chapter 12 stoichiometry the study of the quantitative, or measurable, relationships that exist in...
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Chapter 12Chapter 12
StoichiometryStoichiometry
The study of the quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions.
StoichiometryStoichiometry
2 H2 H22 + O + O22 2 H 2 H22OO
Interpreting Balanced Chemical EquationsInterpreting Balanced Chemical Equations
• Based on the mole ratio
Mole RatioMole RatioMole RatioMole Ratio
• indicated by coefficients in a balanced equation
Molar ratio of H2 to H2O is 2:2
(Simplify 1:1)
Molar ratio of O2 to H2O is 1:2
2 H2 H22 + O + O22 2 H 2 H22OO
2 moles of hydrogen react with
1 mole of oxygen to produce 2
moles of water.
2 H2 H22 + O + O22 2 H 2 H22OO
In terms of molesIn terms of moles
4 grams of hydrogen react with
32 grams of oxygen to produce
36 grams of water.
2 H2 H22 + O + O22 2 H 2 H22OO
In terms of massIn terms of mass
2 hydrogen molecules react with 1 oxygen
molecule to produce 2 water molecules.
• Notice that the number of molecules is
NOT the same on each side of the arrow.
2 H2 H22 + O + O22 2 H 2 H22OO
In terms of moleculesIn terms of molecules
4 atoms of hydrogen react with 2
atoms of oxygen to produce 2
water molecules, which are 4
atoms of hydrogen and 2 atoms
of oxygen.
2 H2 H22 + O + O22 2 H 2 H22OO
In terms of atomsIn terms of atoms
44.8 L of hydrogen gas react with 22.4 L
of oxygen gas to produce 44.8 L of water
vapor.
• Notice that the number of liters of gas is
NOT the same on each side of the arrow.
2 H2 H22 + O + O22 2 H 2 H22OO
In terms of volumesIn terms of volumes
Law of Conservation of Matter & Law of Conservation of Matter & MassMassLaw of Conservation of Matter & Law of Conservation of Matter & MassMass
2 moles H2 react with 1 mole of
O2 to form 2 moles of H2O.
2 H2 H22 + O + O22 2 H 2 H22OO
HH22 OO22 HH22OO2 mol 1 mol 2 mol2 g
1 mol
32 g
1 mol
18 g
1 mol
4 g 32 g 36 g++ ==
VERIFIED!VERIFIED!VERIFIED!VERIFIED!
Lead will react Lead will react with hydrochloric with hydrochloric acid to produce acid to produce lead (II) chloride lead (II) chloride and hydrogen. and hydrogen. How many moles How many moles of hydrochloric of hydrochloric acid are needed acid are needed to completely to completely react with 0.36 react with 0.36 moles of lead?moles of lead?
Practice Problem #1Practice Problem #1
Oh, no … Oh, no … where do I where do I start?start?
Lead will react with hydrochloric acid Lead will react with hydrochloric acid to produce lead chloride and to produce lead chloride and hydrogen. hydrogen.
Write an equation and balance it.Write an equation and balance it.
Pb + HCl Pb + HCl PbCl PbCl22 + H + H2222
How many moles of How many moles of hydrochloric acidhydrochloric acid are needed to completely react with are needed to completely react with 0.36 moles of 0.36 moles of leadlead??
Determine Mole RatioDetermine Mole Ratio
Pb + HCl Pb + HCl PbCl PbCl22 + H + H2222
hydrochloric hydrochloric acidacid
leadlead
Coefficient = 1Coefficient = 1Coefficient = 2Coefficient = 2
““wanted”wanted”““given”given”
How many moles ofHow many moles of hydrochloric acidhydrochloric acid are needed to completely react with are needed to completely react with 0.36 moles of0.36 moles of leadlead??
