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12.1 STOICHIOMETRY Deals with amounts of reactants
used & products formed.
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What is Stoichiometry?
• The study of the relationship between amounts of reactants used and products formed in a chemical reaction
• Based on the Law of Conservation of Mass and Matter– Matter is neither created nor destroyed– Mass of reactants equals the mass of the
products
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MOLE-MASS RELATIONSHIPS
• Balanced equation: 4Fe (s) + 3O2(g) 2Fe2O3(s)
• Interpret: 4 atoms 3 molecules 2 formula units (Particles)
• Mole ratio: 4 moles 3 moles 2 moles (Coefficients)
Note: Particles of ionic compounds are called “formula units”
Calculate the mass of each reactant and product by multiplying the number of moles by the molar mass
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SHOW MASS IS CONSERVED4Fe (s) + 3O2(g) 2Fe2O3(s)
• Mass reactants:
• 4 mol Fe (55.8g Fe) = 223.2 g Fe
(1 mole Fe)
• 3 mol O2 (32.00 g O2) = + 96.0 g O2
(1 mol O2) ____________
• 319.2 g Total
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MASS OF PRODUCTS:4Fe (s) + 3O2(g) 2Fe2O3(s)
• 2 mol Fe2O3 (159.6 g Fe2O3) = 319.2 g (1 mol Fe2O3)
• Equals the mass of the reactants
(319.2g)
• Law of Conservation of Matter
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PROBLEM
• Interpret in terms of particles, moles, and mass. Show that mass is conserved:
• (Hint: look at coefficients for particles & moles)
4 Al + 3O2 2 Al2O3
• Particles:
• Moles:
• Mass:
• Conserved?
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SOLUTION: 4Al + 3O2 2 Al2O3
• Particles: 4 molecules 3 molecule 2 molecule
• Moles: 4 mole 3 mole 2 moles
• Mass: 4 (27.0 g) + 3 (26.0 g) = 2 (102.0 g)
• Conserved? 204.0 g = 204.0 g YES!
• Law of Conservation of Matter shown.
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MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced
chemical equation
• There are six mole ratios for the following:
• Ex. 4 Al(s) + 3 O2(g) 2 Al2O3(s)
• Note: 4 moles 3 moles 2 moles
• 4 mol Al and 3mol O2
3 mol O2 4 mol Al
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MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced
chemical equation
• 4 mol Al and 2 mol Al2O3
2 mol Al2O3 4 mol Al
• 3 mol O2 and 2 mol Al2O3
2 mol Al2O3 3 mol O2
• All stoichiometry calculations begin with a balanced equation and mole ratios!!
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PROBLEM - FIND MOLE RATIOS FOR:
• 2 NH3 N2 + 3 H2
• 2 mol 1 mol 3 mol
• Ans. 2 mol NH3 or 1 mol N2
1 mol N2 2 mol NH3
• 2 mol NH3 or 3 mol H2
3 mol H2 2 mol NH3
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Ans. 2 NH3 N2 + 3 H2
• 1 mole N2 or 3 mole H2
3 mole H2 1 mole N2
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12.2 STOICHIOMETRIC CALCULATIONS
There are 3 Basic Stoichiometry Calculations
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1. Mole to Mole Conversions
A piece of magnesium burns in the presence
of oxygen forming magnesium oxide (MgO).
How many moles of oxygen are needed to produce 12 moles of magnesium oxide?
Step 1: Write a balanced equation• 2 Mg (s) + O2 (g) 2 MgO (s)
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Write mole ratios
2
2
1 mol O 2 mol Mgand
2 mol Mg 1 mol O
Choose the correct mole ratio needed for this problem
21 mol O
2 mol MgO
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Mole to Mole Conversion cont’d
Multiply the known number of moles of MgO by the mole ratio
6 mols of oxygen is needed to produce 12 mols of magnesium oxide
22
12 mol MgO 1 mol O= 6 mol O
2 mol MgO
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2. Mole to Mass Conversions
• The following reaction occurs in plants undergoing photosynthesis
• CO2(g) + H2O(l) C6H12O6(s)+ O2(g)
• How many grams of glucose (C6H12O6) are produced when 24.0 mols CO2 reacts in excess water?
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Mole to Mass Conversion cont’d Write a balanced equation
• 6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g)
Use mole ratios to determine the number of moles of glucose produced by the given amount of carbon dioxide
6 12 626 12 6
2
1 mol C H O24 mol CO4.00 mol C H O
6 mol CO
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Mole to Mass Conversion cont’d
Multiply by the molar mass
721 g glucose is produced from 24.0 moles carbon dioxide
6 12 6 6 12 66 12 6
6 12 6
4 mol C H O 180.2g C H O721g C H O
1 mol C H O
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3. Mass to Mass ProblemsThe only new step is first step:
Convert grams of given substance to moles!
Massgiven Molesgiv Molesdesired Massdes
• ÷ molar massgiven X mole ratio X molar massdesired
• 3 step problem!
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Label: 25.0 g ? g
NH4NO3 N2O + 2H2O Mole Ratio: 1 mol 1 mol 2 mol
• Ammonium nitrate (NH4NO3) produces N2O gas and H2O when it decomposes. Determine the mass of water produced from the decomposition of 25.0 g of NH4NO3.
• 1) Find moles NH4NO3 (molar mass): 80.04 g/mol
• Use the inverse of the molar mass to convert grams of NH4NO3 to moles of NH4NO3
4 3 4 3
4 3
25.0g NH NO 1 mol NH NO 0.312 mol
80.04g NH NO
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Label: 25.0 g ? g
NH4NO3 N2O + 2H2OMole ratio: 1 mol 1 mol 2 mol
• Determine the mole ratio of mol H2O to mol NH4NO3 from the chemical equation. The desired substance is the numerator.
