Stoichiometry

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Stoichiometry

• By Ms. Michelle Buroker
• Scott High School
• Taylor Mill, Kentucky
• Go Eagles!

Lets Get Started!
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A Brief Introduction.

• Stoichiometry is the mathematics of chemical
  reactions. This is a tutorial meant to introduce
  you to the basics so that you will be able to
  calculate with confidence! ?
• There is a brief quiz at the end of the lesson,
  you need to take it, print out the certificate at
  the end, and bring it to me for credit.

Here We Go!
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Have you ever wondered when you were watching
fireworks exactly how they make that happen?

• Its Stoichiometry!
• Stoichiometry is the study of quantitative
  relationships between amounts of reactants used
  and products formed by a chemical reaction. The
  purpose of this learning module is for you to
  interactively learn Stoichiometry mathematical
  operations which are very important to
  scientists.
• When you follow the lessons laid out in this
  assignment, hopefully you will gain an
  understanding of the process by which you can
  solve questions such as if I start with so many
  grams of reactants, how many grams of product
  will I get?

Main Menu
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Main Menu

• If at any time, you need to return to the Main
  Menu, click on the home icon at the bottom left
  of each screen.
• The Mole (a.k.a. not the animal)
• Mole to Mole Relationships
• Mass to Mass Relationships
• Limiting Reagents
• Percent Yield

Return to the Beginning of The SIM
Are you Ready To Begin?!?
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The Mole
Click Here To Watch a Video about The Mole

• The SI base unit used to measure the amount of a
  substance!
• One mole of anything is 6.02 x 1023 particles
• Remember that a particle can be atoms,
  molecules, or formula units ?
• Ex 1 mole of C 6.02 x 1023 atoms
• 1mole of NaCl 6.02 x 1023 formula units
• 1 mole of H2O 6.02 x 1023 molecules
• The mass of any pure substance is equal to its
  molar mass
• Remember that the molar mass of any element is
  equal to its atomic weight and the molar mass of
  any molecule or formula unit is equal to the sum
  of the atomic weights of the individual atoms
• Ex 1 mole of Ca weighs 40.08g
• 1 mole of H2O weighs 18.0g

Conversion Factors
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Conversion Factors

• The following are the conversion factors you need
  to do the math
• 1.) To go from moles to grams
• number of moles given X atomic weight or
  molar mass
• 1 1 mole
• 2.) To go from grams to moles
• number of grams given X 1 mole
• 1 atomic weight or molar mass

Lets Try Some Practice Problems
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Conversion Factors

• The following are the conversion factors you need
  to do the math
• 1.) To go from moles to grams
• number of moles given X atomic weight or
  molar mass
• 1 1 mole
• 2.) To go from grams to moles
• number of grams given X 1 mole
• 1 atomic weight or molar mass

Lets Try Some Practice Problems
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Lets Practice!!!!

• 1 How much does 5 moles of AgNO3
• weight?
• Click on your choice!
• A. 169.91g
• B. 849.40g
• C. 339.76g

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Try Again

• Take a look at the conversation factor page again
  use those to help you. ?

Conversation Factor Page
Back to the Problem
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Try Again

• Take a look at the conversation factor page again
  use those to help you. ?

Conversation Factor Page
Back to the Problem
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Lets Practice!!!

• 2. How many moles are 250g of NaCl?
• Click on your choice!
• A. 58.45 moles
• B. 1.46 x 104moles
• C. 4.28moles

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Mole to Mole Relationships

• Now that you have mastered basic conversions
  between moles and grams youre ready to move on
  to calculations within chemical reactions!
• The mole is how we relate reactants to other
  reactants products to other products or
  reactants to products. Its called the Mole
  Ratio!

Say What???
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The Mole Ratio

• Mole ratios are used as conversion factors to
  convert a known number of moles of one substance
  to moles of another substance in the same
  chemical reaction.
• Example
• 2K(s) 2H2O(l) ? 2KOH(aq) H2(g)
• For every 2 moles of potassium, you have 2 moles
  of H2O, so the mole ratio would look like this
• 2 moles K
• 2 moles H2O

Lets Try A Practice Problem!
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Lets Practice!

• One disadvantage to burning propane (C3H8) is
  that carbon dioxide (CO2) is one of the products.
  The release of carbon dioxide increases the
  growing concentration of CO2 in the atmosphere.
  How many moles of carbon dioxide are produced
  when 10.0 moles of propane are burned in excess
  oxygen in a gas grill?

Check Your Answer
Having Trouble Getting Started and Need a Hint
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Heres a Hint .

• You need to write a balanced chemical equation
  before you can solve this problem!
• Hopefully, you recognize this as a combustion
  reaction. ?
• C3H8 5O2 ? 3CO2 4H2O

On To The Answer!
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Answer

• Your first step is to write the balanced chemical
  equation for the problem
• C3H8 5O2 ? 3CO2 4H2O
• If you start with 10.0moles of C3H8, then
• (10.0moles C3H8) X (3molesCO2/1mole C3H8)
  30.0moles CO2
• You needed the mole ratio between C3H8 and CO2 to
  solve the problem!

