ib chemistry limiting, excess, theoretical and percentage yield

http://lawrencekok.blogspot.com

Prepared by Lawrence Kok

IB Chemistry on Limiting, Excess, Percentage Yield and Ionic Equations

http://lawrencekok.blogspot.com/

1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq)

Lead + Potassium → Lead + Potassium Nitrate iodide iodide nitrate

Chemical Eqn

Reactant – Left Product – Right Conservation MassTotal Mass reactant = Total Mass product

Mole Ratio - Coefficient of reactant/product

1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq)1 : 2 → 1 : 2

Word equation Chemical formula

Calcium + hydrochloric → Calcium + carbon + watercarbonate acid chloride dioxide

1CaCO3(s) + 2HCI(aq) → 1CaCI2(aq) + 1CO2(g)

+ 1H2O(l)

Product – Right Reactant – Left Conservation MassTotal Mass reactant = Total Mass product

1 : 2 → 1 : 1 : 1 Mole Ratio - Coefficient of reactant/product

Physical state/ symbol

(s) – solid(I) - liq(g) – gas(aq) – aqueous ∆ - heatppt –

precipitate/solid↔ - reversible

Physical state/ symbol

(s) – solid(I) - liq(g) – gas(aq) – aqueous ∆ - heatppt –

precipitate/solid↔ - reversible

Mass reactants (Pb(NO3)2 + KI) = 15.82

Mass products (PbI3 + KNO3) = 15.82

AfterBefore

Chemical rxn Matter is neither created nor destroyed Undergoes physical/chemical change.

LAW of conservation of mass.

http://www.youtube.com/watch?v=J5hM1DxaPLw

http://www.youtube.com/watch?v=J5hM1DxaPLw

http://www.schoolsworld.tv/node/3259?terms=64,87,99

Mole proportion/ratio (reactant) →

(product) 1 : 2 → 1

: 2

Concept Map

represented by

1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)

Limiting reactant Use up first

Limit products formRxn stop if all used up

Excess reactant left over

remains behind

Percentage Yield mass of Actual Yield x 100%mass of Theoretical Yield

Theoretical yieldMax amt prod form if rxn complete

Stoichiometry ratio Assume all limiting reagent used

up

Actual yieldAmt of prod formed

experimentallyLess than theoretical yield due to experimental error

Rxn StoichiometryQuantitative relationship bet quantities react/

prodFind quantities/amt (mass, mole, vol)

Predict how much react and amt prod formChemical rxn react in definite ratio

Chemical Change

Chemical Equation

Balanced Chemical equation

Molecular Eqn

CompleteIonic Eqn

Net Ionic Eqn

1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)

1Pb2+(aq) + 2NO3

-(aq) + 2Na+

(aq) + 2I-(aq) → 1PbI2(s) + 2Na+

(aq)

+ 2NO3-(aq)

1Pb2+(aq) + 2CI-

(aq) → 1PbCI2(s)

http://www.youtube.com/watch?v=LQq203gyftA

http://www.youtube.com/watch?v=LicEaaXhlEY

Limiting and Excess

Which is limiting and excess ?

How many hot dog with 6 bun and 3 hot dog?

Stoichiometric ratio 1 mol (bun) : 1 mol (hot dog) → 1 mol

+5 5 5

+

No Excess No limiting

Excess - BunLimiting - Hot dog are used up

Both hot dog and bun used up

How many burger with 12 bun and 6 patties?

+ +

Stoichiometric ratio 2 mol (bun) : 1 mol (burger) → 1 mol

No Excess No limiting

Limiting reactant Use up first, limit the prod

formRxn stop if all used up

Excess reactant Left over, remain behind

1Zn (s) + 2HCI (aq) → 1ZnCI2(aq) + 1H2(g)

Mole ratio1 : 2 → 1: 1

Moles reactant given, which is limiting and excess ?


