Download 9 SHS CH 9 LECTURE shs_ch_9_lecture

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Stoichiometry
STOICHIOMETRY
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Quantitative measurements are those that
involve actual number relationships,
relationships, not just
human judgment.
Ex. 3.21 cm, 5.42 moles, 342.65 g, 24.1 L
not just, “It was really, really a lot”
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STOICHIOMETRY
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We will look at quantitative
measurements in chemical reactions.
Mass and mole relationships in
chemical reactions
Limiting Reactants in a reaction
Theoretical and percent yield
Solution concentrations and
preparations (acid/base reactions)
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Hamburger Analogy
Recipe for a bacon double cheeseburger is:
1 hamburger bun
2 hamburger patties
2 slices of cheese
4 strips of bacon
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If I have five bacon double cheeseburgers:
•How many hamburger buns do I have?
•How many hamburger patties do I have?
•How many slices of cheese do I have?
•How many strips of bacon do I have?
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How many bacon double cheeseburgers can
you make if you start with:
a) 1 bun,
bun, 2 patties
patties,, 2 slices cheese,
cheese, 4 strips bacon
b) 2 bun,
bun, 4 patties
patties,, 4 slices cheese,
cheese, 8 strips bacon
c) 1 dozen bun,
bun, 2 dz. patties
patties,, 2 dz. slices cheese,
cheese,
4 dozen strips bacon
d) 1 mole bun,
bun, 2 mole patties
patties,, 2 mole slices cheese,
cheese,
4 mole strips bacon
e) 10 bun,
bun, 20 patties
patties,, 2 slices cheese,
cheese, 40 strips bacon
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10 bun,
bun, 20 patties
patties,, 2 slices cheese,
cheese, 40 strips bacon
1 BacDbl Cheeseburger
If you had fixings for 100 bacon double cheeseburgers,
but when you were cooking you burned 20 paddies.
What percentage of the bacon double cheeseburgers
do you actually make?
90% BacDbl Cheeseburger
At this point what is the limiting ingredient?
Why?
DCBURGERS AND CHEMISTRY
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How many sandwiches can you make?
12
____
slices of bread
10
+ ____
burger patties
5
= ____ burgers
What is left over?
2 slices of bread
What is the limiting ingredient? burger
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EQUATION REVIEW
C2H5OH + 3O2 → 2CO2 + 3H2O
reactants
products
EQUATION REVIEW
C2H5OH + 3O2 → 2CO2 + 3H2O
reactants
When the equation is balanced it has
quantitative significance:
1 mole of ethanol reacts with 3 moles of oxygen
to produce 2 moles of carbon dioxide and 3
moles of water
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BURGERS AND CHEMISTRY
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The math and the concepts are identical to the
Previous slide.
The only difference between burger and formula is the
names in the recipe.
Here are two examples of chemical recipes:
1.
2.
Na+ + SO42- → Na2SO4
1 mole of H2SO4 + 2 mole NaOH →
1 mole Na2SO4 + 2 mole H2O
PROPORTIONAL RELATIONSHIPS
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Stoichiometry
mass relationships between substances in a
chemical reaction based on the mole ratio
Mole Ratio
indicated by coefficients in a balanced equation
2 Mg2+ + O22- → 2 MgO
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2Na+ + Cl2- → 2NaCl
If I have 1 mole of NaCl
How many moles of sodium created it?
How many moles of diatomic chloride
created it?
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0.5
1 mole of H2SO4 + 2 mole NaOH →
1 mole Na2SO4 + 2 mole H2O
If I want to make 5 moles of Na2SO4:
How many moles of H2SO4 do I need?
How many moles of NaOH do I need?
MOLE RATIO CONVERSIONS
C2H5OH + 3O2 → 2CO2 + 3H2O
1
2
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How many moles of CO2 can I make if I have:
12 mole of C2H5OH
2moleCO2
24 mole of CO2
1moleC2 H 5OH
.90 mole of O2
2moleCO2
3moleO2
0.6 mole of CO2
How many moles of C2H5OH did I need to make
1moleC2 H 5OH
10 mole of H2O
3.3 mole of C2H5OH
3moleH 2O
0.65 mole of CO2
1moleC2 H 5OH
2moleCO2
3.3 mole of C2H5OH
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MOLE RATIO CONVERSIONS
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1 mole of H2SO4 + 2 mole NaOH →
1 mole Na2SO4 + 2 mole H2O
How many moles of Na2SO4 can I make if I have:
1 mole of H2SO4
1moleNa2 SO4
1 mole of Na2SO4
1moleH 2 SO4
1moleNa2 SO4 0.1 mole of Na SO
2
4
2moleNaOH
How many moles of H2SO4 did I need to make
17 mole of H2O
1moleH 2 SO4
8.5 mole of Na2SO4
2moleH 2O
0.65 mole of Na2SO4
0.65 mole of Na2SO4
1moleH 2 SO4
.20 mole of NaOH
1moleNa2 SO4
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1 mole of H2SO4 + 2 mole NaOH →
1 mole Na2SO4 + 2 mole H2O
How many moles of Na2SO4 can I make if I have:
1 mole of H2SO4 and 2 moles of NaOH
10 mole of H2SO4 and 20 mole of NaOH
0.1 mole of H2SO4 and 0.2 mole of NaOH
1 mole of H2SO4 and 20 mole of NaOH
0.42 mole of H2SO4 and 0.65 mole of NaOH
1
10
.1
1
.325
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The key to understanding…
The coefficients in a
balanced equation give the relative
amounts (in moles)
of reactants and products.
The coefficients represent moles…
2NaCl → 2Na + Cl2
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There are ____ moles of salt.
There are ____ moles of sodium.
There is ____ mole of diatomic Cl.
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MOLAR MASS
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A substance’s molar mass (molecular
weight) is the mass in grams of one mole
of the compound.
