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TRANSCRIPT
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EXTRA CREDIT
REMINDER
Due Tonight at Midnight (January 21 at 11:59 pm) via email
*** Kinesthetic: If you do not know how to use Prezi you may
do a power point otherwise email your Prezi link.
This will count as a bonus in the test category –
approximately 10 points averaged with your other grades in
the category.
It can only HELP! Not harm your grade.
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Important things to Remember
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The Mole
A counting unit
Similar to a dozen, except instead
of 12, it’s
602,000,000,000,000,000,000,000
6.02 X 1023 (in scientific notation)
This number is named in honor of
Amedeo Avogadro (1776 – 1856)
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The Mass of 1 mole (in grams)
Equal to the numerical value of the average
atomic mass (get from periodic table), or add
the atoms together for a molecule
1 mole of C atoms = 12.0 g
1 mole of Mg atoms = 24.3 g
1 mole of O2 molecules = 32.0 g
Diatomic elements are: H, N, O, F, Cl, Br, I
Molar Mass
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Molar Mass of Compounds
The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2
Avg. Atomic mass of Calcium = 40.08g
Avg. Atomic mass of Chlorine = 35.45g
Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl2
20
Ca 40.08 17
Cl
35.45
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Flowchart Atoms or
Molecules
Moles
Mass
(grams)
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Multiply by
atomic/molar mass
from periodic table Divide by
atomic/molar mass
from periodic table
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Practice
Calculate the Molar Mass of
calcium phosphate
Formula =
Masses elements: Ca: 3 Ca’s X 40.1 =
P: 2 P’s X 31.0 =
O: 8 O’s X 16.0 =
Molar Mass =
Ca3(PO4)2
120.3 g
62.0 g
128.0 g
120.3g + 62.0g +128.0g 310.3 g/mol
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molar mass Avogadro’s number Grams Moles particles
Everything must go through
Moles!!!
Calculations
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Atoms/Molecules and Grams
How many atoms of Cu are
present in 35.4 g of Cu?
35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu
63.5 g Cu 1 mol Cu
= 3.4 X 1023 atoms Cu
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Chocolate Chip 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
How many eggs are needed to make 3 dozen cookies?
How much butter is needed for the amount of chocolate chips used?
How many eggs would we need to make 9 dozen cookies?
How much brown sugar would I need if I had 1 ½ cups white sugar?
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Cookies and Chemistry…Huh!?!?
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
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Chemistry Recipes
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?
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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 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 mol Na = 2.5 moles Cl2
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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?
2 Na + Cl2 2 NaCl
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Mole-Mass Conversions
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 70.90g Cl2
2 mol Na 1 mol Cl2 = 177g Cl2
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Practice
Calculate the mass in grams of Iodine
required to react completely with 0.50
moles of aluminum.
2 Al + 3 I2 2 AlI3
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Mass-Mole
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
18.0 g H2O 6 mol H20
= 0.185
mol C2H6
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Practice Calculate how many moles of oxygen are
required to make 10.0 g of aluminum oxide
4 Al + 3 O2 2 Al2O3
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Practice
Write the balanced reaction for hydrogen gas reacting with oxygen gas.
H2 + O2 H2O
How many moles of each reactants are needed?
What if we wanted 4 moles of water how many moles of each reactant would you need?
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?
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Practice
Write the balanced reaction for hydrogen gas
reacting with oxygen gas.
2 H2 + O2 2 H2O
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?
2 mol H2
1 mol O2
4 mol H2
2 mol O2
6 mol H2, 6 mol H2O
25 mol O2, 50 mol H2O
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Practice
A2 + 2B 2AB
What is the mole ratio of substance A to substance AB?
What is the mole ratio of substance A to substance B?
What is the mole ratio of substance B to substance AB?
Show the work for the problems below:
If you have 4 moles of substance A, how many moles of substance AB can you produce?
If you have 10 moles of substance B and an excess of substance A, how many moles of substance AB can you
produce?