• Multiply mol NH4NO3 by the mole ratio.
• Calculate the mass of H2O using the molar mass.
2
4 3
2 mol H O
1 mol NH NO
4 3 22
4 3
0.312 mol NH NO 2 mol H O = 0.624 mol H O
1 mol NH NO
2 22
2
0.624 mol H O 18.02g H O = 11.2 g H O
1 mol H O
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Label: 5.6g ? g
PROBLEM: N2 + 3H2 2 NH3
• If 5.6 g nitrogen reacts completely with hydrogen, what mass of ammonia is formed?
• 1) Find moles of nitrogen (given) – molar mass:
• 5.6 g N2|_____ mass N2 = 2 x 14.0 = 28.0 g/mol
• 1 | g N2
• 5.6 mol N2|_1mol N2
• 1 | 28.0 g N2
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Label: 5.6 g ? g N2 + 3H2 2 NH3
Mole Ratio: 1mol 3mol 2mol
• 2) Find moles ammonia (desired) - mole ratio:
• 5.6 g N2 |1mol N2 |_2 mol NH3___• 1 | 28 g N2 | 1 mol N2
• 3) Find grams ammonia (desired) – • molar mass: NH3 = 14 + 3 = 17 g/mole
• 5.6 g N2 |1mol N2|_2 mol NH3 | 17.0 g NH3
• 1 | 28g N2 | 1 mol N2 | 1 mol NH3
• given ÷ MM given mole ratio x MM desired
• = 6.8 g NH3
• Mass H2 = 6.8 g – 5.6 g = 1.2 g
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LIMITING REACTANTS
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Limiting and Excess Reactants• When a chemical reaction occurs, the reactants
are not always present in the exact ratio indicated by the balanced equation.
• What usually happens is that a chemical reaction will run until the reactant that is in short supply is used up.
Which reactant will be used up first?
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What is a Limiting Reactant?
• A limiting reactant is the reactant that limits (stops) a reaction and determines the amount of product
• An excess reactant is any reactant that is left over after the reaction stops (all reactants except the limiting reactant)
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IN ORDER FOR CHEMICAL REACTION TO OCCUR, YOU MUST HAVE A COMPLETE
SET OF REACTANTS (REAGENTS):
• You don’t always have the exact amounts.• Ex. Let’s make some McBurgers!!!
• Ingredients:– 2 buns; 1 beef patty; 1cheese
slice; 1 tomato slice; 1 lettuce leaf; 3 pickles
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YOU HAVE AVAILABLE:
• 6 buns• 3 burger patties• 5 cheese slices• 6 tomato slices• 5 lettuce leafs• 6 pickles
RECIPE CALLS
FOR: 2 buns
1 burger patty
1 cheese slice
1 tomato slice
1 lettuce leaf
3 pickles
How many McBurgers can you make?
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Based on the individual ingredients you have available:
You have enough:• Buns for (3)
• Burger Patties for (3)• Cheese Slices for (5)• Tomato Slices for (6)• Lettuce Leafs for (5)• Pickles for (2)
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The Limiting Reactant (LR) for the McBurger making process in the _______!
• This is the ingredient we ran out of first and were unable to continue with the McBurger making process
• We were only able to make 2 McBurger from the 6 pickles we had available to use–(Remember each McBurger requires
3 pickles)
pickles!
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Excess reactants (XR) are the ingredients not used in the McBurger making
process:• 2 buns
• 1 burger patty
• 3 cheese slices
• 4 tomato slices
• 3 lettuce leafs
• Note: You made 2 McBurgers
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N2 + 3H2 2 NH3
Using the above equation you are given 3.0 mols N2 and 5.0 mols H2
• Determine the limiting reactant• Determine the excess reactant
• Determine the amount of NH3 produced
PRACTICE PROBLEM:
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• Draw 2 dimensional analysis problems. Smaller product is your answer. (Always use L.R. to find answer!)
3.0 mol N2 x 2 mol NH3 = 6.0 mol NH3
1 mol N2
5.0 mol H2 x 2 mol NH3 = 3.3 mol NH3
3 mol H2 .
Label: 3.0mol 5.0 mol ?mol N2 + 3H2 2 NH3
M.R. 1 mol 3 mol 2 mol
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ANSWER: 3.3 mol ammonia formed. Hydrogen (H2) is L.R. Nitrogen (N2) is the XS.
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IF A PAPER BURNS IN A ROOM:
• What is the limiting reactant?
• What is in excess?
• What would happen if the paper burned in a closed jar?
• LR?• XS?
+ +
+ + +
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4.0mol 7.0mol
PROBLEM: 2 Al + 3 Cl2 2 AlCl3
• Given 4.0 mol Al and 7.0 mol Cl2, what is the maximum amount of aluminum chloride formed?
• Step 1: 2 set ups:
• 4.0 mol Al _____________ mol Al
• 7.0 mol Cl2 ________
mol Cl2
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Label: 4.0mol 7.0 mol ? Mol
2 Al + 3 Cl2 2 AlCl3
• 4.0 mol Al 2 mol AlCl3 = 4.0 mol AlCl3
2 mol Al
• 7.0 mol Cl2 2 mol AlCl3 = 4.7 mol AlCl3 3 mol Cl2
• Which is the limiting reactant?• *Al is L.R. Ans. 4.0 mol AlCl3
• *Cl2 is XS
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THE END