1st method

2nd method

0.30 mol Zn + 0.52 mol HCl add

1 mol Zn → 2 mol HCI0.30 mol Zn → 2 x 0.30 mol HCI = 0.60 mol HCI needed = 0.52 mol HCI added (HCI – limiting)

0.52 mol HCI0.30 mol Zn

Reactant that produce least amt product → will be limiting

Assume Zn limiting 1 mol Zn → 1 mol H2 gas0.3 mol Zn → 1 x 0.3 = 0.3 mol H2

Assume HCI limiting 2 mol HCI → 1 mol H2 gas0.52 mol HCI → 1 x 0.52 = 0.26 mol H2 2

Simulation on limiting/excess

1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) Mole ratio1 : 2 → 1: 2


10 g Pb(NO3)2 + 10 g NaI added

0.0302 mol Pb(NO3)2 + 0.0667 mol NaIWhich is limiting and excess ?

1 mol Pb(NO3)2 → 2 mol NaI0.0302 mol Pb(NO3)2 → 2 x 0.0302 mol NaI = 0.0604 mol NaI needed = 0.0667 mol NaI add (NaI excess)

Mass = 10.0RMM 149.9 = 0.0667 mol

Mass = 10.0RMM 331.2= 0.0302 mol

1st method

2nd methodReactant that produce least

amt product → will be limiting

Assume Pb(NO3)2 limiting 1 mol Pb(NO3)2→ 1 mol PbI20.0302 mol Pb(NO3)2→ 1x0.0302 mol PbI2

= 0.0302 mol PbI2 Assume NaI limiting 2 mol NaI → 1 mol PbI2 0.0667 mol NaI → 1 x 0.0667 = 0.0334 mol PbI2 2

3rd methodMole ratio method

1 : 2

57.052.03.0

).().(

5.021

).().(

HCIMoleZnMoleHCIMoleZnMole

from eqn

given mass

Ratio higher ↓

Zn excess/HCI limit

1 : 2 → 1 : 1

http://phet.colorado.edu/en/simulation/reactants-products-and-leftovers

http://chemcollective.org/activities/tutorials/stoich/limiting-reagents



Limiting and excess ?

Which is limiting and excess ?1st method

2nd method

1 mol Mg → 2 mol HCI0.0256 mol Mg → 0.0512 mol HCI = 0.0512 mol HCI needed = 0.0341 mol HCI added (HCI – limiting)

27.3 ml, 1.25M HCI0.623 g Mg


Assume Mg limiting 1 mol Mg → 1 mol H2 0.0256 mol Mg → 0.0256 mol H2

Assume HCI limiting 2 mol HCI → 1 mol H2 0.0341 mol HCI → 0.01705 mol H2



0.02 mol NaOH + 0.025 mol H2SO4


2 mol NaOH → 1 mol H2SO4 0.02 mol NaOH → 0.01 mol H2SO4

= 0.01 mol H2SO4 needed = 0.025 mol H2SO4 add (H2SO4 excess)

1st method



Assume NaOH limiting 2 mol NaOH → 1 mol H2O 0.02 mol NaOH → 0.01 mol H2O

Assume H2SO4 limiting 1 mol H2SO4 → 1 mol H2O 0.025 mol H2SO4 → = 0.025 mol H2O

Mg(s) + 2HCI(aq) → MgCI2(aq) + H2 (g)

0.623 g Mg + 27.3 ml, 1.25M HCI add

Mass = 0.623RMM 24= 0.0256 mol

Mol = M x V 1000= 1.25 x 0.0273= 0.0341 mol

0.0256 mol Mg + 0.0341 mol HCI

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

1 : 2 → 1 : 1 2 : 1 → 1 : 1

100 ml, 0.2M, NaOH + 50 ml, 0.5M H2SO4 add

Mol = M x V 1000= 0.2 x 0.1= 0.02 mol

Mol = M x V 1000= 0.5 x 0.05= 0.025 mol


1 : 2

75.00341.00256.0

).().(

5.021

).().(

HCIMoleMgMoleHCIMoleMgMole

from eqn

given mass

Ratio higher ↓

Mg excess/HCI limit





Limiting and excess ?