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Mass-Mass Conversions
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
gA mol A mol B gB
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Mass-Mass Conversion Ex. Calculate how many grams of
ammonia are produced when you react
2.00g of nitrogen with excess hydrogen.
N2 + H2 NH3
1. Make sure you have a balanced equation.
2. Convert grams of nitrogen to moles of nitrogen
3. Convert moles of nitrogen to moles of ammonia
4. Then, convert moles of ammonia to grams of
ammonia
With a partner discuss which values you will need for
each step above. Then, work the problem.
Hint: gA mol A mol B gB
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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
gA mol A mol B gB
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Practice
How many grams of calcium nitride are
produced when 2.00 g of calcium reacts
with an excess of nitrogen?
Ca + N2 Ca3N2
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Theoretical, Actual, and Percent
Yield
Theoretical yield:
the maximum amount of product, which is calculated
using the balanced equation.
Actual yield:
the amount of product obtained when the reaction
takes place
Percent yield:
the ratio of actual yield to theoretical yield
Percent yield = actual yield (g) x 100
theoretical yield (g)
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Percent Yield
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Calculating Percent Yield
Suppose you have prepared cookie dough to make 5
dozen cookies. The phone rings and you answer. While
you talk, a sheet of 12 cookies burns, and you have to
throw them out. The rest of the cookies you make are
okay. What is the percent yield of edible cookies?
Theoretical yield: 60 cookies possible
Actual yield: 48 cookies to eat
Percent yield: 48 cookies x 100% = 80.% yield
60 cookies
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Learning Check
With a limited amount of oxygen, the reaction of
carbon and oxygen produces carbon monoxide.
2C(g) + O2(g) 2CO(g)
What is the percent yield if 40.0 g of CO are
produced when 30.0 g of O2 are used?
1) 25.0%
2) 75.0%
3) 76.2%
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Solution
STEP 1 Given: 40.0 g of CO produced (actual)
30.0 g of O2 used
Need: percent yield of CO
STEP 2 Write a plan to calculate % yield of CO:
g of O2 moles of moles of g of CO
O2 CO (theoretical)
Percent yield of CO = g of CO (actual) x 100%
g of CO (theoretical)
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Solution (continued)
STEP 3 Write conversion factors:
1 mole of O2 = 32.0 g of O2
1 mole O2 and 32.0 g O2
32.0 g O2 1 mole O2
1 mole of O2 = 2 moles of CO
1 mole O2 and 2 moles CO
2 moles CO 1 mole O2
1 mole of CO = 28.0 g of CO
1 mole CO and 28.0 g CO
28.0 g CO 1 mole CO 31
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Solution (continued)
STEP 4 Setup to calculate theoretical yield in g of O2:
30.0 g O2 x 1 mole O2 x 2 moles CO x 28.0 g CO
32.0 g O2 1 mole O2 1 mole CO
= 52.5 g of CO (theoretical)
Setup to calculate percent yield:
40.0 g CO (actual) x 100 = 76.2% yield (3)
52.5 g CO (theoretical)
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Learning Check
When N2 and 5.00 g of H2 are mixed, the reaction
produces 16.0 g of NH3. What is the percent yield
for the reaction?
