Download Ch. 12 Stoichiometry

Ch. 12 Stoichiometry
Objective: To learn how to use a complete
chemical equation to calculate quantities
of a substance
Chemical Equations
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A balanced chemical equation provides the
same kind of quantitative information that a
recipe does.
If we were to make a bicycle:
/
/
Frame + Seat + 2 Wheels  Bike
F
+ S +
2W
 FSW2
/ Mr. Focht can make 345 bicycles in a week. How
many wheels should I go out and buy on Sunday to
be prepared for the work week
Balanced Equations
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Balanced chemical equations tell what amounts
of reactants to mix and what amounts of
products to expect. The quantity is usually
expressed in moles.
/
Can be expressed in liters, grams, or molecules
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Why Would we need to know what
amounts of reactants to mix or
how much product is going to be
produced?
Stoichiometry
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Calculations using balanced equations
are called Stoichiometric calculations
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Calc. chemical quantities within reaction
Form of Bookkeeping
Interpreting Chemical Equations
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A balanced chemical equation can be interpreted in
terms of different quantities such as number of
atoms, molecules or moles; mass; and volume.
/
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Atoms: both number and types of atoms are not changed
Molecules: Indicates amt of molecules reacting/ being
produced
Moles: Coefficients of balanced equation indicate relative
# of moles of reactant and products
/ Most important info from equation
Mass: Obeys law of conservation of mass
/
Mass and atoms are conserved in every chemical reaction,
however; molecules, formula units, moles, and volume may
not be
N2
(g)
+ 3H2
Quantity
Amount of
Reactants
Number of Atoms
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Number of
Molecules
moles
Mass(g)
(g)
 2NH3
(g)
Amount of Quantity remains
Products
the same or is
different
On your own:
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Hydrogen sulfide, which smells like rotten eggs, is found
in volcanic gases. The balanced equation for the burning
of hydrogen sulfide is:
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2H2S (g) + 3O2 (g)  2SO2 (g) + 2H2O (g)
Determine the number of representative particles and
moles for the reactants and products.
Determine the masses of the reactants and products.
Writing and using mole ratio
Writing and using mole ratios:
N2 + 3H2 --> 2NH3
“1 mol of nitrogen reacts with 3 mol of hydrogen to
form 2 mol ammonia.”
Mole ratio: tells the ratio between 2 substances in
a balanced chemical equation.
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Using mole ratio
How many moles of ammonia are produced when
0.60 mol of nitrogen reacts with hydrogen?
N2 + 3H2 --> 2NH3
Step 1: identify your known and unknown:
Known: 0.60 mol N2
Unknown: ? mol NH3
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N2 +
3H2 --> 2NH3
Step 2: Use the unknown and known in the equation
to develop a mole ratio:
2 mol NH3 (unknown from equation)
1 mol N2 (known from equation)
Step 3: multiply the known by the mole ratio:
0.60 mol N2
x 2 mol NH3
1 mol N2
=
1.2 mol NH3
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4Al + 3O2 --> 2Al2O3
How many mol of aluminum are needed to
form 7.8 mol aluminum oxide?
Known: 7.8 mol Al2O3
Unknown: ? mol Al
Mole ratio:
4 mol Al
(unknown from equation)
*
2 mol Al2O3 (known from equation)
7.8 mol Al2O3 x 4 mol Al = 16 mol Al
2 mol Al2O3
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Practice
Identify all the mole ratios possible from the following equation
a.) 5F2 + 2NH3
b.) 2Al + 3Cl2
N2F4 + 6HF
2AlCl3
Using Equation “a” answer the following questions
1.) How many moles of HF will be produced from 6.7
moles of NH3.
2.) How many moles of F2 are needed to produce 3.4
moles of N2F4.
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Mass to Mole/ Mole to Mass
4Al + 3O2 --> 2Al2O3
When 5.00g of Al reacts with Oxygen, how many
moles of Aluminum Oxide is formed?
Step 1: Identify known and unknown
Known: Al= 5.00g
Unknown: Al2O3= ? moles
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4Al + 3O2 --> 2Al2O3
Step 2: Set up the problem
Known x molar mass x mole ratio
5.00g Al x 1 mole Al x 2 mol Al2O3
27.0g Al
4 mol Al
= 0.0926 mol Al2O3
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4Al + 3O2 --> 2Al2O3
How many grams of Oxygen are needed to
produce 1.23 moles of Aluminum Oxide
Step 1: Identify known and unknown
Know: 1.23 mole aluminum oxide
Unknown: ? grams Oxygen
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4Al + 3O2 --> 2Al2O3
Step 2: Set up the equation:
Known x Mole ratio x Molar mass
1.23 mol Al2O3 x 3 mole O2 x 32.0 g O2
2 mole Al2O3 1 mole O2
= 59.0g O2
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Mass to Mass Problems
Solid lithium hydroxide is used in space vehicles to
remove exhaled carbon dioxide. The lithium
hydroxide reacts with gaseous carbon dioxide to
form solid lithium carbonate and liquid water. How
many grams of carbon dioxide can be absorbed by
1.00g of lithium hydroxide?
