Download Cookies and Chemistry…Huh!?!?

Cookies and Chemistry…Huh!?!?
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Just like chocolate chip
cookies have recipes,
chemists have recipes as well
Instead of calling them
recipes, we call them reaction
equations
Furthermore, instead of using
cups and teaspoons, we use
moles
Lastly, instead of eggs, butter,
sugar, etc. we use chemical
compounds as ingredients
Chemistry Recipes

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Looking at a reaction tells us how much of
something you need to react with
something else to get a product (like the
cookie recipe)
Be sure you have a balanced reaction
before you start!
Example: 2 Na + Cl2  2 NaCl
 This reaction tells us that by mixing 2 moles of
sodium with 1 mole of chlorine we will get 2 moles
of sodium chloride
 What if we wanted 4 moles of NaCl? 10 moles?
50 moles?

2H2 + O2  2H2O


Two molecules of hydrogen and one
molecule of oxygen form two molecules of
water.
2 Al2O3 Al + 3O2
2 formula units Al2O3 form 4 atoms Al
and 3 molecules O2
Now try this: 2Na + 2H2O  2NaOH + H2
Practice

Write the balanced reaction for hydrogen gas
reacting with oxygen gas.
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2 H2 + O 2  2 H2 O
How many moles of reactants are needed?
What if we wanted 4 moles of water?
What if we had 3 moles of oxygen, how much hydrogen
would we need to react and how much water would we
get?
What if we had 50 moles of hydrogen, how much oxygen
would we need and how much water produced?
Mole Ratios


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 are
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 mol Na
= 2.5 moles Cl2
Mole-Mole Conversions

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?
Mole-Mole Conversions
2.6 moles Cl2 2 mol NaCl
1 mol Cl2
5.2 mol NaCl
Mole to Mole conversions

2 Al2O3 Al + 3O2

each time we use 2 moles of Al2O3 we will
also make 3 moles of O2
2 moles Al2O3
3 mole O2
or
3 mole O2
2 moles Al2O3
These are the two possible
conversion factors
Mole to Mole conversions
 How
many moles of O2 are
produced when 3.34 moles of
Al2O3 decompose?
 2 Al2O3 Al + 3O2
3.34 mol Al2O3
3 mol O2
2 mol Al2O3
= 5.01 mol O2
Mole-Mass Conversions
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
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
70.90g Cl2
1 mol Cl2
= 177g Cl2
Practice

Calculate the mass in grams of Iodine
required to react completely with 0.50
moles of aluminum.
Practice: Mole Mass Conversions
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
Balanced Equation:
2 Al + 3 I2 ---> 2 AlI3
Stoichiometry:
.5 mol Al
3 mol I2
2 mol Al
252g I
1 mol I2
Mass-Mole
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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
18.0 g H2O 6 mol H20
mol C2H6
Mass-Mass Conversions
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Most often we are given a starting mass
and want to find out the mass of a
product we will get (called theoretical
yield) or how much of another reactant we
need to completely react with it (no
leftover ingredients!)
Now we must go from grams to moles,
mole ratio, and back to grams of
compound we are interested in
Mass-Mass Conversion
Ex. 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 17.06g NH3

28.02g N2 1 mol N2
= 2.4 g NH3
1 mol NH3
Mass-Mass Practice

Calculate how many grams of oxygen are
required to make 10.0 g of aluminum oxide
Balanced Equation:
3 O2 + 4 Al -> 2 Al2O3
Mole Ratio Needed:
2 mole
Al2O3 to 3 mol O2
Stoichiometry:
10.0g
3 mol
O2
2 mol
Al2O3
32 g
O2
1 mol
O2
Practice:
2C2H2 + 5 O2  4CO2 + 2 H2O
• If 3.84 moles of C2H2 are burned, how
many moles of O2 are needed? (9.6 mol)
•How many moles of C2H2 are needed to
produce 8.95 mole of H2O? (8.95 mol)
•If 2.47 moles of C2H2 are burned, how
many moles of CO2 are formed? (4.94 mol)
Mass-Mass Problem:
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al + 3 O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
= ? g Al2O3
(6.50 x 2 x 101.96) ÷ (26.98 x 4) = 12.3 g Al2O3
Another example:
 If
10.1 g of Fe are added to a
solution of Copper (II) Sulfate,
how much solid copper would
form?
 2Fe + 3CuSO4  Fe2(SO4)3 + 3Cu
Answer = 17.2 g Cu
Volume-Volume Calculations:


