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Chapter 12 - Stoichiometry. “SUPER DIMENSIONAL ANALYSIS”. Review of moles…. 1 mole = 6.022 x 10 23 particles Molar mass =. ( ). X grams. X = molar mass of substance. 1 mole. Use atomic mass. Calculate molar mass of CaBr 2. Ca:. x 1. = 40.1 g/mol. 40.1. Br:. x 2. - PowerPoint PPT Presentation
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Chapter 12 - Stoichiometry
“SUPER
DIMENSIONAL
ANALYSIS”
Review of moles…..
• 1 mole = 6.022 x 1023 particles
• Molar mass = X grams
1 moleX = molar mass of substance( )
Use atomic mass
Calculate molar mass of CaBr2
Ca: 40.1 x 1 = 40.1 g/mol
Br: 79.9 x 2 = 159.8 g/mol
= 199.9 g/mol
What is mass of 0.250 mol CO2?
C: 12 x 1 = 12 g/mol
O: 16 x 2 = 32 g/mol
= 44 g/mol
0.250 mol CO2
1 mol CO2
44 g CO2 = 11 g CO2
Stoichiometry
• Defn – study of mass relationships between reactants and products in a chemical reaction
• What does this mean?– How much of something (products) can be
made?– How much of starting materials were there?
Stoichiometry
You must have a _____________________
to do stoichiometry calculations.
balanced chemical reaction
To make a basic cheeseburger:
2 buns + 1 meat patty + 1 cheese
1 cheeseburger
1 c-burger
2 buns
1 c-burger
1 meat patty
1 c-burger
1 cheese
1 meat patty
2 buns
1 cheese
1 meat patty
2 buns
1 cheese
Mole Ratio
• Defn – ratio btwn # of moles of any two substances in a balanced chem rxn
Ex reaction
2 Al + 3 Br2 2 AlBr3
• give all the mole ratios
2 mol Al
3 mol Br2
2 mol Al
2 mol AlBr3
2 mol Al
3 mol Br2
3 mol Br2
2 mol AlBr3
2 mol Al
2 mol AlBr3
3 mol Br2
2 mol AlBr3
4 steps to a basic stoichiometry problem
1) Balance the equation
2) Identify the given, convert to moles
3) Identify unknown, do a mole to mole ratio between given and unknown (KEY STEP)
4) Convert unknown to unit specified in problem
Stoichiometry Flow Chart
grams A grams Bmoles A moles B
MolarMass A
MolarMass B
MoleRatio
BA
examples
• How many moles of H2 are formed when 65 g HCl is used?
• If 3.7 mol KBr reacts with calcium, how many moles of CaBr2 are formed?
A
A
B
B
Mole A to Mole B Conversions
grams Bmoles A moles B
MolarMass A
MolarMass B
MoleRatio
BA
grams A
__ C3H8 + __ O2 __ CO2 + __ H2O
• How many moles of CO2 are produced when 10.0 moles of O2 is used?
1 45 3
10.0 mol ? mol
10.0 mol O2
MoleRatio
BA
5 mol O2
3 mol CO2
A B
= 6 mol CO2
Mole A to Mole B Conversions
Mole A to Mass B Conversion
grams Bmoles A moles B
MolarMass A
MolarMass B
MoleRatio
BA
grams A
___ Na + ___ Cl2 ___ NaCl
• How many grams of sodium chloride is formed when 1.25 moles of sodium react w/ chlorine gas?
Mole A to Mass B Conversion
2 21
1.25 mol ? g
1.25 mol Na 2 mol NaCl
2 mol Na 1 mol NaCl
58.5 g NaCl
AB
MoleRatio
BA
MolarMass B
= 73.1 g NaCl
Mass A to Mole B Conversion
grams Bmoles A moles B
MolarMass A
MolarMass B
MoleRatio
BA
grams A
___ Na + ___ Cl2 ___ NaCl
• How many moles of chlorine gas is needed to make 50 g NaCl?
