Download 12.1 The Arithmetic of Equations

12.1 The Arithmetic of Equations >
Chapter 12
Stoichiometry
12.1 The Arithmetic of
Equations
12.2 Chemical Calculations
12.3 Limiting Reagent and
Percent Yield
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12.1 The Arithmetic of Equations > Using Equations
Do Now:
Ammonia, NH3, is one of the most
highly produced inorganic chemicals.
It is used for fertilizing crops and to
produce plastics, fibers, explosives,
nitric acid, dyes, and pharmaceuticals.
Balance the equation for producing
ammonia.
__N2(g) + __H2(g)  __NH3(g)
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12.1 The Arithmetic of Equations > Using Equations
Do Now:
Ammonia, NH3, is one of the most
highly produced inorganic chemicals.
It is used for fertilizing crops and to
produce plastics, fibers, explosives,
nitric acid, dyes, and pharmaceuticals.
Balance the equation for producing
ammonia.
N2(g) + 3H2(g)  2NH3(g)
3
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12.1 The Arithmetic of Equations > Using Equations
Stoichiometry
Calculation of quantities of reactants
and products in chemical reactions
From the Greek words
stoicheion – element
metron - measure
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12.1 The Arithmetic of Equations > Chemical Equations
Balanced equations can be interpreted in
terms of
• numbers of atoms, molecules, or moles
• mass
• volume
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12.1 The Arithmetic of Equations > Chemical Equations
N2(g)
+
3H2(g)
2NH3(g)
+
1
2 atoms N
+
6 atoms H
2 atoms N and 6 atoms H
1 molecule N2
+
3 molecules H2
2 molecules NH3
10 molecules N2
+
30 molecules H2
20 molecules NH3
(
6.02  1023
molecules N2
)
3
(
)
2
(
6.02  1023
molecules NH2
1 mol N2
+
3 mol H2
2 mol NH3
28 g N2
+
3  2 g H2
2  17 g NH3
34 g reactants
34 g products
Assume
STP
22.4
L
22.4 L N2
6
+
6.02  1023
molecules H2
+
22.4
L
22.4
L
67.2 L H2
22.4
L
22.4
L
)
22.4
L
44.8 L NH3
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12.1 The Arithmetic of Equations > Chemical Equations
• MASS & ATOMS are conserved in EVERY
chemical reaction
• Molecules, formula units, moles, and
volumes are not necessarily conserved—
although they may be.
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12.1 The Arithmetic of Equations > Chemical Equations
Combustion of Methane
CH4(g) + 2O2(g)  CO2(g) + H2O(g)
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∆H = -810 kJ
Interpret this equation in terms of
- moles of reactants and products
- mass of reactants and products
- volume at STP of reactants and
products
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12.2 Chemical Calculations >
Writing and Using Mole Ratios
mole ratio: a conversion factor derived from
the coefficients of a balanced chemical
equation interpreted in terms of moles
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12.2 Chemical Calculations >
Writing and Using Mole Ratios
In chemical calculations, mole ratios
are used to convert between a given
number of moles of a reactant or product
to moles of a different reactant or product.
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12.2 Chemical Calculations >
Writing and Using Mole Ratios
N2(g) + 3H2(g)  2NH3(g)
The three mole ratios derived from the balanced
equation above are:
1 mol N2
3 mol H2
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2 mol NH3
1 mol N2
3 mol H2
2 mol NH3
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12.2 Chemical Calculations >
Sample Problem 12.3
Calculating Moles of a Product
How many moles of NH3 are
produced when 0.60 mol of
nitrogen reacts with
hydrogen?
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12.2 Chemical Calculations >
Sample Problem 12.3
Multiply the given quantity of N2 by the mole
ratio in order to find the moles of NH3.
0.60 mol N2 
13
2 mol NH3
1 mol N2
= 1.2 mol NH3
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12.1 The Arithmetic of Equations > Chemical Equations
Combustion of Gasoline
2C8H18 + 25O2  16CO2 + 18H2O
∆H = -10,224 kJ
How many moles of CO2 will be produced from
the combustion of 30 moles of C8H18?
(note: 30 moles C8H18 is ~ 1 gallon of gasoline)
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12.1 The Arithmetic of Equations > Chemical Equations
Combustion of Hydrogen / Electrolysis
2H2(g) + O2(g) ↔ 2H2O(l)
∆H = -482 kJ
How many moles of O2 are required to react
with 10 moles of H2?
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12.1 The Arithmetic of Equations >
END OF 12.1
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