chemical reaction balancing
TRANSCRIPT
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Balancing EquationStrategy
Balance elements that occur in only onecompound on each side first.
Balance free elements last.
Balance unchanged polyatomics asgroups.
Fractional coefficients are acceptable and
can be cleared at the end by multiplication.
Please practice on your own.
Tons of exercises are available at
the end of the chapter
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FeSO4 Fe2O3 + SO2 + O2
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STOICHIOMETRY- the study of the
quantitativeaspects ofchemicalreactions.
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PROBLEM: If 454 g of NH4NO3decomposes,how much N
2
O and H2
O are formed? What isthe theoretical yield of products?
STEP 1
Write the balanced chemicalequation
NH4
NO3
N2
O + H2
O
2
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454 g of NH4NO3 N2O + 2 H2O
STEP 2 Convert mass reactant(454 g) moles
STEP 3 Convert moles reactant (5.68 mol) molesproduct
Relate moles NH4NO3to moles product
expected: 1 mol NH4NO3 2 mol H2O
2 mol H2O made
1 mol NH4NO2
Express as STOICHIOMETRIC FACTOR
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454 g of NH4NO3 N2O + 2 H2O
= 11.4 mol H2O produced
STEP 3 Convert moles reactant (5.68 mol)
moles product
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454 g of NH4NO3 N2O + 2 H2O
STEP 4 Convert moles product (11.4mol) mass product
This is called the THEORETICALYIELD
ALWAYS FOLLOW THESE STEPS INSOLVING STOICHIOMETRY
PROBLEMS!
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454 g of NH4NO3 N2O + 2 H2O
STEP 5 How much N2O is formed?
Total mass of reactants =
total mass of products
454 g NH4NO3= ___ g N2O + 204 g H2O
mass of N2O = 250. g
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454 g of NH4NO3 N2O + 2 H2O
When you decomposed 454 g NH4NO3you
obtained 131g of N2O. What is the percent
yield of N2O?
This compares the theoretical (250. g)
and actual (131 g) yields.
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454 g of NH4NO3 N2O + 2 H2O
Calculate the percent yield
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GENERAL PLAN FORSTOICHIOMETRY
CALCULATIONS
Mass
reactant
Stoichiometric
factorMolesreactant
Molesproduct
Mass
product
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Example
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Al + HCl ! AlCl3+ H2
Write and Balance the Chemical
Equation:
Example Experimental Setup
2 6 2 3
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2 Al + 6 HCl ! 2 AlCl3+ 3 H2
Example
Plan the strategy:
cm3alloy!g alloy!g Al!mole Al!mol H2!g H2
density % A.M. stoich. M.W.
We need 5 conversion factors!
! !
Write the Equation
mH2= 0.691 cm3alloy ! ! !2.85 g alloy
1 cm3
93.7 g Al100 g alloy
1 mol Al26.98 g Al
3 mol H22 mol Al
2.016 g H21 mol H2
= 0.207 g H2
and Calculate:
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2 Al + 6 HCl ! 2 AlCl3+ 3 H2
Example
Plan the strategy:
cm3alloy!g alloy!g Al!mole Al!mol H2!g H2
density % A.M. stoich. M.W.
We need 5 conversion factors!
! !
Write the Equation
mH2= 0.691 cm3alloy ! ! !2.85 g alloy
1 cm3
93.7 g Al100 g alloy
1 mol Al26.98 g Al
3 mol H22 mol Al
2.016 g H21 mol H2
= 0.207 g H2
and Calculate:
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Trouble in understanding?
Try this then:
For the reaction A B 1 mole A will give 1 mole B
For the reaction A 3B 1 mole A will give 3 mole B
For the reaction 2A B 1 mole A will give 1/2 mole B
For the reaction 2A 3B 1 mole A will give 3/2 mole B
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Close contact between reagents isnecessary for chemical reaction to occur
can be achieved by using solutions
Solution: solute dissolved in solvent. Solute: present in smallest amount.
Water as solvent = aqueous solutions.
The amount of solute in a solution is
given by itsconcentration.
Molarity: Moles of solute per liter of
solution.
Concentrations of Solutions
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Molarity
Concentrations of Solutions
If we know molarity and liters of
solution, we can calculate moles
(and mass) of solute.
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Calculating the Mass of Solute in a Solution of Known
Molarity.
We want to prepare exactly 0.2500 L (250 mL) of an 0.250
M K2CrO4solution in water. What mass of K2CrO4should
we use?
Plan strategy:
Example
Volume!moles!mass
We need 2 conversion factors!