1st method

2nd method

2 mol CO → 1 mol O2

2 mol CO → 1 mol O2 = 1 mol O2 needed = 0.5 mol O2 added (O2 – limiting)


Assume CO limiting 2 mol CO → 2 mol CO2 2 mol CO → 2 mol CO2

Assume O2 limiting 1 mol O2 → 2 mol CO2 0.5 mol O2 → 1 mol CO2



0.02 mol NaOH + 0.025 mol H2SO4


2 mol NaOH → 1 mol H2SO4 0.02 mol NaOH → 0.01 mol H2SO4

= 0.01 mol H2SO4 needed = 0.025 mol H2SO4 add (H2SO4 excess)

1st method



Assume NaOH limiting 2 mol NaOH → 1 mol H2O 0.02 mol NaOH → 0.01 mol H2O

Assume H2SO4 limiting 1 mol H2SO4 → 1 mol H2O 0.025 mol H2SO4 → = 0.025 mol H2O

Mol = Volume Molar vol = 45.42 = 2 mol 22.4

Mol = Volume Molar vol= 11.36 = 0.5 mol 22.42 mol CO + 0.5 mol O2

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

2 : 1 → 2 2 : 1 → 1 : 1

100ml, 0.2M, NaOH + 50 ml, 0.5M H2SO4 add

Mol = M x V 1000= 0.2 x 0.1= 0.02 mol

Mol = M x V 1000= 0.5 x 0.05= 0.025 mol

2CO(g) + 1O2(g) → 2CO2 (g)

45.42 L CO + 11.36 L O2 add


2 : 1

45.02

).().(

212

).().(

2

2

OMoleCOMoleOMoleCOMole

from eqn

given mass

Ratio higher ↓

CO excess/O2 limit






1st method

2nd method

0.30 mol Zn + 0.52 mol HCl add

1 mol Zn → 2 mol HCI0.30 mol Zn → 2 x 0.30 mol HCI = 0.60 mol HCI needed = 0.52 mol HCI added (HCI – limiting)

0.52 mol HCI0.30 mol Zn


Assume Zn limiting 1 mol Zn → 1 mol H2 gas0.3 mol Zn → 1 x 0.3 = 0.3 mol H2

Assume HCI limiting 2 mol HCI → 1 mol H2 gas0.52 mol HCI → 1 x 0.52 = 0.26 mol H2 2




1 : 2

57.052.03.0

).().(

5.021

).().(

HCIMoleZnMoleHCIMoleZnMole

from eqn

given mass

Ratio higher ↓

Zn excess/HCI limit

Theoretical, Actual and % Yield

1Zn (s) + 2HCI (aq) → 1ZnCI2(aq) + 1H2(g)

1 : 2 → 1 : 1


Find theoretical yield, cm3 for H2 gasFind % yield if expt yield is 5800 cm3

%2.98%10059025800%

%100.

.exp%.

yield

yieldltheoreticayieldtyield

HCI limiting ↓

Mole ratio2 mol HCI : 1 mol H2

1Zn + 2HCI → 1ZnCI2 + 1H2

2 mol HCI → 1 mol H2

0.52 mol HCI→ 0.26 mol H2

Theoretical yield 1 mol H2 – 22700 cm3

0.26 mol H2 – 5902 cm3

Expt yield = 5800 cm3

Click here tutorial on austute



http://www.ausetute.com.au/exceslim.html




Which is limiting and excess ?1st method

2nd method

1 mol Mg → 2 mol HCI0.0256 mol Mg → 0.0512 mol HCI = 0.0512 mol HCI needed = 0.0341 mol HCI added (HCI – limiting)

27.3 ml, 1.25M HCI0.623 g Mg


Assume Mg limiting 1 mol Mg → 1 mol H2 0.0256 mol Mg → 0.0256 mol H2

Assume HCI limiting 2 mol HCI → 1 mol H2 0.0341 mol HCI → 0.01705 mol H2



Mg(s) + 2HCI(aq) → MgCI2(aq) + H2 (g)