CO2 = 44.01 grams per mole
H2O = 18.02 grams per mole
Ca(OH)2 = 74.10 grams per mole
STOICHIOMETRY
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STEPS TO STOICHIOMETRY
STEP 1
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Write the balanced equation!
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STEP 2
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Find the number of moles of
the given substance
STEP 3
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Use the balanced equation to
find the mole ratio of given
substance to moles of unknown
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STEP 4
Convert moles of unknown
to whatever quantity is specified
STOICHIOMETRY
N2 + 3 H2
→
S
2NH3
In a balanced chemical equation, the coefficients
allow us to calculate how much of a substance is
used, produced, or needed in the reaction.
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STOICHIOMETRY
N2(g) + 3H2 (g)
→
S
2NH3
From this equation we can determine that 1 mole of
Nitrogen gas (28.02 g ) reacts with 3 moles of
Hydrogen gas (6.06 g) to produce 2 moles of
ammonia (28.02g + 6.06g=34.08g of ammonia)
Always remember the conservation of matter
STOICHIOMETRY RECAP
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How many moles of NaOH will react with 3.21 moles
of sulfuric acid?
Step 1. Balanced the chemical equation.
Step 2. Write down the given numeric info. If it is
not in mole units convert it to moles using the
molecular mass as a conversion factor.
Step 3. Now convert from the moles of starting
substance to the moles of the desired substance by
using a mole ratio from the chemical reaction.
reaction.
Step 4. If the problem is to find grams of the
substance, convert the moles to grams using the
molecular mass conversion factor. If the problem is to
find atoms of the substance, convert moles to atoms.
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STOICHIOMETRY
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How many moles of NaOH will react with 3.21 moles of
sulfuric acid (H2SO4)?
Now lets work out the example problem:
Step 1 H2SO4 + 2NaOH → Na 2SO4 + 2H2O
Step 2 (Given in moles; so is not needed)
Step 3 (Note that step 2 is not needed)
3.21 moles H2SO4 x 2 moles NaOH
1 mol H2SO4
= 6.42 moles NaOH
Step 4 (Asked for moles; so is not needed)
MOLE RATIOS
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These mole ratios can be used to calculate the
moles of one chemical from the given amount
of a different chemical
Example: How many moles of chlorine is needed
to react with 5 moles of sodium (without any
sodium left over)?
2 Na + Cl2 → 2 NaCl
5 moles Na 1 mol Cl2
= 2.5 moles
2 mol Na Cl2
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MOLE RATIO - THE BRIDGE S
EVERYTIME YOU GO ACROSS A BRIDGE I EXPECT
YOU TO THINK ABOUT ME IN THIS
CLASSROOM STILL TEACHING CHEMISTRY AS
YOU DRIVE/WALK ACROSS.
IF YOU TAKE PHYSICS WITH ME YOU WILL GET
TO BUILD BRIDGES.
THE BRIDGE IS THE MOST IMPORTANT THING I
CAN TEACH YOU TO UNDERSTANDING
CHEMICAL CALCULATIONS.
MOLE RATIO
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Shows the molemole-to
to--mole ratio between
two of the substances in a balanced
equation
Derived from the coefficients of any two
substances in the equation
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MOLE RATIO
→
3 H2(g) + N2(g)
2 NH3(g)
A. A mol ratio for H2 and N2
1)
3 mol N2
2) 1 mol N2
1 mol H2
3) 1 mol N2
3 mol H2
2 mol H2
B. A mol ratio for NH3 and H2 is
1)
1 mol H2
2) 2 mol NH3
2 mol NH3
3) 3 mol N2
3 mol H2
2 mol NH3
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MOLE RATIO
→
4 Fe + 3 O2
2 Fe2O3
What is the mole ratio of the compounds?
Fe and O2
4 mol Fe
3 mol O2
or
3 mol O2
4 mol Fe
Fe and Fe2O3
4 mol Fe
2 mol Fe2O3
or
2 mol Fe2O3
4 mol Fe
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MOLE RATIO
4 Fe +
3 O2
→
2 Fe2O3
O2 and Fe2O3
3 mol O2
2 mol Fe2O3
and 2 mol Fe2O3
3 mol O2
Which MR we use depends on what numbers we
Know and which we are trying to get rid of.
Mole--Mole Conversions
Mole
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How many moles of sodium chloride
will be produced if you react 2.6
moles of chlorine gas with an excess
(more than you need) of sodium
metal?
2 Na +
Cl2 → 2 NaCl
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MOLE TO MASS
&
MASS TO MOLE
MASS--MASS Conversions S
MASS
Most of the time in chemistry, the amounts are
given in grams instead of moles
We still go through moles and use the mole ratio,
but now we also use molar mass to get to
grams
Example: How many grams of chlorine are required to
react completely with 5.00 moles of sodium to produce
sodium chloride?
2 Na + Cl2 → 2 NaCl
5.00 moles Na 1 mol Cl2
2 mol Na
70.90 g Cl2
1 mol Cl2
= 177g Cl2
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MOLE--MASS
MOLE
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How many moles of Fe2O3 are produced when
6.0 moles O2 react with Iron?
→
4 Fe + 3 O2
2 Fe2O3
How many grams of Fe2O3 would that be?
MOLE--MASS
MOLE
4 Fe
+
6.0 mol O2 x
2
3
3 O2
→
S
2 Fe2O3
mol Fe2O3= 4.0 mol Fe2O3
mol O2
4.0 mol Fe2O3 x 159.6 g Fe2O3 = 640 g Fe2O3
mol Fe2O3
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MASS--MASS
MASS
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How many grams of Fe are needed to react
with 12.0 mol of O2?
4 Fe
+
3 O2
→ 2 Fe2O3
How many grams of Fe are needed?
MASS--MASS
MASS
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How many grams of Fe are needed to react
with 48g of O2?
Fe
+
O2
→
Fe2O3
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MOLE--MASS
MOLE
4 Fe
+
3 O2 →
S
2 Fe2O3
How many grams of O2 are needed to produce 0.400
mol of Fe2O3?
