Stoichiometry • APPLICATION: •We are applying concepts learned in Chapters 5, 6 & 7 » Naming/Formulas » Molar Conversions » Balancing Chemical Equations • PURPOSE: •To understand how chemical equations are used to express ratios of reactants and products 1
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Stoichiometry APPLICATION: We are applying concepts learned in Chapters 5, 6 & 7 » Naming/Formulas » Molar Conversions » Balancing Chemical Equations PURPOSE:
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Stoichiometry• APPLICATION:
•We are applying concepts learned in Chapters 5, 6 & 7
»Naming/Formulas»Molar Conversions»Balancing Chemical Equations
• PURPOSE:• To understand how chemical equations are used to express ratios of reactants and products
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Stoichiometry• Stoichiometry:– From Greek: – “Stoichio” = matter or chemicals– “Metry” = measure
=The calculation of quantities in a chemical reaction.
• Kinds of quantities measured:• Particles (atoms, molecules, formula units)• Moles• Mass• Volume
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StoichiometryWhy is it useful?
• To determine the mass, volume or number of particles needed for a particular reaction.• To determine the mass, volume or number
of particles produced by a particular reaction.
Who uses stoichiometry?• Chemists,• Chemical engineers,• Engineers,• Rocket scientists,• Environmental scientists….• + criminals making illegal drugs, terrorists making bombs…
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Stoichiometry• Why are balanced chemical equations
useful?• Look at the ratios of all reactants and products
• One molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia
»1:3:2 ratio
• What happens if you have 10 molecules of N2?
1 2N2(g) + H2(g)
NH3 (g) 310
30 20
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Stoichiometry• Mass:
• All balanced chemical reactions must follow the Law of Conservation of Mass
• 1 mol N2 = 28.0 g
• 3 mol H2 = 6.0 g
• 2 mol NH3 = 34.0 g
1 2N2(g) + H2(g)
NH3 (g) 3
Reactants = 34.0 g
Products = 34.0 g
MASS IS ALWAYS CONSERVED IN CHEMICAL REACTIONS!
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Stoichiometry• Volume:
• At STP (standard temperature and pressure); 1 mol of any gas = 22.4 L
• Is volume conserved?• 1 mol N2 = 22.4 L
• 3 mol H2 = 67.2 L (3 x 22.4)
• 2 mol NH3 = 44.8 L (2 x 22.4)
1 2N2(g) + H2(g)
NH3 (g) 3
Reactants = 89.6 L
Products = 44.8 L
VOLUME IS NOT USUALLY CONSERVED!
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Stoichiometry• Moles:
• The coefficients in a balanced chemical equation represent the number of moles of that substance
• Are # of Moles conserved?• 1 mol N2
• 3 mol H2
• 2 mol NH3
1 2N2(g) + H2(g)
NH3 (g) 3
Reactants = 4 mol
Products = 2 molMOLES ARE NOT USUALLY
CONSERVED!
KEY CONCEPT: Coefficients show relative numbers of moles involved
in a reaction = the molar ratio
Stoichiometry• Moles:
• All coefficients are related to the number of moles required for a reaction.
•One mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia
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1 2N2(g) + H2(g)
NH3 (g) 3
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Stoichiometry• Mole-Mole Calculations
• Balanced chemical equations are valuable because they shoe the relative ratio of reactants to products.• You can now relate the # of moles of a reactant(s) to find the # of moles of a product(s)
1 2N2(g) + H2(g)
NH3 (g) 3
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Stoichiometry• Sample Problem 1:
–How many moles of ammonia are produced when 1.2 moles of nitrogen reacts with hydrogen?
1 2N2(g) + H2(g)
NH3 (g) 3
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Stoichiometry• Example:
–Step 1:• Look at balanced chemical equation
» 1 mol of N2 produces 2 mol NH3
–Step 2:• Use a conversion factor
»Convert mol N2 to mol NH3
1 2N2(g) + H2(g)
NH3 (g) 3
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Stoichiometry• Example: How much NH3 can be
made from 1.2 mol N2?