How many liters of CH4 at STP are required to
completely react with 17.5 L of O2 ?
CH4 + 2O2  CO2 + 2H2O
1 mol O2 1 mol CH4 22.4 L CH4
17.5 L O2
22.4 L O2 2 mol O2 1 mol CH4
= 8.75 L CH4
Notice anything concerning these two steps?
Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and
solid NaHCO3 (baking soda) is to be used to neutralize the
acid. How many grams of NaHCO3 must be used?
H2SO4(aq) + 2NaHCO3  2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 mL
?g
6.0 M
=
6.0 mol
L
H2SO4
50.0 mL 6.0 mol H 2SO 4
1000mL
H 2SO 4
NaHCO3 NaHCO3
84.0 g
2 mol
= 50.4 g NaHCO3
mol
1 mol
NaHCO3
H2SO4
gram to gram conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Units match
3.45 g Al
mol Al
27.0 g Al
2 mol AlCl 3 133.3 g AlCl 3
mol AlCl 3
2 mol Al
Now
We must
Now
use
Let’s
the
always
use
work
molar
thethe
convert
molar
mass
problem.
ratio.
to
toconvert
moles.
to grams.
=
17.0 g AlCl3
Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
2HCl(aq) +
Ba(OH)2(aq) 
0.40 M
47.1 mL
0.75 M
? mL
Ba(OH)2
47.1 mL
0.75mol Ba(OH)2
2H2O(l) + BaCl2
Units match
HCl
2 mol
1000 mL Ba(OH)2 1 mol
Ba(OH)2
HCl
1000 mL
0.40 mol
HCl
= 176 mL HCl
Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution.
If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of
the barium hydroxide solution, what was the concentration of the barium
hydroxide solution in moles per liter (M)?
2
1
2 2O(l) + ____BaCl
1
____HCl(aq)
+ ____Ba(OH)
2(aq)  ____H
2(aq)
25.00 mL
23.28 mL
Units match on top!
? mol
0.135 mol
L
L
23.28 mL HCl
25.00 x 10-3 L
Ba(OH)2
0.135 mol HCl l mol Ba(OH)2
= 0.0629 mol Ba(OH)2
1000 mL HCl 2 mol HCl
L Ba(OH)
2
Units Already Match on Bottom!
Limiting Reactant: Cookies
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
If we had the specified amount of all ingredients listed, could we make 4
dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make 9
dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?
Limiting Reactant
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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.
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Limiting
Reactant
To find the correct answer, we have to try all of
the reactants. We have to calculate how much
of a product we can get from each of the
reactants to determine which reactant is the
limiting one.
The lower amount of a product is the correct
answer.
The reactant that makes the least amount of
product is the limiting reactant. Once you
determine the limiting reactant, you should
ALWAYS start with it!
Be sure to pick a product! You can’t compare to
see which is greater and which is lower unless
the product is the same!
Limiting Reactant: Example
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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
27.0 g Al

2 mol AlCl3 133.5 g AlCl3
2 mol Al
1 mol AlCl3
= 49.4g AlCl3
Now Cl2:
35.0g Cl2
1 mol Cl2
71.0 g Cl2
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 43.9g AlCl3
LR Example Continued

We get 49.4g of aluminum chloride from the given
amount of aluminum, but only 43.9g of aluminum
chloride from the given amount of chlorine.
Therefore, chlorine is the limiting reactant. Once
the 35.0g of chlorine is used up, the reaction
comes to a complete
.
If 10.6 g of copper reacts with 3.83 g
sulfur, how many grams of product (copper
Cu is Limiting
Reagent
(I) sulfide) will be formed?
 2Cu + S  Cu2S
1 mol Cu2S 159.16 g Cu2S
1
mol
Cu
10.6 g Cu
63.55g Cu 2 mol Cu
1 mol Cu2S

= 13.3 g Cu2S
1
mol
S
3.83 g S
32.06g S
1 mol Cu2S 159.16 g Cu2S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S
Limiting Reactant Practice

15.0 g of potassium reacts with 15.0 g of
iodine. Calculate which reactant is limiting
and how much product is made.
Finding the Amount of Excess
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
By calculating the amount of the excess
reactant needed to completely react with
the limiting reactant, we can subtract that
amount from the given amount to find the
amount of excess.
Can we find the amount of excess
potassium in the previous problem?
Finding Excess Practice

15.0 g of potassium reacts with 15.0 g of iodine.
2 K + I2  2 KI

We found that Iodine is the limiting reactant, and 19.6 g of
potassium iodide are produced. We must now find the amound of K
actually used.
15.0 g I2
1 mol I2
2 mol K
39.1 g K
254 g I2
1 mol I2
1 mol K
= 4.62 g K
USED!
15.0 g K – 4.62 g K = 10.38 g K EXCESS
Given amount
of excess
reactant
Amount of
excess
reactant
actually
used
Note that we started with
the limiting reactant! Once
you determine the LR, you
should only start with it!
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15.15
mol
? moles
0.10 mol
Based on:
0.15 mol KO2
KO2
3mol O 2
4mol KO 2
Based on: 0.10 mol H2O 3mol O 2
H2 O
2mol H O
2
= 0.1125 mol O2
= 0.150 mol O2
Limiting Reactant: Recap
1.
2.
3.
4.
5.
6.
7.
You can recognize a limiting reactant problem because
there is MORE THAN ONE GIVEN AMOUNT.
Convert ALL of the reactants to the SAME product (pick
any product you choose.)
The lowest answer is the correct answer.
The reactant that gave you the lowest answer is the
LIMITING REACTANT.
The other reactant(s) are in EXCESS.
To find the amount of excess, subtract the amount used
from the given amount.
If you have to find more than one product, be sure to
start with the limiting reactant. You don’t have to
determine which is the LR over and over again!
Percent Yield

To determine percentage yield, you need
two pieces of information:
1) Theoretical yield =
Comes from the stoichiometry
2) Actual yield =
Comes from experimental data
Percent Yield =
Actual Yield
Theoretical Yield
100%