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• LIMITING REACTANTLIMITING REACTANT

The limiting reactant is used up first

The limiting reactant determines the

amount of product

Need balanced equation to proceed

• Limiting ReagentsLimiting Reagents

2H2 + O2 2H2O

10 H2 and 7 O2 10 H2O and 2 O2

reaction stops when one of the reactants is depleted

What if you had only 220.0 g Pb(NO3)2?How much PbI2 would precipitate?

• PERCENT YIELDPERCENT YIELD

PercentYield =

actual yieldtheoretical yield

x 100%

Calculation is just one more stepbeyond a standard stoichiometrycalculation

• PERCENT YIELDPERCENT YIELD

EXAMPLESilicon carbide (SiC) is made from sand (silicon dioxide, SiO2) and carbon at high T. CO is alsoformed. If 100.0 kg of sand are reacted and 55.0 kg SiC are formed, what is the percent yield?

SiO2(s) + 3 C(s) SiC(s) + 2 CO(g)

Convert kg SiO2 to moles:

100.0 kg x 103 g/kg x 1 mol60.09 g

= 1664 mol SiO2

moles SiO2 = moles SiC

Convert moles SiC to kg:

1664 mol x 40.10 g/mol x 10-3 = 66.73 kg SiC

% yield = 55.0 kg66.73 kg

x 100% = 82.4%

• TITRATIONTITRATION

Goal: find conc. of unknown React solution of known conc.

(std. soln.) with soln. of unknown conc. Find equivalence point (vol.)

soln. of unknown conc.

Find volume at equivalence point

(need way to signal the equiv. pointsuch as indicator)

• ACIDSACIDS--BASEBASENEUTRALIZATIONNEUTRALIZATION

ACID + BASE SALT + WATER

HNO3 + KOH KNO3 + H2O

H+(aq) + NO3(aq) + K+(aq) + OH(aq)

H2O(l) + K+(aq) + NO3(aq)

completeionic

equation

H+(aq) + OH(aq) H2O(l) net ionicequation

eliminate spectator ions

• Strong Acids and BasesStrong Acids and BasesTable 4.2

Strong Acids

Hydrochloric, HClHydrobromic, HBrHydroiodic, HIChloric, HClO3Perchloric, HClO4Nitric, HNO3Sulfuric, HSO4

Strong Bases

Group 1A metal hydroxides(LiOH, NaOH, KOH, RbOH, CsOH)

Heavy Group IIA metal hydroxides(Ca(OH)2, Sr(OH)2, Ba(OH)2)

• TITRATION EXAMPLETITRATION EXAMPLEA flask contains an unknown amount of HCl.This solution is titrated with 0.101 M NaOH.It takes 23.35 mL of NaOH to complete thereaction. How many grams of HCl were there?

HCl + NaOH NaCl + H2O

(1) Find moles of NaOH used

(23.35 x 103 L NaOH)(0.101 mol/L) = 2.36 x 103 mol

(2) Find moles of HClsame as NaOH

(3) Find g of HCl

(2.36 x 103 mol)(36.5 g/mol) = 86.1 x 103 g

86.1 mg HCl

If we knew the volume of HCl soln., we couldcalc. the M of the HCl soln.

• TITRATING HTITRATING H22SOSO44 with with NaOHNaOH

2 H+(aq) + 2 OH (aq) 2 H2O(l) net ionicequation

H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l)

Begin with 20 mL of 0.1 M H2SO4(aq), titrate with0.1 M NaOH(aq)

As NaOH(aq) is added, H+(aq) ions are used to make H2O(l). When enough NaOH is added, all of the H+ ions are consumed and the indicator turns pink.

What volume of 0.1 M NaOH is required?What is final volume of solution?What is final conc. Of H+(aq) and OH(aq)?What is final conc. of Na+(aq) and SO42(aq)

• STOICHIOMETRYSTOICHIOMETRY

When chemical reactions involve gases,the balanced equation provides the numberof moles of reactants and products.

The ideal gas equation provides the linkbetween number of moles and P, V, Tof gases.

How much gas is produced or consumedby a chemical reaction?

find n and then V or P

• Air BagAir Bag2 NaN3(s) 2 Na + 3 N2(g)

sodium azide gas in air bag

How many L of N2 at 735 mm Hg and26 C are produced from 125 g NaN3?

Moles NaN3125 g

65.01 g/mol= 1.92 mol

Moles N2 (1.92 mol NaN3)(3/2) = 2.88 mol N2

V = (2.88)(0.0821)(299)

0.967= 73.1 L

29 NaN3 + 4 Fe2O3 + NaNO3 15 Na2O + 8 Fe + 44 N2

2/3 = 0.67 29/44 = 0.66

(40 msec)

• IDEAL GAS LAW EXAMPLEIDEAL GAS LAW EXAMPLECalculate pressure change in cylinder of a cars engine when 0.250 g of octane [C8H18(l)] is burned with a stoichiometric amount of O2assuming complete combustion.

Cylinder volume = 0.100 L

Initial Temp = 80 oC; final Temp = 700 oC

• ENERGYENERGY

KINETICENERGY

POTENTIALENERGY

mechanical(moving mass mv2)

[ joule = kg m2/s2]

mechanical(mass in a placewhere force can act)

chemical (bonds)

nuclear(binding energy)

electrical(moving charge)

light (photons)

sound (moleculesmoving uniformly)

heat (moleculesmoving randomly)

• ENERGY CONVERSIONENERGY CONVERSION

Energy can be converted from oneform to another

When it is converted, the total energyremains constant

Law of Conservation of EnergyFirst Law of Thermodynamics

All energy lost by a system underobservation is gained by the surroundings

During energy conversion, some heatis always produced

• The energy of the universe is constant.