0.623 g Mg + 27.3 ml, 1.25M HCI add

Mass = 0.623RMM 24= 0.0256 mol

Mol = M x V 1000= 1.25 x 0.0273= 0.0341 mol

0.0256 mol Mg + 0.0341 mol HCI

1 : 2 → 1 : 1


1 : 2

75.00341.00256.0

).().(

5.021

).().(

HCIMoleMgMoleHCIMoleMgMole

from eqn

given mass

Ratio higher ↓

Mg excess/HCI limit

Theoretical, Actual and % YieldFind theoretical yield, cm3 for H2 gasFind % yield if expt yield is 300 cm3

Mg + 2HCI → MgCI2 + H2

HCI limiting ↓

Mole ratio2 mol HCI : 1 mol H2

2 mol HCI → 1 mol H20.0341 mol HCI→ 0.01705 mol H2

Theoretical yield 1 mol H2 – 22700 cm3

0.01705 mol H2 – 387 cm3

Expt yield = 5800 cm3

%5.77%100387300%

%100.

.exp%.

yield


Click here tutorial on chemwiki



http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Limiting_Reagents



http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Limiting_Reagents


1st method

2nd method

2 mol CO → 1 mol O2

2 mol CO → 1 mol O2 = 1 mol O2 needed = 0.5 mol O2 added (O2 – limiting)


Assume CO limiting 2 mol CO → 2 mol CO2 2 mol CO → 2 mol CO2

Assume O2 limiting 1 mol O2 → 2 mol CO2 0.5 mol O2 → 1 mol CO2



Mol = Volume Molar vol = 45.42 = 2 mol 22.4

Mol = Volume Molar vol= 11.36 = 0.5 mol 22.42 mol CO + 0.5 mol O2

2 : 1 → 2

2CO(g) + 1O2(g) → 2CO2 (g)

45.42 L CO + 11.36 L O2 add


2 : 1

45.02

).().(

212

).().(

2

2

OMoleCOMoleOMoleCOMole

from eqn

given mass

Ratio higher ↓

CO excess/O2 limit

Theoretical, Actual and % YieldFind theoretical yield, g for CO2 gas

Find % yield if expt yield is 30 g

2CO + 1O2 → 2CO2

O2 limiting ↓

Mole ratio1 mol O2 : 2 mol CO2

1 mol O2 → 2 mol CO2 0.5 mol O2 → 1 mol CO2

Theoretical yield 1 mol CO2 – 44 g

%1.68%1004430%

%100.

.exp%.

yield


Expt yield = 30 g

Click here tutorial on chemtamu



http://www.chem.tamu.edu/class/majors/tutorialnotefiles/stochiometryproblems.htm



http://www.chem.tamu.edu/class/majors/tutorialnotefiles/stochiometryproblems.htm

1st method

2nd method

1 mol CH3COOH → 1 mol C5H11OH0.0596 CH3COOH → 0.0596 C5H11OH = 0.0596 C5H11OH need = 0.0539 C5H11OH add (C5H11OH – limiting)

0.0539 mol C5H11OH0.0596 mol

CH3COOHSimulation on limiting/excess


Mole ratio method1 : 1

1.10539.00596.0

).().(11

).().(

115

3

115

3

OHHCMoleCOOHCHMoleOHHCMole

COOHCHMolefrom eqn

given mass

Ratio higher ↓

CH3COOH excess/C5H11OH limit


CH3COOH + C5H11OH → Ester + H2O1 : 1 → 1 : 1


Find mass of ester forms if it has 45% yield

gyieldtyieldltheoretica

yieldtyield

15.3%1007.exp%45

%100.

.exp%.

C5H11OH limiting ↓

Mole ratio1 mol C5H11OH : 1 mol Ester

1 mol C5H11OH → 1 mol Ester0.0539 mol C5H11OH→ 0.0539 Ester

Theoretical yield 1 mol Ester - 130 g0.0539 mol Ester – 7 gExpt yield = ????