0.400 mol Fe2O3
3 mol O2
2 mol Fe2O3
32.0 g O2
1 mol O2
= 19.2 g O2
MASS TO MASS
S
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Mass--Mass Conversions
Mass
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Usually we are given a starting mass and want to
find out the mass of a product.
This calculated mass is called theoretical yield.
Theoretical yield is how much of another reactant
we need to completely react with it (no leftover
ingredients!)
Now we go from grams to moles, mole ratio, and
back to grams of compound.
Many have trouble at conversions. So I have
a visual model to help us, it is called the horseshoe.
Mass--Mass Conversions
Mass
aX + bY → cZ
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1. Balance an equation in moles
2. Convert given grams/atoms X to moles X
HERE
IS WHAT THE CONVERSION
3. Use mole factor to give desired moles Y/Z
LOOKS
4. Convert moles
Y/Z toLIKE
grams/atoms Y/Z
atoms/grams (given) X
atoms/grams (desired) Y/Z
X MOLAR MASS
/ MOLAR MASS
or Avogadro's #
moles (given)
X
MOLE RATIO
moles (desired) Y/Z
atoms/g X x 1 mol X x b/c mol Y/Z x atoms/grams Y/Z
atoms / mm X
a mol X
1 mol Y/Z
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HORSESHOE ANALOGY
atoms/grams
(given)
(X)
Molar
Mass
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atoms/grams
(desired)
The horseshoe pictorially
shows the stoichiometric
process. The only thing you
will really mess up is the
MOLE RATIO.
RATIO. Make to use
(/)
Molar
Mass
Want / Have.
Have.
moles
(given)
moles
(desired)
mole ratio moles (desired)
moles (given)
“HOW TO” HORSESHOE
Grams N2
How many moles of Hydrogen are
required to react with 103.25 g of
Nitrogen (AKA mole ratio) in the
Haber Process to produce ammonia?
(/) Molar
Mass N2
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Grams H2
(X) Molar
Mass H2
moles
N2
moles
H2
mole ratio moles H2
moles N2
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MASS--MOLE
MASS
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How many moles of Hydrogen are required to react
with 103.25 g of Nitrogen (AKA mole ratio) in the
Haber Process to produce ammonia?
H2 + N2 → NH3
Step 1. Balance
3H2 +
N2
→
2NH3
Step 2. Determine Mole of Known
103.25 g N2 x 1mole N2 = 3.68 moles of N2
28.02 g N2
Step 3. Multiply Mole Ratio by Known Moles
3.68 moles N2 x 3 moles H2 = 11.4mole H2
1 mol N2
MASS--MOLE
MASS
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How many moles of Hydrogen are required to
react with 103.25 g of Nitrogen (AKA mole ratio)
in the Haber Process to produce ammonia?
3 H2 +
N2
→ 2 NH3
103.25 g N2 1mole N2 3 moles H2 =11.4mol H2
28.02 g N2 1 mol N2
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MASS--MASS
MASS
S
The reaction between H2 and O2 produces 13.1 g of
water.. How many grams of O2 reacted?
water
H2 (g)
+
O2 (g)
→
2 H2O (g)
HORSESHOE
Grams H2O
The reaction between H2 and O2
produces 13.1 g of water. How
many grams of O2 reacted?
(/) Molar
Mass H2O
S
Grams O2
(X) Molar
Mass O2
moles
H2O
moles
O2
mole ratio moles O2
moles H2O
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MASS TO MASS
S
The reaction between H2 and O2 produces 13.1 g of
water.. How many grams of O2 reacted?
water
2 H2 + O2
→
2 H2O
13.1 g H2O x 1 mole H2O x 1 mol O2 x 32.0 g O2 = _11.6_ g O
2
18.0 g H2O
2 mol H2O 1 mol O2
HORSESHOE
Grams CO2
2
S
Acetylene gas C2H2 burns in the
oxyactylene torch for welding.
How many grams of C2H2 are
Grams C2H2
burned if the reaction produces
75.0 g of CO2?
C2H2+ 5O2 →4 CO2 + 2 H2O
(/) Molar
Mass CO2
(X) Molar
Mass C2H2
moles
CO2
moles
C2H2
mole ratio moles C2H2
moles CO2
27
S
MASS TO MASS
Acetylene gas C2H2 burns in the oxyactylene
torch for welding. How many grams of C2H2 are
2
4 75.0 g 2
burned
if the 5
reaction produces
of CO2?
C2 H2
+
O2
→
CO2 +
H2O
75.0 g CO2 x 1 mol CO2 x 2 mol C2H2 x 26.0 g C2H2
44.0 g CO2
4 mol CO2
1 mol C2H2
= 22.2 g C2H2
HORSESHOE
Grams N2
Calculate how many grams of
ammonia are produced when you
react 2.00g of nitrogen with excess
hydrogen.
S
Grams NH3
N2 + 3 H2 → 2 NH3
(/) Molar
Mass N2
(X) Molar
Mass NH3
moles
N2
Moles
NH3
mole ratio moles NH3
moles N2
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S
MASS--MASS
MASS
Calculate how many grams of ammonia are
produced when you react 2.00g of nitrogen
with excess hydrogen.
N2 + 3 H2 → 2 NH3
2.00g N2
1 mol N2
2 mol NH3
28.02g N2 1 mol N2
17.06g NH3
1 mol NH3
= 2.4 g NH3
HORSESHOE
Grams H2O
Calculate how many grams of
Oxygen produces 13.1 g H2O in the
following reaction.
S
Grams O2
2H2 + O2 → CO2 + 2H2O
(/) Molar
Mass H2O
(X) Molar
Mass O2
moles
H2O
Moles
O2
mole ratio moles O2
moles H2O
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MASS--MASS
MASS
S
Calculate how many grams of Oxygen
produces 13.1 g H2O in the reaction.