Write equation and
calculate:
mK2CrO
4= 0.2500 L! ! = 12.1 g0.250 mol
1.00 L194.02 g
1.00 mol
mass = V x M x Mw =
moles/V = M moles = MV
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Introduction to Reactions inAqueous SolutionsChapter 5
This Chapter is a general overview to thetypes of reactions that we will see for the
remainder of the course
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IONIC COMPOUNDS
Compounds in Aqueous Solution
Many reactions involve ionic compounds, especially
reactions in water aqueous solutions.
.
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Aqueous Solutions
How do we know ions are present in aqueoussolutions?
The solutions conduct electricity!
They are called ELECTROLYTES
HCl, KMnO4, MgCl2, and NaCl are strongelectrolytes.They dissociate completely
(or nearly so) into ions.
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General Properties of Aqueous
Solutions
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Aqueous Solutions
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Representation of Electrolytesusing Chemical Equations
MgCl2(s)!Mg2+(aq) + 2 Cl-(aq)
A strong electrolyte:
CH3OH(aq)
A non-electrolyte:
A weak electrolyte:
CH3CO2H(aq) CH3CO2-(aq) + H+(aq)
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The Nature of the Solutions(The example of KMnO4)
KMnO4(aq)!K+(aq) + MnO4-(aq)
If you make a solution that is 0.30 M in KMnO4, thismeans that
[K+] = [MnO4-] = 0.30 M
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The Nature of a Na2CO3Solution
This water-soluble compound is ionic
Na2CO3(aq)!2 Na+(aq) + CO3
2-(aq)
If [Na2CO3] = 0.100 M, then
[Na+] = 0.200 M
[CO32-] = 0.100 M
3 Na2CO3
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Step 1: Calculate moles of acid required.
(0.0500 mol/L)(0.250 L) = 0.0125 mol
Step 2: Calculate mass of acid required.
(0.0125 mol )(90.00 g/mol) = 1.13 g
USING MOLARITY
moles = M V
What mass of oxalic acid, H2C2O4,isrequired to make 250 mL of a 0.0500 Msolution?
Conc (M) = moles/volume = mol/V
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Dilution
We recognize that the number of moles are the
same in dilute and concentrated solutions.
So we can dilute a concentrated solution to get onethat is less concentrated:
MdiluteVdilute= moles = MconcentratedVconcentrated
Concentrations of Solutions
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PROBLEM: You have 50.0 mL of 3.0 MNaOH and you want 0.50 M NaOH. Whatdo you do?
Add water to the 3.0 M solution to lowerits concentration to 0.50 M
Dilute the solution!
How much water is added?
The important point is that!
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution
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PROBLEM: You have 50.0 mL of 3.0 MNaOH and you want 0.50 M NaOH. Whatdo you do?
Moles of NaOH in original solution =
M V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Therefore, moles of NaOH in final solution must
also equal 0.15 mol NaOH
(0.15 mol NaOH) / (0.50 mol) = 0.30 L
or 300 mL = volume of final solution
add enough waterto make the initial 50.0 mL of
3.0 M NaOH to a final volume of 300 mL and the
final molarity will be 0.50 M NaOH.
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Minitial Vinitial = Mfinal Vfinal
Preparing Solutions byDilution
A shortcut
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A particular analytical chemistry procedure requires
0.0100 M K2CrO4. What volume of 0.250 M K2CrO4should we use to prepare 0.250 L of 0.0100 M K2CrO4?
Calculate:
VK2CrO4= 0.2500 L " " = 0.0100 L0.0100 mol
1.00 L1.000 L0.250 mol
Preparing a solution by dilution.
Plan strategy: Mf= MiViVf
Vi= VfMfMi
Minitial Vinitial = Mfinal Vfinal
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Zinc reacts with acidsto produce H2gas.
If you have 10.0 g ofZn, what volume of2.50 M HCl isneeded to convert
the Zn completely?
SOLUTION STOICHIOMETRYAn Example
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Step 1: Write the balanced equation
Zn(s) + 2 HCl(aq) ZnCl2
(aq) + H2
(g)
Step 2: Calculate moles of Zn
Zinc reacts with acids to produce H2gas. If youhave 10.0 g of Zn, what volume of 2.50 M HCl isneeded to convert the Zn completely?
2 mol HCl/1 mol Zn
Step 3: Use the stoichiometric factor tocalculate moles of HCl
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Step 3: Use the stoichiometric factor tocalculate moles of HCl
Zinc reacts with acids to produce H2gas. If youhave 10.0 g of Zn, what volume of 2.50 M HCl isneeded to convert the Zn completely?
Step 4: Calculate volume of HCl required
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Reactions Involving aLIMITING REACTANT
Definition: In a given reaction, there is
not enough of one reagent to use up
the other reagent completely.
