chemical reactions chapter 4

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CHEMICAL REACTIONSCHEMICAL REACTIONSChapter 4Chapter 4

Reactants: Zn + IReactants: Zn + I22 Product: Zn IProduct: Zn I22

22

Chemical Chemical EquationsEquations

Depict the kind of Depict the kind of reactantsreactants and and

productsproducts and their relative amounts and their relative amounts in a reaction.in a reaction.

4 Al(s) + 3 O4 Al(s) + 3 O22(g) ---> 2 Al(g) ---> 2 Al22OO33(s)(s)

The numbers in the front are calledThe numbers in the front are called

stoichiometric coefficientsstoichiometric coefficientsThe letters (s), (g), and (l) are the physical The letters (s), (g), and (l) are the physical

states of compounds.states of compounds.

33

Chemical EquationsChemical EquationsChemical EquationsChemical Equations4 Al(s) + 3 O4 Al(s) + 3 O22(g) (g)

---> 2 Al---> 2 Al22OO33(s)(s)

This equation meansThis equation means

4 Al atoms + 3 O4 Al atoms + 3 O22 molecules molecules

---give--->---give--->

2 molecules of Al2 molecules of Al22OO33

4 moles of Al + 3 moles of O4 moles of Al + 3 moles of O22

---give--->---give--->

2 moles of Al2 moles of Al22OO33

44

Chemical EquationsChemical EquationsChemical EquationsChemical Equations• Because the same atoms Because the same atoms

are present in a reaction are present in a reaction at the beginning and at at the beginning and at the end, the amount of the end, the amount of matter in a system does matter in a system does not change. not change.

• The The Law of the Law of the Conservation of Conservation of MatterMatter

Demo of conservation of matter, See Screen 4.3.

55

Because of the principle of the Because of the principle of the

conservation of conservation of mattermatter, ,

an an equation must be equation must be balancedbalanced..

It must have the same It must have the same number of atoms of the number of atoms of the

same kind on both same kind on both

sides.sides.

Chemical EquationsChemical Equations

Lavoisier, 1788Lavoisier, 1788

66

Balancing Balancing EquationEquationss

Balancing Balancing EquationEquationss

___ Al(s) + ___ Br___ Al(s) + ___ Br22(liq) ---> ___ Al(liq) ---> ___ Al22BrBr66(s)(s)

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Balancing Balancing EquationsEquationsBalancing Balancing EquationsEquations

____C____C33HH88(g) + _____ O(g) + _____ O22(g) ---->(g) ---->

_____CO_____CO22(g) + _____ H(g) + _____ H22O(g)O(g)

____B____B44HH1010(g) + _____ O(g) + _____ O22(g) ---->(g) ---->

___ B___ B22OO33(g) + _____ H(g) + _____ H22O(g)O(g)

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STOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRY

- the study of the - the study of the quantitative quantitative aspects of aspects of chemical chemical reactions.reactions.

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STOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYIt rests on the principle of the It rests on the principle of the conservation of matterconservation of matter..

2 Al(s) + 3 Br2(liq) ------> Al2Br6(s)

1010

PROBLEM: PROBLEM: If 454 g of NHIf 454 g of NH44NONO33 decomposes, how decomposes, how much Nmuch N22O and HO and H22O are formed? What O are formed? What is the theoretical yield of products?is the theoretical yield of products?

PROBLEM: PROBLEM: If 454 g of NHIf 454 g of NH44NONO33 decomposes, how decomposes, how much Nmuch N22O and HO and H22O are formed? What O are formed? What is the theoretical yield of products?is the theoretical yield of products?

STEP 1STEP 1

Write the balanced Write the balanced chemical equationchemical equation

NHNH44NONO33 ---> --->

NN22O + 2 HO + 2 H22OO

1111

454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22O O 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22O O

STEP 2 STEP 2 Convert mass reactant Convert mass reactant (454 g) --> moles(454 g) --> moles

454 g • 1 mol

80.04 g = 5.68 mol NH4NO3

STEP 3 STEP 3 Convert moles reactant Convert moles reactant (5.68 mol) --> moles product(5.68 mol) --> moles product

1212


STEP 3 STEP 3 Convert moles reactant --> Convert moles reactant --> moles productmoles product

Relate moles NHRelate moles NH44NONO33 to moles to moles product expected. product expected.