CH3COOH + C5H11OH → Ester + H2O







1st method

2nd method

1 mol CH3COOH → 1 mol C5H11OH0.0596 CH3COOH → 0.0596 C5H11OH = 0.0596 C5H11OH need = 0.0539 C5H11OH add (C5H11OH – limiting)

0.0539 mol C5H11OH0.0596 mol

CH3COOHSimulation on limiting/excess


Mole ratio method1 : 1

1.10539.00596.0

).().(11

).().(

115

3

115

3

OHHCMoleCOOHCHMoleOHHCMole

COOHCHMolefrom eqn

given mass

Ratio higher ↓

CH3COOH excess/C5H11OH limit


CH3COOH + C5H11OH → Ester + H2O1 : 1 → 1 : 1


Find mole of C5H11OH used if 0.888 mol Ester is needed with a 65% yield

molx


37.1%100888.0%65

%100.

.exp%.

Mole ratio1 mol C5H11OH : 1 mol Ester

1 mol C5H11OH → 1 mol Ester0.888 mol C5H11OH → 0.888 Ester

Expt yield = 0.888 mol



CH3COOH + C5H11OH → Ester + H2O







Aspirin, widely used drugs, prepared below Student reacted salicylic acid with excess ethanoic anhydride. Impure solid aspirin was obtained by filtering. Pure aspirin was obtained by

recrystallization.

Find amt, mol, of salicylic acid, C6H4(OH)COOH, used.

Find theoretical yield, in g, of aspirin, C6H4(OCOCH3)COOH.

Find % yield of pure aspirin.

Find % uncertainty in mass of aspirin.

Salicylic acid ethanoic anhydride aspirin

Mass salicylic acid

3.15 ± 0.02g

Mass pure aspirin

2.50 ± 0.02g

%8.60%10011.450.2%

%100.

.exp%.

yield


molMole

MgMassMole

r

0228.013.13815.3

).(

Mole ratio1 mol salicylic acid : 1 mol aspirin

0.0228 mol salicylic acid : 0.0228 mol aspirin

gMassMMoleMass

MgMassMole

r

r

11.47.1800228.0

).(

%80.0int%.

%10050.202.0int%.

yuncerta

yuncerta

limiting

100 g zinc react with 100 g of iodine producing zinc iodide.

Zn + I2 → ZnI2

Find mass of ZnI2 produced

Find, amt of Zn and I2, and determine which reactant is in excess

molZnMole

MgMassZnMole

r

53.137.65

100).(

).().(

molIMole

MgMassIMole

r

394.08.253

100).(

).().(

2

2

Mole ratio1 mol I2 : 1 mol Zn

0.394 mol I2 : 0.394 mol Zn

I2 – limitingZn - excess

Mole – Zn Mole I2

Mole ratio 1 mol I2 : 1 mol ZnI2

0.394 mol I2 : 0.394 mol ZnI2

gZnIMassMMoleZnIMass

MgMassZnIMole

r

r

8.125319394.0).().(

).().(

2

2

2

Copper metal produced by copper(I) oxide and copper(I) sulfide shown below

Mixture of 10 kg of copper(I) oxide and 5 kg of copper(I) sulfide was heated. 2Cu2O + Cu2S → 6Cu + SO2

Find limiting reagent.Mole – Cu2O Mole Cu2S

molOCuMole

MgMassOCuMole

r

9.6914310000).(

).().(

2

2

molSCuMole

MgMassSCuMole

r

4.311595000).(

).().(

2

2

Find maximum mass Cu produced

Mole ratio 2 mol Cu2O : 1 mol Cu2S

69.9 mol Cu2O: 35 mol Cu2SCu2S – limitingCu2O - excess

Mole ratio1 mol Cu2S : 6 mol Cu

31.4 mol Cu2S : 188 mol Cu

gCuMassMMoleCuMass

MgMassCuMole

r

r

1190055.63188).().(

).().(

Student determine Mr of solid monoprotic acid, HA, by titrating with a known mass of acid.

Data shown below.

Find mass of acid and determine its absolute and % uncertainty

Known mass of acid, HA, was dissolved in water to form a 100ml sol in volumetric flask.

25 ml sample of sol reacted with 12.1 ml of 0.1M NaOH. Find molar mass of acid.

