chapter 4
DESCRIPTION
Chapter 4. Chemical Equations and Stoichiometry. CHEMICAL REACTIONS. Reactants: Zn + I 2. Product: Zn I 2. Chemical Equations. Depict the kind of reactants and products and their relative amounts in a reaction. 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) - PowerPoint PPT PresentationTRANSCRIPT
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Dr. S. M. Condren
Chapter 4
Chemical Equations
and
Stoichiometry
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Dr. S. M. Condren
CHEMICAL REACTIONS
Reactants: Zn + IReactants: Zn + I22 Product: Zn IProduct: Zn I22
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Dr. S. M. Condren
Chemical EquationsDepict the kind of reactants and
products and their relative amounts in a reaction.
4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
The numbers in the front are called
stoichiometric coefficientsThe letters (s), (g), and (l) are the physical
states of compounds, (aq) refers to an aqueous or water solution.
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Dr. S. M. Condren
The Mole and Chemical Reactions:
The Nano-Macro Connection
2 H2 + O2 -----> 2 H2O 2 H2 molecules 1 O2 molecule 2 H2O molecules
2 H2 moles molecules 1 O2 mole molecules 2 H2O moles molecules
4 g H2 32 g O2 36 g H2O
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Dr. S. M. Condren
Stoichiometry
stoi·chi·om·e·try noun1. Calculation of the quantities of reactants
and products in a chemical reaction.2. The quantitative relationship between
reactants and products in a chemical reaction.
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Dr. S. M. Condren
4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
This equation means
4 Al atoms + 3 O2 molecules ---give--->
2 “molecules” of Al2O3
4 moles of Al + 3 moles of O2 ---give--->
2 moles of Al2O3
Chemical Equations
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Dr. S. M. Condren
• Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change.
• The Law of the Conservation of Matter
Demo of conservation of matter
2HgO(s) ---> 2 Hg(liq) + O2HgO(s) ---> 2 Hg(liq) + O22(g)(g)
Chemical Equations
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Dr. S. M. Condren
Combination Reaction
PbNO3(aq) + K2CrO4(aq) PbCrO4(s) + 2 KNO3(aq)
Colorless yellow yellow colorless
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Dr. S. M. Condren
Because of the principle of the conservation of matter,
an equation must be balanced.
It must have the same number of atoms of the same kind on both sides.
Chemical Equations
Lavoisier, 1788Lavoisier, 1788
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Dr. S. M. Condren
___ Al(s) + ___ Br2(liq) ---> ___ Al2Br6(s)
BalancingEquations
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Dr. S. M. Condren
____C3H8(g) + _____ O2(g) ---->
_____CO2(g) + _____ H2O(g)
____B4H10(g) + _____ O2(g) ---->
___ B2O3(g) + _____ H2O(g)
BalancingEquations
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Dr. S. M. Condren
EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2?
H2 + O2 -----> H2O
#mol H2O =
22
(3.3 mol O2)(2 mol H2O)
(1 mol O2)
Now we need the stoichiometric factor
= 6.6 mol H2O
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Dr. S. M. Condren
Stoichiometric Roadmap
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Dr. S. M. Condren
EXAMPLE How much H2O, in grams results from burning an excess of H2 in 3.3 moles of O2?
H2 + O2 -----> H2O
#g H2O =
22
(3.3 mol O2)(2 mol H2O)
(1 mol O2)
= 1.2x102 g H2O
(18.0 g H2O)
(1 mol H2O)
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Dr. S. M. Condren
EXAMPLE How much H2O, in grams results from burning an excess of H2 in 3.3 grams of O2?
H2 + O2 -----> H2O#g H2O =
22
(1 mol O2) (2 mol H2O)
(1 mol O2)
= 3.7 g H2O
(18.0 g H2O)
(1 mol H2O)
(3.3 g O2)
(32.g O2)
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Dr. S. M. Condren
Thermite Reaction
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Thermite Reaction
Fe2O3(s) + 2Al(s) Al2O3(s) + 2 Fe(l)
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Dr. S. M. Condren
Thermite Reaction
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Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?
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Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron.
The mass of iron in 1 inch of this rail is:#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)
= 1.67 X 103 g/in
The mass of iron in a weld adding 10% mass:#g = (1.67 X 103 g) (0.10) = 167 g
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Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?
The mass of iron in a weld adding 10% mass:#g = (1.67 X 103 g) (0.10) = 167 g
Balanced chemical equation:Fe2O3 + 2 Al ---> 2 Fe + Al2O3
What mass of Fe2O3 is required for the thermite process?
#g Fe2O3 = (167 g Fe) (1 mol Fe)---------------- (55.85 g Fe)
(1 mol Fe2O3)------------------- (2 mol Fe)
(159.7 g Fe2O3)---------------------- (1 mol Fe2O3)
= 238 g Fe2O3
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Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?
