# STOICHIOMETRY by xiaopangnv

VIEWS: 28 PAGES: 36

• pg 1
```									STOICHIOMETRY
USING THE REACTION
EQUATION LIKE A RECIPE
USING EQUATIONS
• Nearly everything we use is manufactured
from chemicals.
– Soaps, shampoos, conditioners, cd’s,
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.
– Eqs 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 rxn, you can calculate
the quantity of any other substance
consumed or created in the rxn.
– Quantity meaning the amount of a
substance in grams, liters, molecules,
or moles.
USING EQUATIONS
• The calculation of quantities in chemical
reactions is called stoichiometry.
• Imagine you are in charge of manu-
facturing for Rugged Rider Bicycle
Company.
• The business plan for Rugged Rider
requires the production of 128 custom-
• 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 workweek, 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 (balanced eqn)
• What is unknown?
–# of wheels = ? 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
= 1280
1 FSW2HP2        Wheels
• We can make the same kinds of
connections from a chemical rxn eqn.
N2(g) + 3H2(g)  2NH3(g)
• The key is the “coefficient ratio”.
– The coefficients of the balanced chemical
equation indicate the numbers of moles
of reactants and products in a chem rxn.
• 1 mole of N2 reacts with 3 moles of H2 to
produce 2 moles of NH3.
– N2 and H2 will always react to form
ammonia in this 1:3:2 ratio of moles.
• So if you started with 10 moles of N2 it
would take 30 moles of H2 and would
produce 20 moles of NH3
• Using the coefficients, from the balanced
rxn equation as ratios to make
connections between reactants
and products, is the most important
information that a rxn equation provides.
– 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 rxn 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)

2 mol Al2O3
1.8 mol Al                  = 0.90mol
4 mol Al            Al2O3

Mole Ratio
MOLE – MOLE EXAMPLE 2
• The following rxn 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)
4 mol Al
24 mol Al2O3                 = 48 mol Al
2 mol Al2O3

3 mol O2
24 mol Al2O3                 = 36 mol O2
2 mol Al2O3
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 CALCULAT’NS
• No lab balance measures moles directly,
generally mass is the unit of choice.
• From the mass of 1 reactant or prod-uct,
the mass of any other reactant or product
in a given chemical equation can be
calculated, provided you have a balanced
rxn equation.
• As in mole-mole calcs, the unknown can
be either a reactant or a product.
MASS – MASS CALCULAT’NS 1
Acetylene gas (C2H2) is produced
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 CALCULAT’NS 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
= .07813 mol
64.0 g CaC2              CaC2
Step 2: Now we are ready for the KEY
step…converting from mols of our given to
mols of unknown using the mole ratio.

.07813 mol   1 mol C2H2
= .07813 mol
CaC2 1 mol CaC
2                 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 mols into grams.

.07813 mol   26.0 g C2H2
= 2.03 g C2H2
C2H2 1 mol C H
2 2

Summary of 3 Steps:
1. Get to Moles
2. Mole Ratio
3. Get to desired final unit
MASS – MASS CALCULAT’NS 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 CALCULAT’NS 2

1mol Pb(NO3)2
25.3 g Pb(NO3)2
331.2g Pb(NO3)2

2mol KI          166 g KI
1mol Pb(NO3)2         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 accor-ding to the
following rxn equation?
BaCl2 + Na3PO4  Ba3(PO4)2 + NaCl
(89.2 g BaCl2)
2. Use the equation to determine what
mass of FeS must react to form 326g of
FeCl2.                              FeS +
HCl  H2S + FeCl2
(226 g FeS)
• 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
eqn is used to convert from the moles of
given to the number of moles of the
unknown
• Then the moles of the unknown are
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 
molecules B
29.2 g 1 mol H2O     1 mol O2
H2O
18.0 g H2O   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 mols B volume B

5.0x1022 mol-     1 mol NO       3 mol NO2
ecules NO
6.02x1023 mol-   1 mol NO
ecules NO
22.4 L NO2
1 mol NO2

= 5.58 L NO2
Aspirin can be made from a chemical rxn
between the reactants salicylic acid and acetic
anhydride. The products of the rxn are acetyl-
salicylic acid (aspirin) and acetic acid (vinegar).
Our factory makes 125,000 100-count bottles
of Bayer Aspirin/day. Each bottle contains 100
tablets, and each tablet contains 325mg of
aspirin. How much in kgs + 10% for production
problems, of each reactant must we have in
order to meet production?

C7H6O3 + C4H6O3  C9H8O4 + HC2H3O2
Salicylic     Acetic       aspirin    vinegar
acid      anhydride
• What do we know?
– Make 125,000 aspirin bottles/day
– 100 aspirin/bottle
– 325 mg aspirin/tablet
– Mole ratio of aspirin to salicylic acid (1:1)
and acetic anhydride (1:1)
– MM aspirin = 180.11g
– MM C7H6O3 = 138.10g
– MM C4H6O3 = 102.06g
• What are we asked for?
– Mass of salicylic acid in kgs + 10%
– Mass of acetic anhydride in kgs + 10%
125,000 100 tablets     325mg asp.
bottles
1 bottle          1 tablet

1g              1mol asp.
1000 mg               180.16g

= 22,549.4 mols aspirin
Salicylic Acid:

22,549.4 1 mol C7H6O3        136.10g C7H6O3
mols
aspirin  1 mol asp          1 mol C7H6O3

1 kg
= 3068.97 kg salicylic
1000 g              acid + (306.897 g)

= 3380 kg of salicylic acid
Acetic Anhydride:

22,549.4 1 mol C4H6O3     102.06g C4H6O3
mols
aspirin  1 mol asp      1 mol C4H6O3

1 kg      = 2301.39 kg
Acetic anhydride
1000 g               + 230.139 kg

= 2530 kg Acetic anhydride

```
To top