Mass bottle 1.737 ± 0.001 g

Mass bottle + acid, HA

2.412 ± 0.001 g

Mass acid = ( 2.412 – 1.737) ± 0.002 = (0.675 ±0.002) g

% uncertainty = (0.002) x 100% = 0.3% 0.675

NaOH + HA → NaA + H2O

NaOH + HA → NaA + H2O M = 0.1M M = ?V = 12.1 ml V = 25 ml

MMM

VMVM

a

a

aa

bb

0484.011

251.121.011

? HA100 ml water added

25 ml transfer

NaOH M = 0.1MV = 12.1 ml

HA M = ?

Amt acid in 1000 ml = 4.84 x 10-2 molAmt acid in 100 ml = 4.84 x 10 -3 mol

1391084.4675.0

).(

3

r

r

M

MgMassMole

Cations/Metals/+ve ions

Gp 1 Gp 2 Gp 3 Transition metals ions ( variable oxidation states)

Oxidation state+1

Oxidation

state+2

Oxidation

state+3

Sc+3

Ti+2+3

V+2+3

Cr+2+3+6

Mn+2+3+6+7

Fe+2+3

Co+2+3

Ni+2

Cu+1+2

Zn+2

Li 1+ Be2+ Sc 3+ Ti 2+

Ti 3+V 2+

V 3+Cr 2+

C r3+

Cr6+

Mn 2+

Mn 3+

Mn 6+

Mn 7+

Fe 2+

Fe 3+

Co2+

Co 3+Ni2+ Cu1+

Cu2+Zn2+

Na 1+ Mg2+ Al 3+

K 1+ Ca2+

Anion/Non metal

Gp 5 Gp 6 Gp 7

Oxidation

state

Oxidation

state

Oxidation

state

-3 -2 -1

N3- O2- F-1

P3- S2- CI-1

Br-1

I-1

Ionic Compound

Li2O MgCI2 Al2O3 FeOIron(II) oxide

NiONickel(II)

oxide

CuOCopper(II) oxide

Li3N Mg3N2 AlN Fe3N2

Iron(II) nitride

Ni3N2

Nickel(II) nitride

Cu3N2

Copper(II) nitride

Oxidation state/Charge ion→ Li1+ O2-

Formula compound Li2 O1

Video on polyatomic ions

Writing Chemical Formula

Step 1 : Write Oxidation state/charge

Step 2 : Balance it, (electrically neutral) by cross multiply – as subscript

Metal/Cations/+ve ionNon Metal/

Anion/-ve ion

http://www.youtube.com/watch?v=URc75hoKGLY

Polyatomic ionsGroup of non-metals bonded

together

Oxidation state

Oxidation state

Oxidation state

-1/+1 -2 -3

(OH)-1

Hydroxide(SO4)2-

Sulphate(PO4)3-

Phosphate

(CN)-1

Cyanide(SO3)2-

Sulphite

(SCN)-1

Thiocyanate

(CO3)2-

Carbonate

(NO3)-1

Nitrate(S2O3)2-

Thiosulphate

(NO2)-1

Nitrite(Cr2O7)2-

Dichromate

(NH4)+1

AmmoniumLi2(CO3) Mg(CO3) Al2(CO3)3 Fe(CO3) Ni(CO3) Cu(CO3)

Li(OH) Mg(OH)2 Al(OH)3 Fe(OH)2 Ni(OH)2 Cu(OH)2

Li2(SO4) Mg(SO4) Al2(SO4)3 FeSO4 Ni(SO4) Cu(SO4) Video on polyatomic ions

Ionic Compound

Cations/Metals/+ve ions

Gp1 Gp 2 Gp3 Transition metals ions ( variable oxidation states)