The mass of iron in a weld adding 10% mass:#g = (1.67 X 103 g) (0.10) = 167 g
Balanced chemical equation:Fe2O3 + 2 Al ---> 2 Fe + Al2O3
What mass of Al is required for the thermite process?
#g Al = (167 g Fe) (1 mol Fe)---------------- (55.85 g Fe)
(2 mol Al)--------------(2 mol Fe)
(26.9815 g Al)------------------- (1 mol Al)
= 80.6 g Al
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Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron.
The mass of iron in a weld adding 10% mass:
#g Fe = 167 g Fe
#g Fe2O3 = 238 g Fe2O3
#g Al = 80.6 g Al
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Dr. S. M. Condren
• In a given reaction, there is not In a given reaction, there is not enough of one reagent to use up enough of one reagent to use up the other reagent completely.the other reagent completely.
• The reagent in short supply The reagent in short supply
LIMITSLIMITS the quantity of product the quantity of product that can be formed.that can be formed.
Reactions Involving aLIMITING REACTANT
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LIMITING REACTANTS
Reactantseactants ProductsProducts
2 NO(g) + O2 (g) 2 NO2(g)
Limiting reactant = ___________Limiting reactant = ___________Excess reactant = ____________Excess reactant = ____________
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Dr. S. M. Condren
EXAMPLEWhat is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
balanced equation relates:2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)
have only:1Fe2S3 (S) <=> 2H2O(l) <=> 3O2(g)
not enough H2O to use all Fe2S3
plenty of O2
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Dr. S. M. Condren
EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all Fe2S3:
(1.0 mol Fe2S3) (4 mol Fe(OH)3) #mol Fe(OH)3 = ------------------------------------------
(2 mol Fe2S3)
= 2.0 mol Fe(OH)3
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Dr. S. M. Condren
EXAMPLE: What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all H2O:
(2.0 mol H2O) (4 mol Fe(OH)3) #mol Fe(OH)3 = -----------------------------------------
(6 mol H2O)= 1.3 mol Fe(OH)3
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Dr. S. M. Condren
EXAMPLE: What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all O2
(3.0 mol O2) (4 mol Fe(OH)3) #mol Fe(OH)3 = ---------------------------------------
(3 mol O2)
= 4.0 mol Fe(OH)3
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Dr. S. M. Condren
EXAMPLE: What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
1.0 mol Fe2S3 => 2.0 mol Fe(OH)3
2.0 mol H2O => 1.3 mol Fe(OH)3
3.0 mol O2 => 4.0 mol Fe(OH)3
least amount
Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3,then H2O is the limiting reactant.
Thus the maximum number of moles of Fe(OH)3 that can be produced by this reaction is 1.3 moles.
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Theoretical Yield
the amount of product produced by a reaction based on the amount of the limiting reactant
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Actual Yield
amount of product actually produced in a reaction
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Percent Yield
actual yield% yield = --------------------- * 100 theoretical yield
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for the overall process
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00 kg Cl2) #kg N2H4 = ---------------------
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00 kg Cl2) (1000 g Cl2)#kg N2H4 = -----------------------------------
(1 kg Cl2) metric conversion
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00) (1000 g Cl2) (1 mol Cl2) #kg N2H4 = -----------------------------------------
(1) (70.9 g Cl2) molar mass
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1 mol Cl2) #kg N2H4 = -----------------------------------------
(1)(70.9)
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1 mol Cl2)(1 mol N2H4) #kg N2H4 = -------------------------------------------------
(1) (70.9) (1 mol Cl2)
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1)(1 mol N2H4) #kg N2H4 = -------------------------------------------------
(1) (70.9)(1)
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1)(1 mol N2H4) (32.0 g N2H4) #kg N2H4 = --------------------------------------------------------
(1)(70.9) (1) (1 mol N2H4)
molar mass
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1)(1) (32.0 g N2H4)(1 kg N2H4)#kg N2H4 = ----------------------------------------------------------
(1)(70.9)(1)(1) (1000 g N2H4)
metric conversion
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)
#kg N2H4 = ----------------------------------------------------------(1)(70.9)(1)(1)(1000)
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation
for the overall process(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)
#kg N2H4 = ----------------------------------------------------------(1)(70.9)(1)(1)(1000)
= 0.451 kg N2H4
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product) # kg N2H4 = --------------------------
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield (0.299 kg product) (98.0 kg N2H4)
# kg N2H4 = -------------------------------------------- (100 kg product)
purity factor
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield (0.299 kg product) (98.0 kg N2H4)
# kg N2H4 = -------------------------------------------- (100 kg product)
= 0.293 kg N2H4
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield # kg N2H4 = 0.293 kg N2H4
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Dr. S. M. Condren
EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield # kg N2H4 = 0.293 kg N2H4
(c) percent yield 0.293 kg
% yield = -------------- X 100 = 65.0 % yield 0.451kg