Set up mole ratio “wanted” to “given”Set up mole ratio “wanted” to “given”
PbPb + + HClHCl PbCl PbCl22 + H + H2222
wantedwanted
leadlead 11
22hydrochloric acidhydrochloric acid
givengiven == ==
0.36 moles of0.36 moles of leadlead
X moles of HClX moles of HCl==
11
22
How many moles ofHow many moles of hydrochloric acidhydrochloric acid are needed to completely react with are needed to completely react with 0.36 moles of0.36 moles of leadlead??
Solve for Solve for “wanted”“wanted”
PbPb + + HClHCl PbCl PbCl22 + H + H2222
0.36 moles of0.36 moles of leadlead
X moles of HClX moles of HCl==
11
22
X = 2 X = 2 (0.36 moles)(0.36 moles)
X = 0.72 moles HClX = 0.72 moles HCl
Umm … It would Umm … It would require 0.72 moles require 0.72 moles of hydrochloric of hydrochloric acid. Is that right?acid. Is that right?
That’s Correct!That’s Correct!
Hem, hem –Hem, hem –you have 9 more you have 9 more problems … problems … ready, go.ready, go.
How to solve, given an amount of one substance
to another substance.
How to solve, given an amount of one substance
to another substance.
““Stoich” ProblemsStoich” Problems
Stoichiometry StepsStoichiometry StepsStoichiometry StepsStoichiometry Steps
1. Write a balanced equation.2. Identify “wanted” & “given”.3. Convert given information to moles.4. Determine Mole Ratio.
(Moles of “wanted” : Moles of “given”)
5. Calculate Moles of “wanted”.6. Convert to required units.
KEY step in all stoichiometry problems!
4. Determine Mole ratio (moles of “wanted” to moles of “given”)
What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide?
Mass-Mass ProblemsMass-Mass Problems
Help … I really need to know this right now!
What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide?
1. Write a balanced equation.1. Write a balanced equation.
Help … I really need to know this right now!
FF22 + 2KBr + 2KBr 2KF + Br 2KF + Br22
What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide?
2. Identify 2. Identify “wanted” “wanted” and and “given”“given”
Help … I really need to know this right now!
FF22 + + 2KBr2KBr 2KF + 2KF + BrBr22
““wanted”wanted”““given”given”
1.72 g of potassium bromide
3. Convert 3. Convert “given” “given” information to molesinformation to moles
Help … I really need to know this right now!
Molar Mass of KBr = 39 + 80 = 119 g/mol
1.72 g
119 g
1 mol= 0.01445 mol KBr
4. Determine Mole Ratio4. Determine Mole Ratio
Help … I really need to know this right now!
FF22 + + 2KBr2KBr 2KF + 2KF + BrBr22
““wanted”wanted”““given”given”
Coefficients tell Mole Ratio Coefficients tell Mole Ratio ““wanted” wanted” toto “given”“given”
11 :: 22
5. Calculate Moles of 5. Calculate Moles of “wanted”“wanted”
Help … I really need to know this right now!
FF22 + + 2KBr2KBr 2KF + 2KF + BrBr22
Moles KBrMoles KBr
Moles BrMoles Br22 11==
22
0.01445 mol KBr==
XX 11
22
2 2 X = X = 0.01445 mol X = X = 0.007225 mol
6. Convert to required units6. Convert to required units
X = X = 0.007225 mol BrBr22
0.007225 mol
Molar Mass BrMolar Mass Br22 80 + 80 = 160 g/mol80 + 80 = 160 g/mol
160 g160 g
1 mol1 mol== 1.16 g Br1.16 g Br22
StoichiometryStoichiometry
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Available IngredientsAvailable Ingredients• 4 slices of bread• 1 jar of peanut butter• 1/2 jar of jelly
Limiting ReactantLimiting Reactant• bread
Excess ReactantsExcess Reactants• peanut butter and jelly
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Available IngredientsAvailable Ingredients• Copper Wire• 0.5 g AgNO3
Limiting ReactantLimiting Reactant• 0.5 grams AgNO3
Excess ReactantsExcess Reactants• Copper Wire
Limiting ReactantLimiting ReactantLimiting ReactantLimiting Reactant
The reactant that limits the The reactant that limits the amount of product that can be amount of product that can be formed.formed.