Oxidation state+1

Oxidation

state+2

Oxidation

state+3

Sc+3

Ti+2+3

V+2+3

Cr+2+3+6

Mn+2+3+6+7

Fe+2+3

Co+2+3

Ni+2

Cu+1+2

Li 1+ Be2+ Sc 3+ Ti 2+

Ti 3+V 2+

V 3+Cr 2+

C r3+

Cr6+

Mn 2+

Mn 3+

Mn 6+

Mn 7+

Fe 2+

Fe 3+

Co2+

Co3+Ni2+ Cu1+

Cu2+

Na 1+ Mg2+ Al 3+

K 1+ Ca2+

Oxidation state/Charge ion → Li1+ (CO3)2-

Formula compound Li2 (CO3)1

Step 1 : Write Oxidation state/charge ion

Step 2 : Balance it, (electrically neutral) by cross multiply – as subscript

Writing Chemical Formula

Metal/Cations/+ve ion Polyatomic ions

http://www.youtube.com/watch?v=MJZeZvDxcx8

Acids AlkaliMetal Hydroxide

Metal oxides Salts Gas

HCIHydrochloric acid

KOHPotassium hydroxide

CuOCopper(II) oxide

CaCO3

Calcium carbonate

COCarbon monoxide

HNO3

Nitric acidNaOH

Sodium hydroxideMgO

Magnesium oxideNa2CO3

Sodium carbonate

CO2

Carbon dioxide

H2SO3

Sulphurous acidCa(OH)2

Calcium HydroxideZnO

Zinc oxideNaHCO3

Sodium bicarbonate

SO2

Sulphur dioxide

HCOOHMethanoic acid

NH3

AmmoniaNa2O

Sodium oxideKNO3

Potassium nitrateSO3

Sulphur trioxide

CH3COOHEthanoic acid

Mg(OH)2

Magnesium hydroxide

Al2O3

Aluminium oxidePb(NO3)2

Lead (II) NitrateNO2

Nitrogen dioxide

H3PO4

Phosphoric acidCu(OH)2

Copper (II) hydroxide

Fe2O3

Iron(III) oxideNaNO3

Sodium nitrateCH4

Methane

H2CO3

Carbonic acidAl(OH)3

Aluminium hydroxide

K2SPotassium sulphide

PbI2

Lead (II) nitrateH2S

Hydrogen sulphide

HNO2

Nitrous acidFe(OH)2

Iron (II) hydroxidePbS

Lead(II) sulphideAgCI

Silver chlorideO2

Oxygen

HFHydrofluoric acid

Fe(OH)3

Iron (III) hydroxideZnS

Zinc sulphideMgSO4

Magnesium sulphate

N2

Nitrogen

HCIOHypochlorous acid

Zn(OH)2

Zinc hydroxideAI2S3

Aluminium sulphide

Na2S2O3

Sodium thiosulphate

CI2

Chlorine

Chemical Formula for common chemicals

Naming chemical compound

Writing chemical formula

Writing chemical formula

VIDEO TUTORIALS

http://www.youtube.com/watch?v=53T44hVKS_Q

http://www.youtube.com/watch?v=bvFylpHrJJY

http://www.youtube.com/watch?v=16agvZ8K2eM

Chemical Eqn

1Pb(NO3)2(aq) + 2NaCI(aq) → 1PbCI2(s) + 2NaNO3(aq)

1Pb2+(aq) + 2NO3

-(aq) + 2Na+

(aq) + 2CI-(aq) → 1PbCI2(s) + 2Na+

(aq) + 2NO3

-(aq)

unchanged

1Pb2+(aq) + 2CI-

(aq) → 1PbCI2(s)

1Pb2+(aq) + 2NO3

-(aq) + 2Na+

(aq) + 2CI-(aq) → 1PbCI2(s )+ 2Na+

(aq) + 2NO3-(aq)

Break aq → ions

Cancel outCancel outSpectators ions- don’t participate in rxnCancel out ions from both sides of eqn

Only ions involved in rxn

Net ionic eqnComplete ionic eqnMolecular eqn

Break down electrolytes, (aq) → ion

Leave sol, liq, gas unchanged

Molecular eqn

Complete ionic eqn

Net ionic eqn

Na2CO3(aq) + 2HNO3(aq) → 2NaNO3(aq) + H2O(l)

+ CO2(g)

2Na+(aq) + CO3

2-(aq) + 2H+

(aq) + 2NO3-(aq) → 2Na+

(aq) + 2NO3-(aq) +

H2O(l) + CO2 (g)