→
2H2 + O2
CO2 +
13.1 g H2O x 1 mol H2O x 1 mol O2 x
18.0 g H2O 2 mol H2O
=
2H2O
32.0 g O2
1 mol O2
11.6 g O2
HORSESHOE
Grams H2
How many grams of Hydrogen are
required to react with 100.00g of Oxygen
in the production of water?
(/) Molar
Mass H2
S
Grams O2
(X) Molar
Mass O2
moles
H2
moles
O2
mole ratio moles O2
moles H2
30
MASS TO MASS
S
How many grams of Hydrogen are required to react with
100.00g of Oxygen in the production of water?
2 H2 + O2 → 2 H2O
100.0 g O2 x 1 mole O2 x 2mol H2 x 2.02 g H2 = 12.60 g H2
32.0 g O2 1 mol O2 1 mol H2
MASS--MASS
MASS
S
If you start with 58.44g of salt (NaCl
(NaCl),
), how many
grams of Cl2 are produced when salt is
decomposed?
2NaCl → 2Na + Cl2
58.44g NaCl
1 mol NaCl
____ gNaCl
=
1 mol Cl2
70.90 g Cl2
2 mol NaCl 1 mol Cl2
Cl2
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MASS--MASS
MASS
Iron combines with Sulfur according to the
following reaction:
Fe
(s)
+
S(l)
→
S
FeS (s)
In an experiment, .25 mol FeS (s) is created in a
reaction of iron and excess S(l) . Calculate the
mass of Fe that was used.
HORSESHOE
Grams FeS
In an experiment, .25 mol
FeS (s) is created in a reaction
of iron and excess S(l) .
(/) Molar
Mass FeS
S
Grams Fe
(X) Molar
Mass Fe
moles
FeS
moles
Fe
mole ratio moles Fe
moles FeS
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MASS--MASS
MASS
0.25 mol FeS x 1 mol Fe x 55.85g Fe
1 mol FeS
1 mol Fe
S
=
14g FeS
“HOW TO” HORSESHOE
Grams N2
How many moles of Hydrogen are
required to react with 103.25 g of
Nitrogen (AKA mole ratio) in the
Haber Process to produce ammonia?
(/) Molar
Mass N2
S
Grams H2
(X) Molar
Mass H2
moles
N2
moles
H2
mole ratio moles H2
moles N2
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MASS--MASS
MASS
S
N2 + 3 H2 → 2 NH3
2.00g N2
1 mol N2
2 mol NH3
17.06g NH3
28.02g N2 1 mol N2
1 mol NH3
= 2.4 g NH3
HORSESHOE
Grams H2O
Calculate how many grams
of Oxygen produces 13.1 g
H2O in the following
reaction.
(/) Molar
Mass Na
moles
H2O
S
Grams O2
(X) Molar
Mass O2
moles
O2
mole ratio moles O2
moles H2O
34
MASS--MASS
MASS
S
Calculate how many grams of Oxygen produces
13.1 g H2O in the following reaction.
2H2 + O2 →
CO2 +
2H2O
13.1 g H2O x 1 mol H2O x 1 mol O2 x 32.0 g O2
18.0 g H2O 2 mol H2O 1 mol O2
= 11.6 g O2
MASS--MASS
MASS
S
We can also start with mass and convert
to moles of product or another reactant
We use molar mass and the mole ratio to
get to moles of the compound of interest
Calculate the number of moles of ethane (C2H6) needed
to produce 10.0 g of water
2 C2H6 + 7 O2 → 4 CO2 + 6 H20
10.0 g H2O 1 mol H2O
2 mol C2H6 = 0.185 mol
18.0 g H2O 6 mol H20
C2H6
= 0.185 mol C2H6 x 30.0g/mol =5.5g C2H6
35
MASS--MASS
MASS
4 Fe
+
48 g O2 x
→
3 O2
mol O2
=
S
2 Fe2O3
1.5 mol O2
32.0g O2
1.5 mol O2 x
4mol Fe=
3
2.0 mol Fe
mol O2
2.0 mol Fe x 55.85 g Fe =
1 mol Fe
111.4 g Fe
MASS--MASS
MASS
4 Fe
+
3 O2
12.0 mol O2 x
→
S
2 Fe2O3
4 mol Fe =
3 mol O2
16.0 mol Fe x 55.85 g Fe =
1 mol Fe
16.0 mol Fe
894g Fe
36
MASS--MASS
MASS
S
How many grams of Fe are needed to react
with 68g of Fe2O3?
4 Fe
→ 2 Fe2O3
3 O2
+
How many grams of Fe are needed?
MASS--MASS
MASS
4 Fe
+
68 g Fe2O3 x
3 O2
→
S
2 Fe2O3
mol Fe2O3 = .425 mol Fe2O3
160.0g Fe2O3
.425 mol Fe2O3 X
4
2
mol Fe = .85 mol Fe
mol Fe2O3
.85 mol Fe x 55.85 g Fe =
1 mol Fe2O3
47.58 g Fe
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S
aX + bY → cZ
mol X
3 mol Y/Z
mol X
= Y/Z 1 mol Y
S
MASS TO ATOM
ATOM,MOLECULE,PARTICLE,
CRIDER
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•Example: Calculate how many molecules of Carbon Dioxide is
produced with 18.0g H2O in the following reaction.
S
•Grams H2O
•CH4 + 2O2 → CO2 + 2H
2H2O
•Grams CO2
•(X) Molar
•(/) Molar
• Mass CO2
• Mass H2O
•
•moles•(X) AV #
•Atoms
•moles
•atoms
•CO2
• H2 O
• H2 O
•
1moles
C
O
2
•(/) AV #
• CO2
• 2moles H2O
S
The reaction between H2 and O2 produces
3.0x1024 atoms of water. How many
atoms of O2 reacted?
39
HORSESHOE
atoms H2O
S
The reaction between H2
and O2 produces 3.0x1024
atoms of water. How many
molecules of O2 reacted?