When quantities of reactants are When quantities of reactants are available in the exact ratio available in the exact ratio described by the balanced described by the balanced equation, they are said to be in equation, they are said to be in Stoichiometric proportions.Stoichiometric proportions.
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Limiting ReactantLimiting Reactant• used up in a reaction• determines the amount of all
products formed
Excess ReactantExcess Reactant• added to ensure that the other
reactant is completely used up• usually cheaper & easier to recycle
Solving Problems – Limiting Solving Problems – Limiting ReactantsReactantsSolving Problems – Limiting Solving Problems – Limiting ReactantsReactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. The reactant that produces the smaller amount of product is the limiting reactant.
Very similar to mass-mass problems!
Step 1: Write a balanced Step 1: Write a balanced equation.equation.Step 1: Write a balanced Step 1: Write a balanced equation.equation.
Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water.
O2 + 2H2 2 H2O
Step 2:Step 2:Step 2:Step 2:
Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water.
O2 + 2H2 2 H2O
For each reactant,
calculate the amount of
product formed.
Step 2: Step 2: Step 2: Step 2:
1.22 g oxygen
O2 + 2H2 2 H2O
wanted
given
wanted
=X
0.038 mol
2
1
given
X = 0.076 mol H2O
32 g
1 mole= 0.038 mol O2
Step 2: Step 2: Step 2: Step 2:
1.05 g H2
O2 + 2H2 2 H2O
wanted
given
wanted
=X
0.525 mol
2
2
given
X = 0.525 mol H2O
2 g
1 mole= 0.525 mol H2
Step 3:Step 3:Step 3:Step 3:
1.22 g of O2 would produce 0.0763 mol H2O
The one that produces the smallest amount is your
limiting reactant.
1.05 g of H2 would produce .525 mol H2O
Oxygen is your limiting reactant!
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.
Step 1: Write a balanced Step 1: Write a balanced equation.equation.Step 1: Write a balanced Step 1: Write a balanced equation.equation.
Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.
2Na + Cl2 2NaCl
Step 2:Step 2:Step 2:Step 2:
Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.
2Na + Cl2 2NaCl
For each reactant,
calculate the amount of
product formed.
Step 2: Step 2: Step 2: Step 2:
1.7 g Na
2Na + Cl2 2NaCl
wanted
given
wanted
=X
0.0739 mol
2
2
given
X = 0.0739 mol NaCl
23 g
1 mole = 0.0739 mol Na
Step 2: Step 2: Step 2: Step 2:
2.6 L Cl2
2Na + Cl2 2NaCl
wanted
given
wanted
=X
0.116 mol
2
1
given
X = 0.232 mol NaCl
22.4 L
1 mole = 0.116 mol Cl2
Step 3:Step 3:Step 3:Step 3:
1.7 g Na would produce 0.0739 mol NaCl
The one that produces the smallest amount is your
limiting reactant.
2.6 L Cl2 would produce 0.232 mol NaCl
Sodium is your limiting reactant!
Click on the real player file called Sodium_Chlorine_2 to see a demo of this reaction
Percent YieldPercent YieldPercent YieldPercent Yield
100yield ltheoretica
yield actualyield %
calculated on paper
measured in lab
Percent YieldPercent YieldPercent YieldPercent Yield
When 45.8 g of K2CO3 react with
excess HCl, 46.3 g of KCl are formed. Calculate the theoretical yield and % yield of KCl.
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Percent YieldPercent YieldPercent YieldPercent Yield
45.8 gK2CO3
1 molK2CO3
138 gK2CO3
= 49.1g KCl
2 molKCl
1 molK2CO3
74g KCl
1 molKCl
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Theoretical Yield:
Percent YieldPercent YieldPercent YieldPercent Yield
Theoretical Yield = 49.1 g KCl
% Yield =46.3 g
49.1 g 100 = 94.3%
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.1 g
actual: 46.3 g