Moving Right Along! ?
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Mass to Mass Relationships

• The coefficients of a balanced equation give the
  relative amounts (in moles) of reactants and
  products. Calculations to find the masses of
  materials involved in reactions are called
  Mass-Mass Problems.

Okay but how do I actually solve a problem?
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The Pathway

• There is a step- by- step method to solving mass-
  mass problems the following steps must be
  followed!
• Grams Reactant ? Moles Reactant ? Moles Product ?
  Grams Product
• Convert Grams to Moles Mole Ratio
  Convert Moles to grams
• Note If you begin with grams of Product, the
  process is simply done in reverse! ?

Lets Look at an Example
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Mass-Mass An Example!

• How many grams of silver chloride can be produced
  from the reaction of 17.0g of silver nitrate with
  excess sodium chloride solution?
• Step 1 Write the chemical equation for the
  reaction.
• AgNO3(aq) NaCl(aq) ? AgCl(s) NaNO3(aq)
• Step 2 Convert grams of reactant given to moles.
• (17.0g AgNO3) X (1mole/ 169.88gAgNO3)
  0.100moles AgNO3
• Step 3 Using the mole relationship, convert
  moles of reactant into moles of product.
• (0.100moleAgNO3) X (1mole AgNO3/ 1mole
  AgCl) 0.100moles AgCl
• Step 4 Convert moles to grams of product.
• (0.100moles AgCl) X (143.32g/1moleAgCl) 14.33g
  AgCl

Ready to try one On your own??
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Hmmm Lets See How You Do On Your Own.

• Ammonium nitrate (NH4NO3), an important
  fertilizer, produces N2O gas and H2O when it
  decomposes. Determine the mass of water produced
  from the decomposition of 25.0g of solid ammonium
  nitrate.
• Try it on your own before checking the answer!

Answer
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The Origins of Fertilizer???

• Fertilizer originated from a process known as the
  Haber Process, developed during WW I in Germany.
  
• (If youre interested in reading a bit more about
  the Haber Process click the link for some brief
  info)
• Before you see the answer, follow these steps to
  solve the problem
• Step 1 Write the balanced chemical equation for
  the
• reaction.
• Step 2 Convert grams of reactant given to moles.
• Step 3 Using the mole relationship, convert
  moles of
• reactant into moles of product.
• Step 4 Convert moles to grams of product.

Now Take a Look at the Answer
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Okay This Time it Really is The Answer!

• NH4NO3 ? N2O 2H2O
• Convert grams of NH4NO3 to moles of NH4NO3.
• (25.0gNH4NO3/1) X (1mol NH4NO3/ 80.06gNH4NO3 )
  0.312mol NH4NO3
• Convert moles of NH4NO3 to moles of H2O.
• (0.312mol NH4NO3 /1) X (2mole H2O/1mole NH4NO3)
  0.624 mole H2O
• Covert moles of H2O to grams of H2O .
• (0.624molH2O/1) X (18.0g H2O/ 1mole H2O) 11.24g
  H2O

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Why do
reactions stop?

• The answer is limiting reagents!
• If you put 10 boys and 6 girls in a room and
  asked them to pair up one boy to one girl, who
  will be left without a partner?
• The boys there will be four boys left with no
  partner. The girls are the limiting factor, in
  other words, they control how many pairs will be
  formed.

Keep Moving!
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Limiting Reagents and Reactions

• Just like the previous example, in the lab,
  reactions are limited by the reactant present in
  the LEAST amount. So, to solve limiting reagent
  problems, you have to first find out which
  reactant is the limiting reagent!!!
• Hint convert grams of reactant to moles!

Lets Take a Look At an Example!
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Air Bags .
Watch Air Bag Deployment

• The reaction between solid sodium and iron (III)
  oxide is one in a series of reactions that
  inflates an automobile airbag.
• 6Na(s) Fe2O3(s) ? 3Na2O(s) 2Fe(s)
• If 100.0g Na and 100.0g Fe2O3 are used in this
  reaction, determine
• a. the limiting reagent
• b. the excess reagent
• c. the mass of solid iron produced
• d. the amount of excess reaction remaining after
  the reaction is complete.

Take a look at How to work it out!
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The Solution 6Na(s) Fe2O3(s) ? 3Na2O(s)
2Fe(s)

• 1.) Determine the Limiting Reagent by Converting
  grams of product into moles of product.
• (100.0gNa/1) X (1mol Na/ 23.0gNa) 4.35mol Na
• (100.0gFe2O3/1) X (1molFe2O3/159.7gFe2O3)
  0.6262mol Fe2O3
• Fe2O3 is present in the least amount, therefore,
  it is the limiting reagent. The excess reactant
  would be Na.
• 2.) Determine the amount of solid Fe produced by
  converting moles of the limiting reagent into
  grams of product.
• (0.6262mol Fe2O3/1) X (2mol Fe/1mol Fe2O3) X
  (55.85g Fe/1mol Fe) 69.95g Fe
• 3.) To determine the amount of excess reactant
  left, convert moles of limiting reagent to grams
  of excess reactant and subtract that number from
  what you started with.
• (0.6262mol Fe2O3/1) X (6mol Na/1mol Fe2O3) X
  (23.0g Na/ 1mol Na) 86.41g Na
• 100.0g Na Started with 86.41g Na used 13.59g
  Na left in excess

Try a Problem on Your Own Now
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Time to Try a Problem on Your Own! Good Luck ?