The reagent in short supply LIMITSthe quantity of product that can be
formed.
The stoichiometric coefficients are
used to determine the limiting reagent
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Rxn 1 Rxn 2 Rxn 3
mass Zn (g) 7.00 3.27 1.31
LIMITING REACTANTS
React solid Zn with 0.100mol HCl (aq)
The reaction:
Zn + 2 HCl ZnCl2 + H2
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Rxn 1 Rxn 2 Rxn 3
mass Zn (g) 7.00 3.27 1.31mol Zn 0.107 0.050 0.020
mol HCl 0.100 0.100 0.100
mol HCl/mol Zn 0.93 2.00 5.00
LIMITING REACTANTSReact solid Zn with 0.100
mol HCl (aq)
The reaction:
Zn + 2 HCl ZnCl2 + H2
Limiting reagent determines reaction yield
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Reaction to be Studied
2 Al + 3 Cl2 Al2Cl6
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PROBLEM: Mix 5.40 g of Al with 8.10 g ofCl2. How many grams of Al2Cl6can form?
Massreactant
StoichiometricfactorMoles
reactantMolesproduct
Massproduct
Convert lab units into chemical units and then back
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2 Al + 3 Cl2 Al2Cl6
The ideal mole ration of reactants is:
Mole Cl2/ mole Al = 3/2
Step 1 of LR problem:
compare actual mole ratio
of reactants to theoreticalmole ratio.
D t i i th Li iti
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Determining the LimitingReactant
If
then there is not enough Al to
use up all the Cl2, and the
limiting reagent is
Al
2 Al + 3 Cl2 Al2Cl6
D t i i th Li iti
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If
then there is not enough Cl2to
use up all the Al, and the limiting
reagent is Cl2
2 Al + 3 Cl2 Al2Cl6
Determining the LimitingReactant
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We have 5.40 g of Al and 8.10 g of Cl2
Step 2 of LR problem: Calculatemoles of each reactant
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Find mole ratio of reactants
Thisshould be 3/2 or 1.5/1 if
reactants are present in the
exact stoichiometric ratio.
Limiting reagent is Cl2
Mix 5 40 g of Al with 8 10 g of Cl
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Mix 5.40 g of Al with 8.10 g of Cl2.What mass of Al2Cl6can form?
Limiting reactant = Cl2
Base all calculations on Cl2
molesCl2
molesAl2Cl6
gramsCl2
gramsAl2Cl6
Another stoichiometric factor
2 Al + 3 Cl2 Al2Cl6
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CALCULATIONS: calculate mass of
Al2Cl6expected.
Step 1: Calculate moles of Al2Cl6
expected based on LR.
Step 2: Calculate mass of Al2Cl6expectedbased on LR.
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Cl2was the limiting reactant. Therefore,
Al was in excess. But how much?
First find how much Al was required.
Then find how much Al is in excess.
How much of which reactant willremain when reaction is complete?
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2 Al + 3 Cl2 products
0.200 mol 0.114 mol = LR
Calculating Excess Al
Excess Al = Al available - Al required
= 0.200 mol - 0.0760 mol= 0.124 mol Al in excess
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Where are we?
Key concepts from Chapters 2 & 3
Definition of atomic mass units (amu)
Conversion between amu and g
Avogadros number
Conversions between mass, moles, and # of atoms
Determination of atomic mass and molar mass
Determination of % composition by mass
Determination of empirical formula from % composition Determination of molecular formula from empirical
formula and molar mass.
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Key concepts from Chapter 4
Writing and balancing chemical equations
Law of conservation of mass
Define stoichiometric coefficient
Determine quantities of products formed when there is alimiting reagent
Calculation of % yield
Calculating stoichiomteric amount in reactions
Using solutions rather than solids
Limiting reagent and calculation of yields.
Where are we?
Calculating Ion Concentrations in a Solution
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Calculating Ion Concentrations in a Solutionof a Strong Electrolyte.
What are the aluminum and sulfate ion concentrations in0.0165M Al2(SO4)3?.
Al2(SO4)3(s) ! 2 Al3+(aq) + 3 SO4
2-(aq)
Balanced Chemical Equation:
0.0330 M Al3+
[Al] = " =1 L
2 mol Al3+
1 mol Al2(SO4)3
0.0165 mol Al2(SO4)3
0.0495 M SO42-
Sulfate Concentration:
[SO42-] = " =
1 mol Al2(SO4)31 L
3 mol SO42-0.0165 mol Al2(SO4)3
Aluminum Concentration:
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COMMON REACTIONS
precipitation
acid-base
oxidation/reduction (redox)