N2(g) + 3H2(g) 2NH3(g)
1) 31.3% of NH3
2) 56.9% of NH3
3) 80.0% of NH3
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Solution
2) 56.9%
STEP 1 Given: 16.0 g of NH3 produced (actual)
5.00 g of H2 used
Need: percent yield of NH3
STEP 2 Write a plan to calculate % yield of NH3:
g of H2 moles of moles of g of NH3
H2 NH3 (theoretical)
Percent yield of NH3 = g of NH3 (actual) x 100%
g of NH3 (theoretical)
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Solution (continued)
STEP 3 Write conversion factors:
1 mole of H2 = 2.02 g of H2
1 mole H2 and 2.02 g H2
2.02 g H2 1 mole H2
1 mole of H2 = 2 moles of NH3
1 mole H2 and 2 moles NH3
2 moles NH3 1 mole H2
1 mole of NH3 = 17.0 g of NH3
1 mole NH3 and 17.0 g NH3
17.0 g NH3 1 mole NH3
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Solution (continued)
STEP 4 Setup to calculate theoretical yield of g of NH3:
5.00 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3
2.02 g H2 3 moles H2 1 mole NH3
= 28.1 g of NH3 (theoretical)
Setup to calculate percent yield:
Percent yield = 16.0 g NH3 x 100 = 56.9% yield (2)
28.1 g NH3
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Guide to Calculations for
Percent Yield
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Limiting Reactant
A limiting reactant in a chemical reaction is the
substance that
is used up
limits the amount of product that can form and stops
the reaction
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Reacting Amounts
39
In a table setting, there is 1plate,
1 fork, 1 knife, and 1 spoon.
How many table settings are
possible from 5 plates, 6 forks,
4 spoons, and 7 knives?
What is the limiting item?
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Reacting Amounts (continued)
Only 4 place settings are possible.
Initially Used Left over
Plates 5 4 1
Forks 6 4 2
Spoons 4 4 0
Knives 7 4 3
The limiting item is the spoon.
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Example 1 of an Everyday
Limiting Reactant
How many peanut butter sandwiches could be made
from 8 slices of bread and 1 jar of peanut butter?
With 8 slices of bread, only 4 sandwiches could be
made.
The bread is the limiting item.
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Example 2 of an Everyday
Limiting Reactant
How many peanut butter sandwiches could be made
from 8 slices bread and 1 tablespoon of peanut butter?
With 1 tablespoon of peanut butter, only 1 sandwich
could be made. The peanut butter is the limiting item.
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Limiting Reactant
When 4.00 moles of H2 is mixed with 2.00 moles of
Cl2, how many moles of HCl can form?
H2(g) + Cl2(g) 2HCl(g)
4.00 moles 2.00 moles ??? Moles
Calculate the moles of product that each reactant, H2
and Cl2, could produce.
The limiting reactant is the one that produces the
smaller number of moles of product.
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Limiting Reactant (continued)
HCl from H2
4.00 moles H2 x 2 moles HCl = 8.00 moles of HCl
1 moles H2
HCl from Cl2
2.00 moles Cl2 x 2 moles HCl = 4.00 moles of HCl
1 mole Cl2
4.00 moles of HCl is the smaller number of moles
produced. Thus, Cl2 will be used up.
The limiting reactant is Cl2.
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Limiting Reactants Using Mass
If 4.80 moles Ca are mixed with 2.00 moles N2, which is
the limiting reactant? 3Ca(s) + N2(g) Ca3N2(s)
Moles of Ca3N2 from Ca
4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles of Ca3N2
3 moles Ca (Ca is used up)
Moles of Ca3N2 from N2
2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles of Ca3N2
1 mole N2
Ca is used up. Thus, Ca is the limiting reactant. 45
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Learning Check
What is the mass of water that can be produced when
8.00 g of H2 and 24.0 g of O2 react?
2H2(g) + O2(g) 2H2O(l)
1) 8.0 g of H2O
2) 27.0 g of H2O
3) 72 g of H2O
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Solution
3) 72 g of H2O
Moles of H2O from H2:
8.00 g H2 x 1 mole H2 x 2 moles H2O = 4.0 moles of H2O
2.0 g H2 2 moles H2
Moles of H2O from O2:
24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles of H2O
32.0 g O2 1 mole O2 Smaller number
of moles of H2O
1.50 moles H2O x 18.0 g H2O = 27.0 g of H2O
1 mole H2O
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Guide to Calculating Product
from a Limiting Reactant
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Check Calculations
Equation
Initially
H2
4.00 moles
Cl2
2.00 moles
2HCl
0 mole
Reacted/
Formed
–2.00 moles –2.00 moles +4.00 moles
Left after
reaction
2.00 moles
(4.00 – 2.00)
Excess
0 moles
(2.00 – 2.00)
Limiting
4.00 moles
(0 + 4.00)
Product
possible
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Limiting Reactant
A limiting reactant in a chemical reaction is the
substance that
is used up
limits the amount of product that can form and stops
the reaction
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Reacting Amounts
51
In a table setting, there is 1plate,
1 fork, 1 knife, and 1 spoon.
