Transcript Stoichiometry

Proportional Relationships
• Stoichiometry
– mass relationships between substances in a chemical
reaction
– based on the mole ratio
• Mole Ratio
– indicated by coefficients in a balanced equation
2 Mg + O2  2 MgO
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Stoichiometry Island Diagram
Mass
Known
Unknown
Substance A
Substance B
Mass
Use coefficients
from balanced
chemical equation
Mole
Mole
Particles
Particles
Stoichiometry Island Diagram
Visualizing a Chemical Reaction
2 Na
10 mole Na
___
+
Cl2
5 mole Cl2
___
2 NaCl
10
? mole NaCl
___
Stoichiometry Problems
• How many moles of KClO3 must decompose in
order to produce 9 moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
9 mol O2
2 mol KClO3
3 mol O2
9 mol
= 6 mol KClO3
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Stoichiometry Problems
• How many grams of silver will be formed from
12.0 g copper?
Cu + 2AgNO3  2Ag + Cu(NO3)2
12.0 g
12.0
g Cu
?g
1 mol
Cu
63.55
g Cu
2 mol
Ag
1 mol
Cu
107.87
g Ag
1 mol
Ag
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= 40.7 g
Ag
Rocket Fuel
The compound diborane (B2H6) was at one time
considered for use as a rocket fuel. How many grams
of liquid oxygen would a rocket have to carry to burn
10 kg of diborane completely?
(The products are B2O3 and H2O).
Chemical equation
B2H6 + O2
Balanced chemical equation B2H6 + 3 O2
10 kg
x g O2 = 10 kg B2H6
B2O3 + 3 H2O
xg
1000 g B2H6 1 mol B2H6
1 kg B2H6
B2O3 + H2O
28 g B2H6
3 mol O2
32 g O2
1 mol B2H6 1 mol O2
X = 34,286 g O2
Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
–
–
–
Mole ratio =
Molar mass =
Avogadro’s number = particles  moles
– Mole ratio =
moles  moles
moles  grams
moles  moles
Core step in all stoichiometry problems!!
4. Check answer.
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Limiting Reactants
• Limiting Reactant
– used up in a reaction
– determines the amount of product
• Excess Reactant
– added to ensure that the other reactant is
completely used up
– cheaper & easier to recycle
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Percent Yield
measured in lab
% yield =
actual yield
theoretical yield
calculated on paper
x 100
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl
are formed. Calculate the theoretical and % yields of KCl.
actual yield
46.3 g
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
excess
?g
theoretical yield
Theoretical yield
x g KCl = 45.8 g K2CO3
% Yield =
1 mol K2CO3
2 mol KCl 74.5 g KCl
= 49.4 g KCl
1
mol
K
CO
1
mol
KCl
138 g K2CO3
2
3
Actual Yield
Theoretical Yield
% Yield =
46.3 g KCl
x 100
% Yield = 93.7% efficient
Exothermic Reaction
Reactants  Products + Thermal Energy
(Enthalpy)
Energy of reactants
Energy
Energy of products
Reactants
-DH
Products
Reaction Progress
Endothermic Reaction
Thermal Energy + Reactants  Products
(Enthalpy)
Energy
Activation
Energy
Reactants
Products
+DH
Reaction progress
Barbecue
An LP gas tank in a home barbecue contains 11.8 X 103g of propane (C3H8).
Calculate the heat (in kJ) associated with the complete combustion of all of
the propane in the tank. The heat of reaction is -2044 kJ/mol C3H8.
__C3H8 + __O2(g)  __CO2(g) + __H2O(g)
kJ = 11.8 X 103 g C3H8
1 mol C3H8
-2044 kJ
44 g C3H8
1 mol C3H8
kJ = -5.47 X 105 kJ
Enthalpy Changes
H2(g) + ½ O2(g)
Energy
DH = +242 kJ
Endothermic
-242 kJ
Exothermic
-286 kJ
Endothermic
DH = -286 kJ
Exothermic
H2O(g)
+44 kJ
Endothermic
-44 kJ
Exothermic
H2O(l)
H2(g) + 1/2O2(g)  H2O(g) + 242 kJ
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211
DH = -242 kJ
Enthalpy Changes
DH3 = DH1 + DH2 (Hess’s Law)
Change in enthalpy does not depend on path of reaction
a) H2(g) + 1/2O2(g)  H2O(g)
b) H2O(g)  H2O(l)
c) H2(g) + 1/2O2(g)  H2O(l)
DH1 = -242 kJ
DH2 = -44 kJ
DH3 = -286 kJ
Hess’s Law
a) N2(g) + O2(g)  2NO(g)
b) 2NO(g) + O2(g)  2NO2(g)
c) N2(g) + 2O2(g)  2NO2(g)
DH1 = +180 kJ
DH2 = -112 kJ
DH3 = +68 kJ
Hess’s Law
Calculate the enthalpy of formation of carbon dioxide from its elements.
C(g) + 2O(g)  CO2(g)
Use the following data:
2O(g)  O2(g)
C(s)  C(g)
CO2(g)  C(s) + O2(g)
DH = - 250 kJ
DH = +720 kJ
DH = +390 kJ
2O(g)  O2(g)
C(g)  C(s)
C(s) + O2(g)  CO2(g)
DH = - 250 kJ
DH = - 720 kJ
DH = - 390 kJ
C(g) + 2O(g)  CO2(g)
DH = -1360 kJ
Hess’s Law
• Calculate DH for the synthesis of diborane from its elements.
2B(s) + 3H2(g)  B2H6(g)
DH = ?
a) 2B(s) + 3/2O2(g)  B2O3(s)
b) B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g)
c) H2(g) + 1/2O2(g)  H2O(l)
d) H2O(l)  H2O(g)
DH
DH
DH
DH
=
=
=
=
-1273 kJ
-2035 kJ
- 286 kJ
+ 44 kJ
Formation of Ammonia
• C3H8 + 5 O2(g)  3 CO2(g) + 4 H2O(g)