1 mol NH1 mol NH44NONO33 --> 2 mol H --> 2 mol H22OO

Express this relation as the Express this relation as the STOICHIOMETRIC STOICHIOMETRIC FACTORFACTOR..

2 mol H2O produced1 mol NH4NO3 used

2 mol H2O produced1 mol NH4NO3 used

1313


= 11.4 mol H= 11.4 mol H22O producedO produced

5.68 mol NH4NO3 • 2 mol H2O produced1 mol NH4NO3 used

STEP 3 STEP 3 Convert moles reactant (5.68 Convert moles reactant (5.68 mol) --> moles productmol) --> moles product

1414


11.4 mol H2O • 18.02 g1 mol

= 204 g H2O

STEP 4 STEP 4 Convert moles product Convert moles product (11.4 mol) --> mass product(11.4 mol) --> mass product

Called the Called the THEORETICAL THEORETICAL YIELDYIELD

ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY SOLVING STOICHIOMETRY

PROBLEMS!PROBLEMS!

ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY SOLVING STOICHIOMETRY

PROBLEMS!PROBLEMS!

1515

GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

Mass reactant

StoichiometricfactorMoles

reactantMoles product

Mass product

1616


STEP 5 STEP 5 How much NHow much N22O is formed?O is formed?

Total mass of reactants = total mass of Total mass of reactants = total mass of

productsproducts

454 g NH454 g NH44NONO33 = ___ g N = ___ g N22O + 204 g HO + 204 g H22OO

mass of Nmass of N22O = 250. gO = 250. g

1717


STEP 6 STEP 6 Calculate the Calculate the percent percent

yieldyieldIf you isolated only 131 g of NIf you isolated only 131 g of N22O, what is O, what is

the percent yield?the percent yield?

This compares the This compares the theoreticaltheoretical (250. g) (250. g)

and and actualactual (131 g) yields. (131 g) yields.

1818


% yield = actual yield

theoretical yield • 100%

STEP 6 STEP 6 Calculate the percent yieldCalculate the percent yield

% yield = 131 g250. g

• 100% = 52.4%

1919PROBLEM: Using 5.00 g PROBLEM: Using 5.00 g of Hof H22OO22, what mass of O, what mass of O22 and of Hand of H22O can be O can be obtained?obtained?

PROBLEM: Using 5.00 g PROBLEM: Using 5.00 g of Hof H22OO22, what mass of O, what mass of O22 and of Hand of H22O can be O can be obtained?obtained?

2 H2 H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(g) + OO(g) + O22(g)(g)

Reaction is catalyzed by MnOReaction is catalyzed by MnO22

Step 1: moles of HStep 1: moles of H22OO22

Step 2: use STOICHIOMETRIC FACTOR to Step 2: use STOICHIOMETRIC FACTOR to calculate moles of Ocalculate moles of O22

Step 3: mass of OStep 3: mass of O22

2020

Reactions Involving aReactions Involving aLIMITING REACTANTLIMITING REACTANTReactions Involving aReactions Involving aLIMITING REACTANTLIMITING REACTANT

• In a given reaction, there is not enough In a given reaction, there is not enough of one reagent to use up the other of one reagent to use up the other reagent completely.reagent completely.

• The reagent in short supply The reagent in short supply LIMITSLIMITS the quantity of product that can be the quantity of product that can be formed.formed.

2121

LIMITING REACTANTSLIMITING REACTANTS

Reactantseactants ProductsProducts

2 NO(g) + O2 (g) 2 NO2(g)

Limiting reactant = ___________Limiting reactant = ___________Excess reactant = ____________Excess reactant = ____________

2222

LIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTS

Demo of limiting reactants on Screen 4.7Demo of limiting reactants on Screen 4.7

2323

Rxn 1: Balloon inflates fully, some Zn leftRxn 1: Balloon inflates fully, some Zn left* * More than enough Zn to use up the 0.100 mol HClMore than enough Zn to use up the 0.100 mol HCl

Rxn 2: Balloon inflates fully, no Zn leftRxn 2: Balloon inflates fully, no Zn left* Right amount of each (HCl and Zn)* Right amount of each (HCl and Zn)

Rxn 3: Balloon does not inflate fully, no Zn left.Rxn 3: Balloon does not inflate fully, no Zn left.* Not enough Zn to use up 0.100 mol HCl* Not enough Zn to use up 0.100 mol HCl

LIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTS

React solid Zn with 0.100 React solid Zn with 0.100 mol HCl (aq)mol HCl (aq)Zn + 2 HCl ---> ZnClZn + 2 HCl ---> ZnCl22 + H + H22

1 2 3

(See CD Screen 4.8)(See CD Screen 4.8)

2424

Rxn 1Rxn 1 Rxn 2Rxn 2 Rxn 3Rxn 3mass Zn (g)mass Zn (g) 7.007.00 3.273.27 1.311.31mol Znmol Zn 0.1070.107 0.0500.050 0.0200.020mol HClmol HCl0.1000.100 0.1000.100 0.1000.100mol HCl/mol Znmol HCl/mol Zn 0.93/10.93/1 2.00/12.00/1 5.00/15.00/1Lim ReactantLim Reactant LR = HClLR = HCl no LRno LR LR = ZnLR = Zn

LIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTSReact solid Zn with 0.100 React solid Zn with 0.100 mol HCl (aq)mol HCl (aq)

Zn + 2 HCl ---> ZnClZn + 2 HCl ---> ZnCl22 + H + H2 2

2525

Reaction to be StudiedReaction to be StudiedReaction to be StudiedReaction to be Studied

2 Al + 3 Cl2 Al + 3 Cl22 ---> Al ---> Al22ClCl66

2626

PROBLEM: PROBLEM: Mix 5.40 g of Al with 8.10 Mix 5.40 g of Al with 8.10 g of Clg of Cl22. What mass of Al. What mass of Al22ClCl66 can can form?form?

PROBLEM: PROBLEM: Mix 5.40 g of Al with 8.10 Mix 5.40 g of Al with 8.10 g of Clg of Cl22. What mass of Al. What mass of Al22ClCl66 can can form?form?

Mass reactant

StoichiometricfactorMoles

reactantMoles product

Mass product

2727

Step 1 of LR problem: Step 1 of LR problem: compare compare actualactual mole mole ratio of reactants to ratio of reactants to theoreticaltheoretical mole mole ratio.ratio.

Step 1 of LR problem: Step 1 of LR problem: compare compare actualactual mole mole ratio of reactants to ratio of reactants to theoreticaltheoretical mole mole ratio.ratio.

2828


Reactants must be in the mole ratioReactants must be in the mole ratio

Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

mol Cl2mol Al

= 32

mol Cl2mol Al

= 32

2929Deciding on the Limiting Deciding on the Limiting ReactantReactant

Deciding on the Limiting Deciding on the Limiting ReactantReactant

IfIf

There is not enough Al to use up all There is not enough Al to use up all

the Clthe Cl22


mol Cl2mol Al

> 32

Lim reag = Lim reag = AlAl

3030

IfIf

There is not enough ClThere is not enough Cl22 to use to use

up all the Alup all the Al


mol Cl2mol Al

< 32

Lim reag = Lim reag =

ClCl22



3131

We have 5.40 g of Al and 8.10 g of ClWe have 5.40 g of Al and 8.10 g of Cl22

Step 2 of LR problem: Step 2 of LR problem: Calculate moles of each Calculate moles of each reactantreactant

Step 2 of LR problem: Step 2 of LR problem: Calculate moles of each Calculate moles of each reactantreactant

5.40 g Al • 1 mol27.0 g

= 0.200 mol Al

8.10 g Cl2 • 1 mol70.9 g

= 0.114 mol Cl2

3232

Find mole ratio of Find mole ratio of reactantsreactants

Find mole ratio of Find mole ratio of reactantsreactants

This This would be 3/2, or 1.5/1, if would be 3/2, or 1.5/1, if reactants are present in the reactants are present in the exact stoichiometric ratio.exact stoichiometric ratio.

Limiting reagent is Limiting reagent is ClCl22

mol Cl2mol Al

= 0.114 mol 0.200 mol

= 0.57


3333Mix 5.40 g of Al with 8.10 g of ClMix 5.40 g of Al with 8.10 g of Cl22. What mass of . What mass of

AlAl22ClCl66 can form? can form?