Euniverse = Esystem + Esurroundings = 0

Energy is neither created nor destroyed,

only converted from one form to another.

Esystem = q + w

q is heat gained or lost by the system

w is work done by or on the system

First Law of Thermodynamics First Law of Thermodynamics Conservation of EnergyConservation of Energy

• E = Efinal Einitial

E is a state function

System Surroundingsenergy E is

Surroundings Systemenergy E is +

E = q + wwork (work done to system +)

heat (heat added to system +)

State function: a function whose value doesnot depend on pathway used to get topresent state.

ENERGY CHANGESENERGY CHANGES

• ENTHALPYENTHALPYWhen changes occur at constant pressure

E = qp + wexpansionthis is negligible

H = qp

H is the enthalpy changeH is the quantity of thermal energy

transferred into a system at constantpressure

Like energy, enthalpy is a state function

• ENTHALPY OF REACTIONENTHALPY OF REACTION

H = H(products) H(reactants)

Endothermic H > 0Exothermic H < 0

2 H2(g)+ O2(g) 2 H2O(g) + heat

H = 483.6 kJ

Characteristics of Enthalpy(1) Enthalpy is an extensive property(2) H for a reaction is equal in magnitude but

opposite in sign to H for reverse reaction(3) H for a reaction depends on states of

reactants and products (gas, liquid)

• 2 H2 + O2 2 H2O H = 483.6 kJ

Is this reaction exothermic or endothermic? ________

How much heat is given off per mole of O2? _____

How much heat is given off per mole of H2? _____

What is H for 2 H2O O2 + 2 H2 ? _______

How many kJ of heat are needed to convert9.0 g of H2O into H2 and O2 ?

2 H2O O2 + 2 H2

THERMOCHEMICAL THERMOCHEMICAL EQUATIONSEQUATIONS

A balanced chemical equation that alsoincludes the enthalpy change.Ho delta H standard

standard P (1 bar) & T (usually 25 C)

• 2 HI(g) products

H2(g) + I2(g) reactants

HH = 53 kJ

ENDOTHERMIC REACTION

CH4(g) and 2 O2(g) reactants

CO2(g) + 2 H2O(l) products

HH = 890 kJ

EXOTHERMIC REACTION

• EXOTHERMIC REACTIONEXOTHERMIC REACTIONDEMONSTRATIONDEMONSTRATION

SO32(aq) + OCl(aq) SO42(aq) + Cl(aq) + heat

sodium sulfite and bleach

Laundry bleach is 5% NaOCl

This is a redox reaction

S4+ S6+ + 2 e

Cl1+ + 2 e Cl

Heat is evolvedExothermicH is negative

• FLAMING COTTONFLAMING COTTON

Na2O2(s) + 2H2O(l) 2Na+ + 2OH + H2O2(aq)

sodiumperoxide

2H2O2(aq) 2 H2O + O2

reaction produces heathave high conc. of O2cotton has low ignition temperature

.so..

flame

• ENDOTHERMIC REACTIONENDOTHERMIC REACTIONDEMONSTRATIONDEMONSTRATION

Ba(OH)28H2O(s) + 2 NH4(SCN)(s) Ba(SCN)2(aq) + 2 NH3(g) + 10 H2O(l)

mix two solids

highly endothermic reaction

increase in entropy is driving force

go from two ordered crystals to ionsand gases in solution

• HEAT CAPACITYHEAT CAPACITY

Experimental measurement of heat flow

q = C m T

temp changemass

specific heat

For H2O: 4.184 J

g C

heat gained or lost

• QUANTITATIVEQUANTITATIVECALORIMETRYCALORIMETRY

EXAMPLEA calorimeter with 200 g H2O is used for areaction. If T rises from 25.0 C to 33.0 C,how much energy is being released?

heat capacity of H2O(l) = 4.184 J/C g

4.184 JC g H2O

(200 g H2O)(8.0 C) = 6694.4 J

6.7 kJ

q = C m T

or

• HESSHESSS LAWS LAW

H for a sum of steps is the sameas for the overall process

Reason: H is a state function

Analogy to location and distance

• HESSHESSS LAW EXAMPLES LAW EXAMPLE

N2(g) + 2 O2(g) 2 NO2(g) H = ?

Two steps

N2(g) + O2(g) 2 NO(g) H = 180 kJ2 NO(g) + O2(g) 2 NO2(g) H = 112 kJ

N2(g) + 2 O2(g) 2 NO2 (g) H = 68 kJ

sum of H values gives H of net reaction

SMOG REACTION

• H OF FORMATIONH OF FORMATION

Hesss law applied to combination reactions

Elementscombine

Compound Hfheat of

formation

When all substances in standardstates (T, P, state) then

Hfo standard heat of formation

Hfo for stable form of element = 0

• ener

gy

reactants

products

-(HFo)R(HFo)P

HRXNo = ( HF)P ( HF)R

elements

HHFFoo to calculate to calculate HHRXNRXNoo

C6H12O6 + 6 O2

C(s) + H2(g) + O2(g)

6 CO2 + 6 H2O

HRXNo

• HHrxnrxnoo

Hrxno = n Hfo (products) m Hfo (reactants)an application of Hesss law

EXAMPLE: What is Hrxno for the

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