How many table settings are
possible from 5 plates, 6 forks,
4 spoons, and 7 knives?
What is the limiting item?
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Reacting Amounts (continued)
Only 4 place settings are possible.
Initially Used Left over
Plates 5 4 1
Forks 6 4 2
Spoons 4 4 0
Knives 7 4 3
The limiting item is the spoon.
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Example 1 of an Everyday
Limiting Reactant
How many peanut butter sandwiches could be made
from 8 slices of bread and 1 jar of peanut butter?
With 8 slices of bread, only 4 sandwiches could be
made.
The bread is the limiting item.
53
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Example 2 of an Everyday
Limiting Reactant
How many peanut butter sandwiches could be made
from 8 slices bread and 1 tablespoon of peanut butter?
With 1 tablespoon of peanut butter, only 1 sandwich
could be made. The peanut butter is the limiting item.
54
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Limiting Reactant
When 4.00 moles of H2 is mixed with 2.00 moles of
Cl2, how many moles of HCl can form?
H2(g) + Cl2(g) 2HCl(g)
4.00 moles 2.00 moles ??? Moles
Calculate the moles of product that each reactant, H2
and Cl2, could produce.
The limiting reactant is the one that produces the
smaller number of moles of product.
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Limiting Reactant (continued)
HCl from H2
4.00 moles H2 x 2 moles HCl = 8.00 moles of HCl
1 moles H2
HCl from Cl2
2.00 moles Cl2 x 2 moles HCl = 4.00 moles of HCl
1 mole Cl2
4.00 moles of HCl is the smaller number of moles
produced. Thus, Cl2 will be used up.
The limiting reactant is Cl2.
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Limiting Reactants Using Mass
If 4.80 moles Ca are mixed with 2.00 moles N2, which is
the limiting reactant? 3Ca(s) + N2(g) Ca3N2(s)
Moles of Ca3N2 from Ca
4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles of Ca3N2
3 moles Ca (Ca is used up)
Moles of Ca3N2 from N2
2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles of Ca3N2
1 mole N2
Ca is used up. Thus, Ca is the limiting reactant. 57
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Learning Check
What is the mass of water that can be produced when
8.00 g of H2 and 24.0 g of O2 react?
2H2(g) + O2(g) 2H2O(l)
1) 8.0 g of H2O
2) 27.0 g of H2O
3) 72 g of H2O
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Solution
3) 72 g of H2O
Moles of H2O from H2:
8.00 g H2 x 1 mole H2 x 2 moles H2O = 4.0 moles of H2O
2.0 g H2 2 moles H2
Moles of H2O from O2:
24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles of H2O
32.0 g O2 1 mole O2 Smaller number
of moles of H2O
1.50 moles H2O x 18.0 g H2O = 27.0 g of H2O
1 mole H2O
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Guide to Calculating Product
from a Limiting Reactant
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Limiting Reactant: Example 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:
Now Cl2:
10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3
27.0 g Al 2 mol Al 1 mol AlCl3 = 49.4g AlCl3
35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3
71.0 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3
Limiting
Reactant
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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 .
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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.
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Finding the Amount of Excess
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?
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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.
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!
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Limiting Reactant: Recap
1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.
2. Convert ALL of the reactants to the SAME product (pick any product you choose.)
3. The lowest answer is the correct answer.
4. The reactant that gave you the lowest answer is the LIMITING REACTANT.
5. The other reactant(s) are in EXCESS.
6. To find the amount of excess, subtract the amount used from the given amount.
7. 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!
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1 gram of Chalk
What is the formula for Chalk?