Limiting reactant = ClLimiting reactant = Cl22

Base all calcs. on ClBase all calcs. on Cl22

Limiting reactant = ClLimiting reactant = Cl22

Base all calcs. on ClBase all calcs. on Cl22

molesCl2

moles Al2Cl6

gramsCl2

grams Al2Cl6

1 mol Al2Cl63 mol Cl2


3434

CALCULATIONS: calculate mass ofCALCULATIONS: calculate mass ofAlAl22ClCl66 expected. expected.CALCULATIONS: calculate mass ofCALCULATIONS: calculate mass ofAlAl22ClCl66 expected. expected.

Step 1: Step 1: Calculate moles of AlCalculate moles of Al22ClCl66

expected based on LR.expected based on LR.

0.114 mol Cl2 • 1 mol Al2Cl6

3 mol Cl2 = 0.0380 mol Al2Cl6

0.0380 mol Al2Cl6 • 266.4 g Al2Cl6

mol = 10.1 g Al2Cl6

Step 2: Step 2: Calculate mass of AlCalculate mass of Al22ClCl66 expected expected

based on LR.based on LR.

3535

• ClCl22 was the limiting reactant. was the limiting reactant.

• Therefore, Al was present Therefore, Al was present

in excess. But how much?in excess. But how much?

• First find how much Al was required.First find how much Al was required.

• Then find how much Al Then find how much Al is in excess.is in excess.

How much of which reactant How much of which reactant will remain when reaction is will remain when reaction is

complete?complete?

How much of which reactant How much of which reactant will remain when reaction is will remain when reaction is

complete?complete?

3636

2 Al + 32 Al + 3 ClCl22 productsproducts

0.200 mol0.200 mol0.200 mol0.200 mol 0.114 mol = LR0.114 mol = LR0.114 mol = LR0.114 mol = LR

Calculating Excess AlCalculating Excess AlCalculating Excess AlCalculating Excess Al

Excess Al = Al available - Al requiredExcess Al = Al available - Al required

0.114 mol Cl2 • 2 mol Al

3 mol Cl2 = 0.0760 mol Al req' d

= 0.200 mol - 0.0760 mol = 0.200 mol - 0.0760 mol

= = 0.124 mol Al in excess0.124 mol Al in excess

3737Determining the Formula of Determining the Formula of a Hydrocarbon by a Hydrocarbon by

CombustionCombustion

CCR, page 138CCR, page 138

3838

Using Stoichiometry to Using Stoichiometry to Determine a FormulaDetermine a Formula

Burn 0.115 g of a hydrocarbon, CBurn 0.115 g of a hydrocarbon, CxxHHyy, and produce , and produce

0.379 g of CO0.379 g of CO22 and and 0.1035 g of H0.1035 g of H22OO. .

CCxxHHy y + some oxygen ---> + some oxygen ---> 0.379 g CO0.379 g CO22 + + 0.1035 g H0.1035 g H22OO

What is the empirical formula of CWhat is the empirical formula of CxxHHyy??

3939


First, recognize that all C in COFirst, recognize that all C in CO22 and all H in H and all H in H22O O

is from Cis from CxxHHyy..

CCxxHHy y + some oxygen ---> + some oxygen --->

0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

Puddle of CxHy

0.115 g

0.379 g CO0.379 g CO22+O2

+O2 0.1035 g H2O1 H2O molecule forms for each 2 H atoms in CxHy

1 CO2 molecule forms for each C atom in CxHy

4040


First, recognize that all C in COFirst, recognize that all C in CO22 and all H in H and all H in H22O is from CO is from CxxHHyy..

1. Calculate amount of C in CO1. Calculate amount of C in CO22

8.61 x 108.61 x 10-3 -3 mol COmol CO22 --> 8.61 x 10 --> 8.61 x 10-3 -3 mol Cmol C

2. Calculate amount of H in H2. Calculate amount of H in H22OO

5.744 x 105.744 x 10-3-3 mol H mol H22O -- >1.149 x 10O -- >1.149 x 10-2 -2 mol Hmol H


0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

4141


Now find ratio of mol H/mol C to find values of “x” and “y” in Now find ratio of mol H/mol C to find values of “x” and “y” in CCxxHHyy..

1.149 x 10 1.149 x 10 -2 -2 mol Hmol H/ / 8.61 x 108.61 x 10-3 -3 mol Cmol C

= = 1.33 mol H1.33 mol H / / 1.00 mol C1.00 mol C

= = 4 mol H4 mol H / / 3 mol C3 mol C

Empirical formula = CEmpirical formula = C33HH44


0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

chemical reactions chapter 4

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