(/) AV #
(X) AV #
Atoms
O2
moles
O2
moles
H2O
mole ratio moles O2
moles H2O
MASS TO MASS
S
The reaction between H2 and O2 produces 3.0x1024 atoms of
water. How many atoms of O2 reacted?
2 H2 + O2
3x1024 atoms H2O
1 mole H2O
6.022x1023 H2O
→
2 H2O
1 mol O2
2 mol H2O
6.022x1023 O2 =
1 mol O2
1.5 x 1024 atoms O2
40
HORSESHOE
Grams Na
S
How many O2 molecules will
react with 505g of Na to
form Na2O?
(/) Molar
Mass Na
(X) AV #
Atoms
O2
moles
O2
moles
Na
mole ratio moles O2
moles Na
MASS TO ATOMS
S
How many O2 atoms will react with 505g of Na
to form Na2O?
4 Na
+
O2
→
2 Na2O
505 g Na x 1 mol Na x 1 mol O2 x 6.02 x 1023 atoms O2
23.0 g Na
4 mol Na
1 mol O2
= 3.30 x 1024 atoms O2
41
HORSESHOE
Grams Fe2O3
S
How many atoms of Fe are
needed to create 2000.g of
Fe2O3?
(/) Molar
Mass Fe2O3
(X) AV #
moles
Fe
moles
Fe2O3
Atoms
Fe
mole ratio moles Fe
moles Fe2O3
MOLE--MASS
MOLE
4 Fe
+
3 O2
→
S
2 Fe2O3
2000. g Fe2O3 x
mol Fe2O3 = 12.50 mol
160.0g Fe2O3
12.50 mol Fe2O3 X 4
2
mol Fe = 25.00 mol Fe
mol Fe2O3
25 mol Fe x 6.022x1023 g Fe = 1.500x1025 Fe
1 mol Fe
42
HORSESHOE
S
How many molecules of
Fe2O3 are needed to react
with 525g of Fe
Fe?
?
Grams
Fe
(/) Molar
Mass Fe
(X) AV #
moles
Fe2O3
moles
Fe
Atoms
Fe2O3
mole ratio moles Fe2O3
moles Fe
ATOMS--MASS
ATOMS
4 Fe
+
525 g Fe x
3 O2
→
S
2 Fe2O3
1 mol Fe = 9.4 mol Fe
55.85g Fe
9.4 mol Fe X
2
4
mol Fe2O3 = 4.7mol Fe2O3
mol Fe
4.7 mol Fe x 160.0 g Fe2O3 =
1 mol Fe2O3
752 g Fe2O3
4.7 mol Fe x 6.022x1023 g Fe = 2.80x1024 molecules
Fe
1 mol Fe
43
HORSESHOE
How many grams Na will react with 5.0
x 1023 of O2 to form Na2O?
S
Grams Na
(X) Molar
Mass Na
(X) AV #
Atoms
O2
moles
O2
moles O2
moles Na
moles
Na
S
MASS TO ATOMS
How many grams Na will react with 5.0 x 1023 of
O2 to form Na2O?
4 Na
+
O2
5.0x1023 O2 x 1 mol O2
6.022x1023 O2
→
2 Na2O
x 4 mol Na x 23.0 g Na
1 mol O2
1 mol Na
= 76 g Na
44
SOLUTION STOICHIOMETRY
CONCENTRATION
S
S
The measure of the
amount of solute in a
certain amount of solution
Solute the lesser quantity which is dissolved in
the substance of greater quantity ,what
,what is
being dissolved
Solvent is the substance in greater quantity,
what is doing the Dissolving
Solution the mixture of solvent and solute
45
Concentration
S
Dilute
small amount of solute in the solution
Concentrated
large amount of solute
in the solution
These are vague terms
without definite
boundaries
They can be used to
compare solutions
MOLARITY
S
Most often used to specify the concentration of
a solution
MOLARITY = number of moles of solute in one
liter of solution
M=
moles of solute
liters of solution
46
MOLARITY
S
What is the molarity of a solution that has
3.21 moles of HCl dissolved in 2.4 L of
solution?
Molarity = moles of solute = 3.21 moles HCl
Liters of solution
2.4 L solution
= 1.34 M
MOLARITY
S
3.7 moles of HCl is added to water to make 500. mL
of solution. What is the molarity?
Molarity = moles of solute = 3.7 moles HCl
Liters of solution
0.500 L solution
= 7.4 M
47
MOLARITY
S
How many liters of 0.20 M solution can be from 3.00
mol of NaCl?
Molarity = moles of solute = 3.0 moles NaCl
Molarity
0.20 mol/L
= 15L
MOLARITY
S
How many moles of NaCl are needed to make 11.0 L
of 0.15 M solution? How many grams of NaCl would
that be?
Moles solute = (Liters) (Molarity) = (11.0L)(
(11.0L)(0.20
0.20 mol/L)
= 1.7 mol NaCl
1.7 Moles
x
= 58.45g
1 mol
= 99. g NaCl
48
MOLARITY
S
How many grams of sodium sulfate (Na2SO4 )
are required to prepare a 250 mL
mL,, 0.683 M
solution?
250 mL x
1L
= 0.250 L
1000mL
0.250 L x 0.683 moles Na 2 SO
1 liter solution
4
= 0.171 moles
Na 2SO4
0.171 moles Na2SO4 x 142.04 g Na 2SO4 = 24.3 g
1 mole Na2SO4
Na2SO4
21.0 g of NaOH is dissolved in enough
water to make 500. mL of solution. What
is the molarity?
S
1mol NaOH
= 0.525 mol NaOH
39.9970g NaOH
moles of solute
0.525 mol
M=
=
= 1.05 M NaOH
liters of solution
0.500 L
21.0gNaOH ×
49
S
•Example: Calculate how many molecules of Carbon Dioxide is
produced with 13.1 g H2O in the following reaction.
