Transcript STOICHIOMETRY

STOICHIOMETRY
USING THE REACTION
EQUATION LIKE A RECIPE
USING EQUATIONS
• Nearly everything we use is manufactured
from chemicals
– Soaps, shampoos, conditioners,
cosmetics, medications, and clothes
• For a manufacturer to make a profit the
cost of making any of these items can’t be
more than the money paid for them
• Chemical processes carried out in industry
must be economical, this is where
balanced equations help
USING EQUATIONS
• Equations are a chemist’s recipe
– Equations tell chemists what amounts
of reactants to mix and what amounts
of products to expect.
• When you know the quantity of one
substance in a reaction, you can calculate
the quantity of any other substance
consumed or created in the reaction
– Quantity meaning the amount of a
substance in grams, liters, molecules,
or moles
STOICHIOMETRY
• Greek for “measuring elements”
• The calculation of quantities in chemical
reactions is called stoichiometry
• Imagine you are in charge of manufacturing for
Rugged Rider Bicycle Company
• The business plan for Rugged Rider requires
the production of 128 custom-made bikes each
day
• You are responsible for insuring that there are
enough parts at the start of each day
USING EQUATIONS
• Assume that the major components of
the bike are the frame (F), the seat (S),
the wheels (W), the handlebars (H), and
the pedals (P)
• The finished bike has a “formula” of
FSW2HP2
• The balanced equation for the production
of 1 bike is:
F +S+2W+H+2P 
FSW2HP2
USING EQUATIONS
• Now in a 5-day work week, Rugged Riders
is scheduled to make 640 bikes. How
many wheels should be in the plant on
Monday morning to make these bikes?
• What do we know?
– Number of bikes = 640 bikes
– 1 FSW2HP2=2W (from balanced equation)
• What is unknown?
–# of wheels for 640 bikes= ? wheels
• The connection between wheels and
bikes is 2 wheels per bike
– We can use this information as a conversion factor
to do the calculation
640 FSW2HP2
2W
1 FSW2HP2
= 1280
Wheels
• We can make the same kinds of
connections from a chemical equation
N2(g) + 3H2(g)  2NH3(g)
• The key is the “coefficient ratio”
•
• Use the coefficients from the balanced
reaction equation to make mole ratios
– Ratios of balanced coefficients = mole ratios
– Makes connections between reactants
and products
• Using this information, you can
calculate the amounts of the reactants
involved and the amount of product
you might expect
– Any calculation done with the next
process is a theoretical value
• The real world isn’t always perfect
MOLE – MOLE EXAMPLE
• The following reaction shows the
synthesis of aluminum oxide:
3O2(g) + 4Al(s)  2Al2O3(s)
• If you only had 1.8 mols of Al, how much
product could you make?
Given: 1.8 moles of Al
Uknown: ____ moles of Al2O3
MOLE – MOLE EXAMPLE
• Solve for the unknown:
3O2(g) + 4Al(s)  2Al2O3(s)
1.8 mol Al
2 mol Al2O3
4 mol Al
Mole Ratio
= 0.90mol
Al2O3
MOLE – MOLE EXAMPLE 2
• The following reaction shows the
synthesis of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
• If you wanted to produce 24 mols of
product, how many mols of each reactant
would you need?
Given: 24 moles of Al2O3
Uknown: ____ moles of Al
____ moles of O2
MOLE – MOLE EXAMPLE 2
• Solve for the unknowns:
3O2(g) + 4Al(s)  2Al2O3(s)
24 mol Al2O3
24 mol Al2O3
4 mol Al
2 mol Al2O3
3 mol O2
2 mol Al2O3
= 48 mol Al
= 36 mol O2
Practice for You…
1. How many mols of hydrogen will be
produced if 0.44 mol of CaH2 reacts
according to the following equation?
CaH2 + 2H2O  Ca(OH)2 + 2H2
(.89 mol H2)
2. Iron will react with oxygen to produce
Iron III oxide. How many moles of Iron III
oxide will be produced if 0.18 mol of Iron
reacts?
(.090 mol Fe2O3)
MASS – MASS CALCULATIONS
• No lab balance measures moles directly
– Generally, mass is the unit of choice
• From the mass of 1 reactant or product,
the mass of any other reactant or product
in a given chemical equation can be
calculated
– You must have a balanced reaction equation!
• As in mole-mole calculations, the
unknown can be either a reactant or a
product
Basic Steps to Solve Mass-Mass
Stoichiometry Problems
1.
2.
3.
4.
