Stoichiometry
Chapter 12
Pages 352 – 383
Stoichiometry
• The study of the quantitative, or
measurable, relationships that exist in
chemical formulas and chemical
reactions
• Stoichio – from Greek
word meaning element
• Metry – to measure
Interpreting Balanced
Chemical Equations
• The
coefficients
represent any
of the
following:
• # particles
– Atoms
– Molecules
– Formula units
• N2H4 + 2 H2O2 N2 + 4 H2O
• 1 mole/liter/particle of
N2H4 reacts with 2
mole/liter/particle of H2O2
to produce 1
mole/liter/particle of N2
and 4 mole/liter/particle
of H2O
• Moles
• Liters
Mole – Mole Problems
• NH4NO3 N2O + 2 H2O
•
• 1 mole NH4NO3
decomposes to produce 1
mole N2O and 2 moles H2O
How many
moles of H2O
are produced if
3.65 moles of
NH4NO3 are
decomposed?
3.65 mol NH4NO3 x 2 mol H2O = 7.30 mol H2O
1 mol NH4NO3
(mole given)
(mole ratio)
The HEART of
the problem!
Solving Stoichiometric
Problems
•
•
•
•
3 major categories
Mass – mass
Mass – volume
Volume – volume
Mass – Mass problems:
Mass (given) x 1 mol given cmpd x #mol unk cmpd x mass unk cmpd
mass given cmpd #mol given cmpd 1 mol unk cmpd
(molar mass)
(mole ratio)
(molar mass)
What is the mass of H2O
produced from 2.5 grams of
glucose?
• The set up is the most important part of this
problem, besides writing a balanced equation!
• C6H12O6 + 6O2 6 CO2 + 6 H2O
• Set it up!
•
2.5 g C6H12O6 x 1 mol C6H12O6 x 6 mol H2O x 18.0153 g H2O = 1.5 g H2O
180.1559 g C6H12O6 1 mol C6H12O6 1 mol H2O
(mass given) (molar mass)
(mole ratio)
(molar mass)
Solving Stoichiometric
Problems
•
•
•
•
3 major categories
Mass – mass
Mass – volume
Volume – volume
Mass – Volume problems:
•
Mass given x 1mol given cmpd x #mol unk cmpd x 22.4 L unk cmpd
mass given cmpd #mol given cmpd 1 mol unk cmpd
(molar mass)
(mole ratio)
(molar volume)
• If an airbag contains 125 g sodium azide,
NaN3, what volume of nitrogen gas is
produced at STP? (0oC and 1 atm)
2NaN3 2Na + 3N2
Mass – Volume problems:
125 g NaN3 x 1 mol NaN3 x 3 mol N2 x 22.4 L N2 = 64.6 L N2
65.00999 g NaN3 2 mol NaN3 1 mol N2
(given)
(molar mass) (mole ratio) (molar volume)
Solving Stoichiometric
Problems
•
•
•
•
3 major categories
Mass – mass
Mass – volume
Volume – volume
Volume – Volume Problems
•
Volume given x 1mol given cmpd x #mol unk cmpd x 22.4 L unk cmpd
22.4 L given cmpd #mol given cmpd 1 mol unk cmpd
(molar volume)
(mole ratio)
(molar volume)
• What volume of H2 gas reacts with N2 to
form 6.5 L of ammonia? (at STP)
N2 + 3H2 2NH3
Volume – Volume Problems
6.5 L NH3 x 1 mol NH3 x 3 mol H2 x 22.4 L H2 = 9.8 L H2
22.4 L NH3
2 mol NH3
1 mol H2
(given) (molar volume) (mole ratio) (molar volume)
Limiting Reactants and
Percent Yield
• Limiting reactants – Reactants that
limit the amount of product formed in
a chemical reaction
• The quantities of products formed in
a reaction are always determined by
the quantity of the limiting reactant!
• 3.5 g Cu reacts with a solution
containing 6.0 g AgNO3. How much
Ag is produced?
Cu + 2AgNO3 2Ag + Cu(NO3)2
• TWO problems must be done for
each of the reactants, in order to
determine which is the limiting
reactant!
3.5 g Cu x
1 mol Cu x 2 mol Ag x 107.87 g Ag = 12 g Ag
63.54 g Cu
1 mol Cu
1 mol Ag
6.0 g AgNO3 x 1 mol AgNO3 x 2 mol Ag x 107.87 g Ag =3.8g Ag
169.84 g AgNO3 2 mol AgNO3 1 mol Ag
Limiting reactant produces the least amount of
product, so the AgNO3 is the limiting reactant
• 2.0 g NaHCO3 reacts with 0.5 g
H3C6H5O7, citric acid, to produce
CO2, H2O, and sodium citrate,
Na3C6H5O7. What volume of CO2 is
produced?
3NaHCO3 + H3C6H5O7 3 CO2 + 3 H2O + Na3C6H5O7
2.0 g NaHCO3 x 1mol NaHCO3 x 3 mol CO2 x 22.4 L CO2 = 0.53 L CO2
84.0066 g NaHCO3 3 mol NaHCO3 1 mol CO2
0.5 g H3C6H5O7 x
1mol H3C6H5O7 x
3 mol CO2 x
192.1235 g H3C6H5O7 1 mol H3C6H5O7
22.4 L CO2 = 0.2 L CO2
1mol CO2
• How much NaHCO3 is left over?
• You must figure out how much was used first
by converting from the citric acid to sodium
bicarbonate, using their original masses.
0.5 g H3C6H5O7 x 1 mol H3C6H5O7 x 3 mol NaHCO3 x 84.0066 g NaHCO3 = 0.7 gNaHCO3
192.1235 g H3C6H5O7 1 mol H3C6H5O7
1 mol NaHCO3
SUBTRACT this amount from original
amount:
2.0 g NaHCO3(original)
– 0.7 g NaHCO3(used)
1.3 g left over
• Percent Yield – how much yield of
products you actually obtain
compared to what is expected
• Expected yield – amount of product
predicted based on calculations
• Actual yield – amount of product
really obtained from a chemical
reaction
• Reasons for differences:
– Some of reactants may not react
– Side reaction that differs from reaction
upon which calculations were based
– Some product lost during recovery or
transfer of containers
• % Yield = actual yield x 100
expected yield
• What is the percent yield for the previous
question, if 0.18L of CO2 are actually
produced in the lab?
• % Yield = 0.18 L CO2 x 100 = 90%
0.2 L CO2