S
•Grams H2O
2H2O
•CH4 + 2O2 → CO2 + 2H
•Grams CO2
•(X) Molar
•(/) Molar
• Mass CO2
• Mass H2O
•
•Atoms
•moles•(X) AV #
•moles
•Molecules
• H2 O
•CO2
• H2 O
• 1moles CO2
•(/) AV #
• CO2
• 2moles H2O
•(X) M H2O
•Liters H2O
•(/) M CO2
•Liters CO2
50
Calculate how many molecules of Carbon Dioxide is produced
with 13.1 g H2O in the following reaction.
CH3 + 2O2 → CO2 + 2H
2H2O
Grams H2O
Atoms
H2O
(/) Molar
Mass H2O
(X) Molar
Mass CO2
moles
H2O
moles
CO2
(/) AV #
1moles CO2
2moles H2O
Grams CO2
(X) AV #
Molecules
CO2
(/)
(X) M H2O
Liters H2O
S
M CO2
Liters CO2
HORSESHOE
S
How many liters of 3.0 M NaOH will react with
500mL of 6.0 M sulfuric acid?
H2SO4 + 2NaOH → Na 2SO4 + 2H2OH
moles
H2SO4
(X) M H2SO4
Liter H2SO4
mole ratio moles NaOH
moles H2SO4
moles
NaOH
(/) M NaOH
Liter NaOH
51
S
How many liters of 3.0 M NaOH will react with
500mL of 6.0 M sulfuric acid?
H2SO4 + 2NaOH → Na 2SO4 + 2H2OH
.5L H2SO4 x 6.0 moles H2SO4 = 3.0 moles H2SO4
1 L H2SO4
3.0 moles H2SO4 x 2moles NaOH = 6.0 moles NaOH
1 mol H2SO4
6.0 moles NaOH x 1 Liters NaOH = 2.0 L NaOH
3.0 mol NaOH
HORSESHOE
S
How many liters of 3.0 M Sulfuric Acid will react
with 3.21 moles of Sodium Hydroxide ? How
many molecules of Sulfuric Acid ?
H2SO4 + 2NaOH → Na 2SO4 + 2H2OH
(X) AV #
moles
NaOH
moles H2SO4
moles NaOH
moles
H2SO4
Molecules
H2SO4
(/) M H2SO4
Liter H2SO4
52
S
How many liters of 3.0 M Sulfuric Acid will react with 3.21
moles of Sodium Hydroxide ? How many molecules of
Sulfuric Acid ?
H2SO4 + 2NaOH → Na 2SO4 + 2H2OH
3.21 moles NaOH x 1 moles H2SO4 = 1.61 moles H2SO4
2 mol NaOH
1.61 moles H2SO4 x 1 Liters H2SO4 =
.537 Liters H2SO4
3.0 mol H2SO4
1.61 moles H2SO4 x 6.022x1023 atoms = 9.69 x1024 atoms
1.0 mol H2SO4
HORSESHOE
grams Zn
S
Zinc reacts with acids to produce
H2 gas. If 10g of Zn reacts with
2.5M HCl, how many Liters of HCl
is needed to convert all Zn?
(/) Molar Mass
Zn
moles
Zn
mole ratio moles HCl
moles Zn
moles
HCl
(/) M HCl
L HCl
53
S
Zinc reacts with acids to produce H2 gas. If 10g of Zn
reacts with 2.5M HCl, how many Liters of HCl is
needed to convert all Zn?
Zn
(s)
+ 2 HCl
(aq
aq))
10.0 g Zn 1 mol Zn
65.39g Zn
→ ZnCl2
(aq
aq))
+ H2
(g)
2 mol HCl
1.00L
1 mol Zn
2.5 mol HCl
= 0.122 L HCl
HORSESHOE
How many grams of silver will be
formed from 1.2 mL of .5M g AgNO3
and excess copper ?
S
Grams Ag
(X) Molar
Mass Ag
moles
AgNO3
moles Ag
moles AgNO3
moles
Ag
(X) Molarity
AgNO3
Liters AgNO3
54
S
How many grams of silver will be formed from 1.2
mL of .5M AgNO3 and excess copper ?
Cu + 2AgNO3 → 2Ag + Cu(NO3)2
.5M
1.0 mol
AgNO3 AgNO
3
2.0 mol
Ag
107.87
g Ag
.0012 L
AgNO3
2 mol
AgNO3
1 mol
Ag
= .06g Ag
HORSESHOE
S
How many grams of Cu are required to
react with 1.5 L of 0.10M AgNO3?
Grams Cu
(X) Molar
Mass Cu
moles
AgNO3
moles Cu
moles AgNO3
moles
Cu
(X) Molarity
AgNO3
Liters AgNO3
55
How many grams of Cu are required to react
with 1.5 L of 0.10M AgNO3?
S
Cu + 2AgNO3 → 2Ag + Cu(NO3)2
1.5
L
.10 mol
AgNO3
1 mol
Cu
63.55
g Cu
1L
2 mol
AgNO3
1 mol = 4.8 g
Cu
Cu
S
LIMITING REACTANTS
56
•Example: Calculate how many molecules of Carbon Dioxide is
produced with 13.1 g H2O in the following reaction.
S
•Grams H2O
•CH4 + 2O2 → CO2 + 2H
2H2O
•Grams CO2
•(X) Molar
•(/) Molar
• Mass CO2
• Mass H2O
•
•moles•(X) AV #
•Atoms
•moles
•Molecules
•CO2
• H2 O
• H2 O
•
1moles
C
O
2
•(/) AV #
• CO2
• 2moles H2O
•(X) M H2O
•Liters H2O
•(/) M CO2
•Liters CO2
LIMITING REACTANTS
S
The limiting reactant is the reactant present in
the smallest stoichiometric amount
Which of these will limit the reaction?
Not Necessarily. Why?
57
LIMITING REACTANTS
S
Most of the time in chemistry we have more of
one reactant than we need to completely use
up other reactant.
