Chemical Reaction Balancing

8/10/2019 Chemical Reaction Balancing

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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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= 11.4 mol H2O produced

STEP 3 Convert moles reactant (5.68 mol)

moles product


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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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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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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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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.


! !

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


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


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



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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)

Chemical Reaction Balancing

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