1 2N2(g) + H2(g)
NH3 (g) 3
1.2 mol N2
mol N2
mol NH3
= 2.4 mol NH3
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Mole ratio
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Stoichiometry• Sample Problem 2:
– Calculate the number of moles of nitrogen required to make 9.54 mol of NH3.
1 2N2(g) + H2(g)
NH3 (g) 3
9.54 mol NH3
2 mol NH3
1 mol N2
= 4.77 mol N2
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Stoichiometry• Sample Problem 3:
– Calculate the number of moles of hydrogen required to make 9.54 mol of NH3.
1 2N2(g) + H2(g)
NH3 (g) 3
9.54 mol NH3
2 mol NH3
3 mol H2
= 14.3 mol H2
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Stoichiometry• Mass-Mass Calculations:–When scientist conduct a lab
experiment, they measure the amount of a particular substance using grams…not moles
– Remember, you can convert the mass of a substance into moles (1-step conversion)
Mass (grams) A moles A
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Stoichiometry• Mass-Mass Calculations:– Additionally, it is possible to convert
between masses of reactants and products.
– The mole ratio (obtained from the balanced chemical equation) is the key to convert from the mass of substance A to mass of substance B
– Remember, the balanced chemical equation provides the mole ratio!
mass A moles A moles B mass B
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Stoichiometry• Mass-Mass Example #1:
– Calculate the number of grams of NH3 produced by the reaction of 6.80 g of hydrogen with an excess of nitrogen. Use the balanced equation above.
1 2N2(g) + H2(g)
NH3 (g) 3
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Stoichiometry• Mass-Mass Example #1:
– Step 1: What are we calculating?• Grams H2 grams NH3
– Step 2: What is the path we need to take?
1 2N2(g) + H2(g)
NH3 (g) 3
mass H2 moles H2 moles NH3 mass NH3
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Stoichiometry• Mass-Mass Example #1:
– Step 3: Complete conversions
1 2N2(g) + H2(g)
NH3 (g) 3
6.80 g H2
2.0 g H2
1 mol H2
3 mol H2
2 mol NH3
1 mol NH3
17.0 g NH3
= 38.5 g NH3
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Stoichiometry• Mass-Mass Example #2:
• How many grams of nitrogen are needed to produce the 38.5 g of NH3 produced in the previous example?
1 2N2(g) + H2(g)
NH3 (g) 3
38.5 g NH3
17.0 g NH3
1 mol NH3
2 mol NH3
1 mol N2
1 mol N2
28.0 g N2
= 31.7 g N2
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Stoichiometry• Mass-Mass Example #3:
• What is the total amount of mass (in grams) of the reactants in the example balanced chemical equation?
Mass of Reactants = Mass of Products
Is mass conserved in a chemical reaction???
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Stoichiometry• Mass-Mass Example #3:
Mass of Reactants = Mass of Products
31.7 g N2 + ___ g H2 38.5 g NH3
ANSWER 38.5 – 31.7 = 6.8 g H2
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Stoichiometry• Other Stoichiometric Calculations:
– You can use the same steps as before to calculate the following…
»Mass-Volume»Volume-Volume»Particle-Mass
– The first step is to ALWAYS convert to moles
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Stoichiometry• Other Stoichiometric Calculations:
– Choose the appropriate road map…
particles A
mass A
volume A
particles B
mass B
volume B
moles A
moles B
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Limiting Reagent• Limiting Reagent
• Chemical equations are like recipes…» The reactants are combined to make the products
• Reactions take place based upon the mole ratios expressed in the balanced chemical equation
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Limiting Reagent• Limiting Reagent
• Limiting Reagent:–Limits how much product will form =
determines the amount of product that can be formed in a reaction
• Excess Reagent:–The substance(s) that is leftover because
they are in excess = there is more than enough to react with the limiting reagent
A reaction can only occur until the limiting reagent is used up!!