That reactant is said to be in excess (there is too
much).
The other reactant limits how much product we
get. Once it runs out, the reaction
s.
This is called the limiting reactant.
reactant.
LIMITING REACTANTS
S
In the example below, the O2 would be the
excess Reactant
58
LIMITING REACTANTS
S
We have to calculate how much (in moles) of a
product we can get from each of the reactants
to determine which reactant is the limiting one.
The lower amount of moles of a product is the
correct answer.
The reactant that makes the least amount (in
moles) of product is the limiting reactant.
reactant.
Once you determine the limiting reactant, you
should ALWAYS start with it!
LIMITING REACTANTS
S
N2 + O2 → 2NO
If you react 1 mole of N2 with 1 mole of O2
you will get 2 moles of NO
What happens if you start with 2 moles of
O2 and 6 moles of N2?
59
LIMITING REACTANTS
S
Balance the reaction of Aluminum with Oxygen:
4 Al (s) + 3 O2 (g) → 2 Al2O3
How many grams of Al (of course in excess O2)
are needed to produce 3.7 moles of Al2O3 ?
3.7 mole AL2O3 x 4 mol Al x 26.98 g Al
2 AL2O3
1 mol Al
199 g Al
2.0 X 102 g Al
LIMITING REACTANTS
S
4Al (s) + 3O2 (g) → 2Al2O3
If you have only 5.0 moles of O2 gas, can you
still produce 3.7 moles of Al2O3?
5.0 mole O2 x 2 mol AL2O3 = 3.3 mol AL2O3
3 mol O2
No!!
60
Iron combines with Sulfur according to the
following reaction:
Fe
(s)
+
S
S(l) → FeS (s)
In the experiment, 7.62 g Fe are allowed to
react with 8.67 g S.
A. Determine which of the reactants is the
limiting reactant.
B. Calculate the mass of FeS formed.
C.How
C. How much excess would remain?
AMOUNT OF EXCESS
S
Find how much excess (in grams) would we need to
react with the full amount of the limited.
We subtract the excess used (in grams) from excess
started with (in grams) .
61
S
Grams 2
(/) Molar
Mass 2
Grams 1
(/) Molar
Mass 1
moles
1
moles
N2
mole ratio moles
moles 12
mole ratio moles
moles 12
moles Product
22
moles Product
22
How many grams of Hydrogen are required
to react with 100.00g of Oxygen in the
production of water? If you started with
15g of Hydrogen and it is left over after the
reaction stops, how much would be left?
S
2H2 + O2 → 2H2O
100.0 g O2
1 mol O2
2 mol H2 2.02 g H2
= 12.60 g H2
32.0 g O2 1 mol O2
1 mol H2
15.0 g H2 – 12.6 g H2 = 2.4g H2 EXCESS
62
LIMITING REACTANTS
S
In the experiment, 7.62 g Fe are allowed to react with 8.67 g S.
Fe (s) + S(l) → FeS (s)
Start with Fe:
7.62 g Fe
1 mol Fe
55.85 g Fe
1 mol FeS
1 mol Fe
= .136 mol FeS
Now S:
8.67g S 1 mol S
1 mol FeS
32.05 g S
7.62 g Fe
1 mol Fe
55.85 g Fe
1 mol S
1 mol S
1 mol Fe
= 0.27 mol FeS
32.05 g S
= 4.4 g S
1 mol S
8.67 g S – 4.4 g S = 4.3 g S EXCESS
LIMITING REACTANTS
S
10.0g of aluminum reacts with 35.0 grams of chlorine
gas to produce aluminum chloride.
QUESTION #1: Which reactant is limiting, which is in
excess, and how much product is produced?
QUESTION #2: How much of the excess is left over.
63
LIMITING REACTANTS
S
10.0g of aluminum reacts with 35.0 grams of chlorine
gas to produce aluminum chloride. Which reactant is
limiting, which is in excess, and how much product is
produced?
2 Al + 3 Cl2 → 2 AlCl3
Start with Al:
10.0 g Al 1 mol Al
2 mol AlCl3
= .37 mol AlCl3
Now Cl2:
27.0 g Al
2 mol
35.0g Cl2 1 mol Cl2
2 mol AlCl3
71.0 g Cl2
35g
Cl2
71g Cl2
2 mol Al
3 mol Cl2
3 mol Cl2
= .32 mol AlCl3
27.0 g Al
1 mol Al
= 8.8 g Al
USED!
10.0 g Al – 8.8 g Al = 1.2 g Al EXCESS
LIMITING REACTANTS
S
15.0 g of potassium reacts with 15.0 g
of iodine. Calculate which reactant is
limiting and how much product is
made. How much excess reactant is
left over? 2 K + I2 → 2 KI
64
LIMITING REACTANTS
15.0 g of potassium reacts with 15.0 g of
2 K + I2 → 2 KI
iodine.
15.0 g K 1 mol K
2 mol KI
= 0.384mol KI
39.1 g K 2 mol K
15.0 g I2
1 mol I2
254 g I2
15.0 g I2
1 mol I2
S
2 mol KI
= 0.12 mol KI
1 mol I2
2 mol K
254.00 g I2 1 mol I2
39.10 g K
1 mol K
= 4.62 g K
USED!
15.0 g K – 4.62 g K = 10.38 g K EXCESS
S
PERCENT YIELD
actual amount of a product expressed as a %
of the theoretical amount of that substance
65
PERCENT YIELD
% Yield =
S
Actual amount
of product from the lab
Expected calculated
amount from
stoichiometry
PERCENT YIELD
S
Actual yield is the amount of product actually
recovered from an experiment
Theoretical (possible) yield is the maximum
amount of product that could be produced
from the reactant.
Percent Yield is the actual yield compared to
the maximum (theoretical yield) possible.
66
PERCENT YIELD
S
You prepared cookie dough to make 5 dozen cookies.