How do we solve this problem?
Step 1: Write a balanced chemical equation:
2 LiOH (s) + CO2 (g) --> Li2CO3 (s) + H2O (l)
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Understand that each coefficient represents the
number of moles of each substance.
2 LiOH (s) + CO2 (g) --> Li2CO3 (s) + H2O (l)
Pay attention to the substance and units asked for in the
problem......we need grams of CO2.
Step 2: Find the ratio between the unknown substance
and the known substance in the problem:
The mole ratio of unknown = 1 mole CO2
known
2 mole LiOH
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Step 3: Write down the quantity and substance you are
starting with in the problem and convert to moles, and
multiply by the mole ratio.
1.00 g LiOH x 1 mol LiOH x 1 mol CO2
23.9 g LiOH 2 mol LiOH
=
19
Step 4:
But we need grams of CO2, so we
must convert moles of CO2 to grams:
1.00 g LiOH x 1 mol LiOH x 1 mol CO2 x 44.0g CO2 =
23.9 g LiOH 2 mol LiOH 1 mol CO2
=0.921 g CO2
Mathematically, the problem is:
1.00/23.9/2 x 44.0 = 0.92
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5F2 + 2NH3
a.
b.
c.
d.
N2F4 + 6HF
How many molecules of NH3 are needed to produce
2.34 x 1022 molecules of N2F4?
How many grams of HF are produced from a reaction of
4.56 x 1023 molecules of F2 with excess NH3?
What volume of HF, at STP, can be produced from 345g
of NH3?
How many molecules of N2F4 can be produce from 45.6L
of F2 , at STP?
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Chocolate chip cookie recipe:--makes 4 dozen.
4 c. flour
2 eggs
2 c. butter
2 tsp. vanilla
2 tsp. baking powder
1 c. milk
1 tsp. salt
2 bags chocolate chips
I only have 2/3 bag of
chocolate chips. What do I
do? How does this affect
my cookie-making?
I don’t have enough chips
to make 4 dozen cookies.
So, it’s my limiting
ingredient. It will control
how many cookies I can
make.
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Original Recipe:
4 c. flour
2 eggs
2 c. butter
2 tsp. vanilla
2 tsp. baking powder
1 c. milk
1 tsp. salt
2 bags chocolate
chips
So, how do I modify this recipe
for 2/3 bag of chips?
How much flour do I need?
0.667 bag x 4 c. flour = 1.33 c. flour
2 bags
How much butter do I need?
0.667 bag x 2 c. butter = 0.667 c. butter
2 bags
How to modify a recipe: Find your limiting ingredient.
Multiply the given amount of that ingredient by a ratio of
the needed ingredient over the given ingredient from the
original recipe.
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Mg + 2HCl --> MgCl2 + H2
Identify the limiting reagent when 6.00g HCl reacts with 5.00g Mg
1. Convert given quantities to moles:
6.00g HCl x 1 mol = 0.164 mol HCl
Given
36.5g
5.00g Mg x 1 mol = 0.206 mol Mg
Given
24.3g
2. Use mole ratios to find needed quantities:
0.164 mol HCl x 1 mol Mg = 0.082 mol Mg
2 mol HCl
HCl is limiting reagent.
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Analyze:
Given:
0.164 mol HCl
0.206 mol Mg
Needed:
0.082 mol Mg
We have more Mg than is needed, so Mg is in
excess. Mg is the excess reagent. Therefore, HCl is
the limiting reagent.
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2C2H2 + 5O2 --> 4CO2 + 2H2O
How many grams of water can be produced by the
reaction of 2.40 mol C2H2 with 7.40 mol O2?
First, identify the limiting reagent:
2.40 mol C2H2 x 5 mol O2 = 6.00 mol O2
2 mol C2H2
7.40 mol O2 x 2 mol C2H2 = 2.96 mol C2H2
given
5 mol O2
needed
C2H2 is limiting reagent.
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C2H2 is limiting reagent, so use that to find grams of water
produced.
2C2H2 + 5O2 --> 4CO2 + 2H2O
2.40 mol C2H2 x 2 mol H2O x 18.0g H20 = 43.2g H2O
2 mol C2H2 1 mole H2O
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Percent Yield
Theoretical Yield - the maximum amount of
product that could be formed from given amounts
of reactants
Actual Yield - The amount of product that will
actually form when the experiment is done
Percent Yield = actual yield
x 100%
theoretical yield
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CaCO3
CaO + CO2
What is the percent yield if 13.1g of CaO is
produced when 24.8 g CaCO3 is heated?
Step 1- Find theoretical yield
24. 8g CaCO3 x 1 mole CaCO3 x 1 mol CaO x 56.1 g CaO =
100.1g CaCO3
1 mol CaCO3 1 mol CaO
=13.9g CaO -------Theoretical Yield
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Step 2- Plug theoretical yield calculated and
actual yield into percent yield equation.
13.1g CaO x 100% = 94.2% ----percent yield
13.9g CaO
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