• If 4.44g of calcium oxide are mixed with 7.77g of
  water, how many grams of calcium hydroxide will
  form?
• Try working it out on your own before looking at
  the answer!!!

Do You Need A Hint?
Answer
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Need a Hint?

• You need to write a balanced chemical equation to
  solve this problem!
• Hopefully you recognized this as a synthesis
  reaction
• CaO H2O ? Ca(OH)2

On To The Answer!
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Answer

• Begin by writing a balanced chemical equation for
  the reaction
• CaO H2O ? Ca(OH)2
• Determine the limiting reagent
• (4.44g CaO/1) X (1mol CaO/ 56.08g CaO) 0.0792mol
  CaO
• (7.77g H2O/1) X (1mol H2O/ 18.0g H2O) 0.432mol
  H2O
• Convert moles of limiting reagent into grams of
  product
• (0.0792mol CaO/ 1) X (1mol Ca(OH)2/ 1mol CaO) X
  (74.10gCa(OH)2/1mol Ca(OH)2) 5.88g Ca(OH)2

Almost Done!
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How accurate Was my Experiment?

• To answer this question you
• need to know about the percent yield.
• Percent yield is essentially how much product you
  got from a chemical reaction vs. how much you
  should have gotten back according to your
  Stoichiometry calculation.

The Formula
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Percent Yield

• Percent Yield Actual Yield X100
• Theoretical Yield
• Actual Yield What you get from the experiment or
  the experimental value
• Theoretical Yield How much product the
  stoichiometric calculation says you should get,
  based on how much reactant you started with.

Lets see how to use This equation
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Heres An Example .

• When potassium chromate (K2CrO4) is added to a
  solution containing 0.500g silver nitrate
  (AgNO3), solid silver chromate (Ag2CrO4) is
  formed.
• A.) Determine the theoretical yield of silver
  chromate precipitate.
• B.) Determine the percent yield if 0.455g silver
  chromate was actually recovered.
• If you follow the steps find the theoretical
  yield first, then the calculation becomes quite
  simple.

Click Here if you Want to see the Problem Worked
Out
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Answer to The Example .

• 1.) As always, begin by writing a balanced
  chemical equation for the reaction
• K2CrO4 2AgNO3 ? Ag2CrO4 2KNO3
• 2.) Figure out the theoretical yield by
  converting 0.500g AgNO3 to grams of Ag2CrO4.
• (0.500gAgNO3/1) X (1mol AgNO3/169.91g AgNO3)
  0.00294mol AgNO3
• (0.00294mol AgNO3/1) X (1mol Ag2CrO4/ 2mol
  AgNO3) 0.00147mol Ag2CrO4
• (0.00147mol Ag2CrO4/1) X (331.8gAg2CrO4/ 1mol
  Ag2CrO4) 0.488g Ag2CrO4
• 3.) Now that you have the theoretical yield,
  0.488g Ag2CrO4, you can use the actual yield of
  0.455g Ag2CrO4 and find the percent yield
• (0.455gAg2CrO4/ 0.488g Ag2CrO4) X 100 93.2

Yeah!!!
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Great Job!!!!!!!

• Finally you have completed this learning module
  over Stoichiometry. My hope for you is that you
  learned something! ?
• The last leg of your journey is to complete a
  questionnaire over this Learning Module. You
  need to Print out the page, answer the questions,
  and then turn the sheet into me for credit.

Click Here to Proceed to the Evaluation
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The Evaluation

• You must take this evaluation in order to receive
  the extra credit for completing this Learning
  Module. Follow the evaluation link, print the
  page, answer the questions, and turn it into me!

The End Thanks for Stopping By
The Evaluation
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Correct!

• Great Job!!!

Moving On!
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Way To Go!!!!!!
Ready for Whats Next???
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References

• Thank You and I hope you enjoyed the Experience!
• Unless designated otherwise, all pictures are
  from clip art.
• Slide 1
• Scientist Picture
• (http//istockphoto.com/imageindex/310/7/310758/T
  he_Mad_Scientist.html)
• Eagle Picture
• (http//www.scott.kenton.kyschools.us/default1.ht
  m)
• Slide 5
• Mole Picture
• (http//www.ringwood.hants.sch.uk/school2/_Subjec
  ts/sciChemistry_ks3/SpencerNewSite/page_7E_acids.h
  tml)
• Resource Book Used
• Dingrando, Laurel, Kathleen V. Gregg, Nicholas
  Hainen, Phillip Lampe, Cynthia Roepcke, and
  Cheryl Wistrom. Chemistry Matter and Change. 1st
  ed. Vol. 1. Columbus Glencoe/ McGraw Hill, 2002.
  

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