The phone rings while a sheet of 12 cookies is
baking. You talk too long and the cookies burn. You
throw them out (or give them to your dog.) The rest
of the cookies are okay.
How many cookies could you have made
(Theoretical yield)?
yield)?
How many cookies did you actually make to eat?
(Actual yield)
PERCENT YIELD
S
What is the percent yield of cookies?
Percent Yield = Actual Yield (g)_ X 100
Theoretical Yield (g)
% cookie yield =
48 cookies x 100 = 80%
60 cookies
67
PERCENT YIELD
S
FeS was calculated to be 12.0 g. In a lab experiment
the actual yield was 10.55 g. Calculate the % Yield.
10.55g x 100 = 87.92 %
12.0 g
PERCENT YIELD
S
454g of NH4NO3 → NO4 + 2H2O
If 454 g of NH4NO3 reacts completely How much N2O is
formed if 204 g of water is produced as well?
Total mass of reactants =
total mass of products
454 g NH4NO3 = ___ g N2O + 204 g H2O
mass of N2O = 250. g
C.O.M.
68
PERCENT YIELD
S
454g of NH4NO3 → NO4 + 2H2O
a. Calculate the % YIELD
b. If you isolated only 131 g of N2O, what is
the percent yield?
c. This compares the theoretical (250. g)
and actual (131 g) yields.
454g of NH4NO3 → NO4 + 2H2O
S
Calculate the percent yield
% yield =
actual yield
• 100%
theoretical yield
% yield =
131 g
• 100% = 52.4%
250. g
% YIELD
69
PERCENT YIELD
S
If we repeated the previous experiment and
improved our yield from 52.4% to 90% how
many grams would that be?
250. g N2O x .90 = 225 g N2O
PERCENT YIELD
S
Mix 5.40 g of Al with 8.10 g of Cl2.
What mass of Al2Cl6 can form?
2 Al + 3 Cl2 →
Al2Cl6
Grams Cl2
grams
Al2Cl6
Moles Cl2
1 mol Al2Cl6
3 mol Cl2
moles
Al2Cl6
70
PERCENT YIELD
S
IF YOU MIX 5.40g OF Al WITH 8.10g OF Cl2. WHAT IS
THE THEORETICAL YIELD OF Al2Cl6.
2 Al + 3 Cl2 →
5 . 40 g Al •
Al2Cl6
1 mol
= 0.200 mol Al
27.0 g
.200 mol Al x (1 mole Al2Cl6 / 2 mol Al) = .100 mol Al2Cl6
8.10 g Cl 2 •
1 mol
= 0.114 mol Cl 2
70.9 g
0.114 mol Cl2 x (1 mole Al2Cl6/ 3 mol Al) = 0.038 mol Al2Cl6
Limiting Reactant = Cl2
PERCENT YIELD
2 Al + 3 Cl2
S
→ 10.1 g Al2Cl6
How many grams of Al2Cl6 will form?
10.1 g Al2Cl6
8.10 g Cl2
1 Mol Cl2
70.9 g Cl2
Horseshoe
266g Mol Al2Cl6
Mol Al2Cl6
.114 mol Cl2
Mol Al2Cl6
0.038 mol Al2Cl6
3 Mol Cl2
71
HOW MUCH OF THE EXCESS WILL REMAIN
AFTER THE REACTION IS COMPLETE
S
Cl2 was the limiting reactant.
Therefore, Al was present in excess. But how
much?
First find how much Al was used
Based on Cl2 limiting production of Al2Cl6
Then find how much Al is in excess.
Based on subtracting used from had of Al
Mix 5.40 g of Al with 8.10 g of Cl2.
What mass of Al remains?
2 Al + 3 Cl2
grams
Al2Cl6
Mol Al2Cl6
→
S
Al2Cl6
grams
Al used
2 Mol Al
Mol Al2Cl6
Mol Al
72
2 Al + 3 Cl2 → 10.1 g Al2Cl6
S
How many grams of Al is Excess?
10.1 g Al2Cl6
2.05 g Al
27.0 g Al
1 Mol Al
Mol Al2Cl6
266g Mol Al2Cl6
0.038 mol Al2Cl6
Excess Al
2 Mol Al
Mol Al2Cl6
0.076 mol Al
= Al available - Al required
= 5.4 g Al - 2 .05 g Al
= 3.4 g Al in excess
How many grams of H2O would be generated
S
if 1.00 g of Glucose reacted in excess oxygen?
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams
73
S
Iron combines with Sulfur according to the
following reaction:
Fe
(s)
+
S(l)
→
S
FeS (s)
In the experiment, 7.62 g Fe are allowed to react
with 8.67 g S.
a. Determine which one of the reactants is
the limiting reactant.
b. Calculate the mass of FeS formed.
74
Fe
(s)
+
S(l)
→
FeS (s)
S
7.62g Fe x 1 mole Fe =
55.85 g Fe
0.136 mol Fe
8.67 g S x 1 mol S =
32.06 g S
0.270 mol S
We don’t have to go all the way around the
horseshoe to find the Limiting Reactant
We can stop at moles to find which runs out
first - to make it easier (or whole thing if it helps)
Theoretical Yield
0.136 mol Fe x 1 mol FeS
1 mol Fe
0.270 mol S x 1 mol FeS
=
=
S
0.136 mol FeS
0.270 mol FeS
1 mol S
Or you can go all the way to grams
if it keeps it straight for you
75
S
Determine the percent yield if 100g is collected
after reaction.
0.136 mol Fe x 1 mol FeS
=
119 g FeS
1 mol FeS
110 g FeS
119 g FeS
=
92.4% yield
92% yield
Calculate the mass of FeS if the % yield is 82%.
119 g FeS x .82 = 15.58 g FeS
15.6 g FeS
S
How many grams of H2O would be generated
if 1.00 grams of Glucose (C
(C6H12O6) reacted in excess O?
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
.0056
.0333
76