Mass A to Mole B Conversion
2 21
50 g? mol
50 g NaCl
58.5 g NaCl
1 mol NaCl
2 mol NaCl
1 mol Cl2
AB
MoleRatio
BA
MolarMass A
= 0.43 mol Cl2
Mass A to Mass B Conversion
grams Bmoles A moles B
MolarMass A
MolarMass B
MoleRatio
BA
grams A
___NH4NO3 ___ N2O + ___ H2O
• Determine the mass of water formed from decomposition of 25.0 g NH4NO3
Mass A to Mass B Conversion
1 1 2
25 g ? g
25 g NH4NO3
80 g NH4NO3
1 mol NH4NO3
1 mol NH4NO3
2 mol H2O
1 mol H2O
18 g H2O
= 11.25 g H2O
A B
Limiting Reactant
• In basic stoichiometry problems, you are provided with one given quantity. In LR problems, you are given both reactants. Before you can solve the problem, you have to determine which of the two given quantities to use as your given.
Limiting Reactant
You need to choose the one that will run out first, known as the __________________. It controls how much product you can make. The other reactant is known as the _________________ because there will be some left over.
limiting reactant
excess reactant
1 cheese + 2 bread slices 1 cheese sandwich
Given: 10 pieces of cheese, 50 bread slices
How many cheese sandwiches can you make?
What is the limiting reactant?What is the excess reactant?
How many pieces of excess reactant used?How many pieces of excess reactant left over?
10
20
30
cheesebread
How to find limiting reactant
1) Convert both amounts of reactants to moles
2) Divide the mole amount of each reactant by its coefficient in the balanced equation
3) Compare two numbers. The one that is smaller is the limiting reactant. Other one is excess reactant.
____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3
Find the limiting reactant if 6.9 g Al and 0.35 mol CuCl2are available.
Reactant #1: Reactant #2:
6.9 g Al 1 mol Al
27 g Al
= 0.256 mol Al = 0.35 mol CuCl2
1) Convert both amounts of reactants to moles.
____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3
Reactant #1: Reactant #2:
0.256 mol Al 0.35 mol CuCl2
2 3
2) Divide each mole amount by its coefficient inthe balanced equation
= 0.128 = 0.117
CuCl2 is the limiting reactant
____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3
Find the limiting reactant if 6.2 g Al and 48.5 g CuCl2are available.
Reactant #1: Reactant #2:
6.2 g Al 1 mol Al
27 g Al
= 0.230 mol Al = 0.360 mol CuCl2
1) Convert both amounts of reactants to moles.
48.5 g CuCl2 1 mol CuCl2134.5 g CuCl2
____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3
Reactant #1: Reactant #2:
0.230 mol Al 0.360 mol CuCl2
2 3
2) Divide each mole amount by its coefficient inthe balanced equation
= 0.115 = 0.120
Al is the LRCuCl2 is excess reactant
Based on the LR in #4, how many grams of copper will be produced?
1 mol Al
2 mol Al
3 mol Cu
1 mol Cu
63.5 g Cu
____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3
= 21.9 g Cu
27 g Al
6.2 g Al
Percent Yield
• You buy a 500 g ketchup bottle, do you ever use all 500 g?
No. Some is still left on the sides youcannot retrieve. Not all 100% is used.
Percent Yield
• Theoretical Yield – max amount of product that can be produced (what you expect to get)
• Actual Yield – amt of product you actually produced (always less than theoretical)
• Percent Yield =actual
theoreticalx 100
Ex problem #1
• Joe does experiment to form carbon dioxide. The maximum he can obtain is 34.5 grams. He performs the experiment and only obtains 18.6 grams. What is his percent yield?
AT
=18.6 g34.5 g
X 100 = 53.9%
Ex problem #2
• What is the theoretical yield of Ag2CrO4 if 0.500 g AgNO3 reacts with excess K2CrO4?
__K2CrO4 + __ AgNO3 __KNO3 + __Ag2CrO42 21 1
0.500 g ? g
0.5 g K2CrO4
170 g K2CrO4
1 mol K2CrO4
1 mol K2CrO4
1 mol Ag2CrO4
1 mol Ag2CrO4
331.7 g
= 0.49 g Ag2CrO4
Ex problem #2
• If 0.455 g of Ag2CrO4 is produced, what is the percent yield?
AT
=0.455 g0.49 g
X 100 = 93.2%