1. Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
12.2 Chemical Calculations
12.3 Limiting Reagent and
Percent Yield
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2. __N2(g) + __H2(g)  __NH3(g)
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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3. Balance the equation for producing ammonia.
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)
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4. 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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5. Balanced equations can be interpreted in terms of
Chemical Equations
Balanced equations can be interpreted in terms of
numbers of atoms, molecules, or moles
mass
volume
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6. ( ) Chemical Equations N2(g) + 3H2(g) 2NH3(g) 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
1 
3 
2 
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 N2
67.2 L H2
44.8 L NH3
6.02  1023
molecules N2 ( )
molecules H2
molecules NH2
22.4 L
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7. MASS & ATOMS are conserved in EVERY chemical reaction
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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8. CH4(g) + 2O2(g)  CO2(g) + H2O(g) ∆H = -810 kJ
Chemical Equations
Combustion of Methane
CH4(g) + 2O2(g)  CO2(g) + H2O(g) ∆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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9. 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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10. 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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11. Writing and Using Mole Ratios
N2(g) + 3H2(g)  2NH3(g)
The three mole ratios derived from the balanced equation above are:
2 mol NH3
1 mol N2
3 mol H2
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12. Calculating Moles of a Product
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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13. 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  = 1.2 mol NH3
2 mol NH3
1 mol N2
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14. Combustion of Gasoline
Chemical Equations
Combustion of Gasoline
2C8H O2  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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15. 2H2(g) + O2(g) ↔ 2H2O(l) ∆H = -482 kJ
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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16. END OF 12.1
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