definitions & applications of actual yield,
theoretical yield, and percent yield.
Practice #2
2 Na3PO4 + 3 CaSO4  3 Na2SO4 + Ca3(PO4)2
Known = 3 moles CaSO4 unknown = ? Moles Na3PO4
3 x 2/3 = 2 moles Na3PO4
Known = 2.0 moles Na3PO4
unknown = ? Moles Na2SO4
2 x 3/2 = 3 moles Na2SO4
Known = 1.5 moles Ca3(PO4)2
unknown = ? Moles CaSO4
1.5 x 3/1 = 4.5 moles CaSO4
Known = 4.4 moles CaSO4
unknown = ? Moles Na3PO4
4.4 x 2/3 = 2.9 moles Na3PO4
2 Na3PO4 + 3 CaSO4  3 Na2SO4 + Ca3(PO4)2
Known = 6.0 moles Na3PO4
unknown = ? Moles Na2SO4
6.0 x 3/2 = 9.0 moles Na2SO4
Known = 5.4 moles CaSO4
unknown = ? Moles Na3PO4
5.4 x 2/3 = 3.6 moles Na3PO4
Known = 0.6 moles Na3PO4
unknown = ? Moles CaSO4
0.6 x 3/2 = 0.9 moles CaSO4
Stoichiometry Practice #3
4 Na3N + 3 O2 
6 Na2O +
2 N2
Assume you have 12.0 moles of Na3N.
1)
How many moles of O2 do you need?
2)
How many moles of Na2O will you get?
3)
How many moles of N2 will you get?
Stoichiometry Practice #3
4 Na3N + 3 O2 
6 Na2O +
2 N2
Assume you have 12.0 moles of Na3N.
How many moles of O2 do you need?
12 x 3/4 = 9 moles O2
How many moles of Na2O will you get?
12 x 6/4 = 18 moles Na2O
How many moles of N2 will you get?
12 x 2/4 = 6 moles N2
Practice #3
4 Na + O2  2 Na2O
If you have 4 moles of Na, how many grams
of O2 will you need?
If you have 64 grams of O2, how many
moles of Na2O will you produce?
If you have 46 grams of Na, how many
grams of O2 will you need?
Practice #3
4 Na + O2  2 Na2O
If you have 4 moles of Na, how many grams
of O2 will you need?
4 x 1/4 x 32 = 32 grams O2
If you have 64 grams of O2, how many
moles of Na2O will you produce?
64/32 x 2/1 = 4 moles Na2O
If you have 46 grams of Na, how many
grams of O2 will you need?
46/23 x 1/4 x 32 = 16 grams O2