Balance the equation
Convert mass to moles
Set up mole ratios
Use mole ratios to calculate moles of
desired compound
5. Convert moles to grams, if necessary
MASS – MASS CALCULATIONS 1
Acetylene gas (C2H2) is produced by adding
water to calcium carbide (CaC2):
CaC2 + 2H2O  C2H2 + Ca(OH)2
How many grams of C2H2 are produced by
adding water to 5.00 g CaC2?
MASS – MASS CALCULATIONS 1
• What do we know?
– Given mass = 5.0 g CaC2
– Mole ratio: 1 mol CaC2 = 1 mol C2H2
– MM of CaC2 = 64.0 g CaC2
– MM of C2H2 = 26.0g C2H2
• What are we asked for?
– grams of C2H2 produced
Step 1 - “Get to Moles!”
In this case that can be done by using the
Molar Mass of your given compound:
5.0 g CaC2
1 mol CaC2
64.0 g CaC2
= .07813 mol
CaC2
Step 2 - Convert from moles of our given to
moles of unknown using the mole ratio:
1 mol C2H2
.07813 mol
CaC2 1 mol CaC
2
= .07813 mol
C2H2
Step 3 - Since we are asked for mass of our
unknown in this problem, we need to use our
molar mass of our unknown and convert our
newly calculated moles into grams:
26.0 g C2H2
.07813 mol
C2H2 1 mol C H
2 2
Summary of 3 Steps:
1. Get to Moles
2. Mole Ratio
3. Get to desired final unit
= 2.03 g C2H2
MASS – MASS CALCULATIONS 2
The double replacement reaction between
lead (II) nitrate and potassium iodide produces
a bright yellow precipitate that can be used as
a color additive in paint. How many grams of
potassium iodide would we need to completely
react 25.3 g of lead (II) nitrate?
Pb(NO3)2 + 2 KI  PbI2 + 2 KNO3
mass A  mols A  mols B  mass B
MASS – MASS CALCULATIONS 2
25.3 g Pb(NO3)2
1mol Pb(NO3)2
331.2g Pb(NO3)2
2mol KI
1mol Pb(NO3)2
166 g KI
1mol KI
= 25.4 g KI
Practice for You…
1. What mass of barium chloride is needed
to react completely with 46.8 g of
sodium phosphate according to the
following reaction equation?
BaCl2 + Na3PO4  Ba3(PO4)2 + NaCl
(89.2 g BaCl2)
2. Use the following equation to determine
what mass of FeS must react to form
326g of FeCl2.
FeS + HCl  H2S + FeCl2
(226 g FeS)
• Recall, a balanced reaction equation
indicates the relative numbers of moles of
reactants and products
• We can expand our stoichiometric
calculations to include any unit of
measure that is related to the mole
– The given quantity can be expressed in numbers of
particles, units of mass, or volumes of gases at STP!
• The problems can include mass-volume,
volume-volume, and particle-mass
calculations
• In any of these problems, the given
quantity is first converted to moles
• Then, the mole ratio from the balanced
equation is used to convert from the
moles of given to the number of moles of
the unknown
• The moles of the unknown are the
converted to the units that the problem
requests
• The next slide summarizes these steps for
all typical stoichiometric problems
MORE MOLE EXAMPLES
How many molecules of O2 are
produced when a sample of 29.2 g
of H2O is decomposed by
electrolysis according to this
balanced equation:
2H2O  2H2 + O2
MORE MOLE EXAMPLES
• What do we know?
– Mass of H2O = 29.2 g H2O
– 2 mol H2O = 1 mol O2
• From balanced equation
– MM of H2O = 18.0 g H2O
– 1 mol O2 = 6.02x1023 molecules of O2
• What are we asked for?
– Molecules of O2
Mass A  Mols A  Mols B 
29.2 g 1 mol H2O
H2O
18.0 g H2O
Molecules B
1 mol O2
2 mol H2O
6.02x1023
molecules O2
1 mol O2
= 4.88 x 1023
molecules O2
MORE MOLE EXAMPLES
The last step in the production of nitric
acid is the reaction of NO2 with H2O:
3NO2+H2O2HNO3 + NO
How many liters of NO2 must react
with water to produce 5.00x1022
molecules of NO?
MORE MOLE EXAMPLES
• What do we know?
– Molecules NO = 5.0x1022 molecules NO
– 1 mol NO = 3 mol NO2
• From balanced equation
– 1 mol NO = 6.02x1023 molecules NO
– 1 mol NO2 = 22.4 L NO2
• What are we asked for?
– Liters of NO2
Molecules A Mols A Mols B Volume B
5.0x1022 molecules
NO
1 mol NO
3 mol NO2
6.02x1023 molecules
NO
1 mol NO
22.4 L NO2
1 mol NO2
= 5.58 L NO2