2Na+(aq) + CO3

2-(aq) + 2H+

(aq) + 2NO3-(aq) → 2Na+

(aq) + 2NO3-(aq) +

H2O(l) + CO2 (g)

CO32-

(aq) + 2H+(aq) →H2O(l) +

CO2(g)

Break aq → ions

Net ionic eqn

Complete ionic eqn

Molecular eqn

Cancel out

Chemical Eqn

Cancel out


Molecular eqn

Complete ionic eqn

Net ionic eqn

Net ionic eqn

Complete ionic eqn

Molecular eqn

2 Na3PO4(aq) + 3CaCI2(aq) → 6NaCI(aq) + Ca3(PO4)2(s)

6Na+(aq) + 2PO4

3-(aq) + 3Ca2

(aq) + 6CI-(aq) → 6Na+

(aq) + 6CI-(aq) +

Ca3(PO4)2(s)

6Na+(aq) + 2PO4

3-(aq) + 3Ca2

(aq) + 6CI-(aq) → 6Na+

(aq) + 6CI-(aq) +

Ca3(PO4)2(s)

2PO43-

(aq) + 3Ca2+(aq) →

Ca3(PO4)2(s)

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

Zn(s) + Cu2+(aq) + SO4

2-(aq) → Zn2+

(aq) + SO42-

(aq) + Cu(s)

Zn(s) + Cu2+(aq) + SO4

2- (aq) → Zn2+

(aq) + SO42-

(aq)

+ Cu(s)

Zn (s) + Cu2+(aq) → Zn2+

(aq) + Cu(s)

Cancel out

SO42-

(aq) + Ba2+(aq) → BaSO4 (s)

2K+(aq) + SO4

2-(aq) + Ba2+

(aq) + 2Cl-(aq) → BaSO4(s) + 2K+

(aq) + 2Cl-(aq)

2K+(aq) + SO4

2-(aq) + Ba2+

(aq) + 2Cl-(aq) →BaSO4(s) + 2K+

(aq) + 2Cl-(aq)

Mg(s) + 2H+(aq) +2Cl-

(aq) → Mg2+(aq) + 2Cl-

(aq)

+ H2(g)

Mg(s) + 2H+(aq) + 2Cl-

(aq) → Mg2+(aq) + 2Cl-

(aq) +H2(g)

Cancel out

Chemical Eqn

Cancel out


Molecular eqn

Complete ionic eqn

Net ionic eqn

Net ionic eqn

Complete ionic eqn

Molecular eqn

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Mg(s) + 2H+(aq) → Mg2+

(aq) + H2(g)

K2SO4(aq) + BaCl2 (aq) → BaSO4(s) + 2KCl(aq)

3CO32-

(aq) + 2Al3+(aq) →

Al2(CO3)3(s)

6NH4+

(aq) + 3CO32-

(aq) + 2Al3+(aq) + 6NO3

-(aq) → 6NH4

+(aq) + 6NO3

-(aq) +

Al2(CO3)3(s)

6NH4+

(aq) + 3CO32-

(aq) + 2Al3+(aq) + 6NO3

-(aq) → 6NH4

+(aq) + 6NO3

-(aq) +

Al2(CO3)3(s)

OH-(aq) + H+

(aq) → H2O(l)

2Na+(aq) + 2OH-

(aq) + 2H+(aq) + SO4

2-(aq) → 2Na+

(aq) + SO42-

(aq) + 2H2O(l)

2Na+(aq) + 2OH-

(aq) + 2H+(aq) + SO4

2-(aq) → 2Na+

(aq) + SO42-

(aq) + 2H2O(l)

Cancel out

Chemical Eqn

Cancel out


Molecular eqn

Complete ionic eqn

Net ionic eqn

Net ionic eqn

Complete ionic eqn

Molecular eqn

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

3(NH4)2CO3(aq) + 2Al(NO3)3(aq) → 6NH4NO3(aq) + Al2(CO3)3(s)

Ca2+(aq) + CO3

2-(aq) → CaCO3 (s)

Ca2+(aq)+ 2Cl- (aq) + 2Na+

(aq)+ CO32- (aq)→ 2Na+

(aq) + 2Cl-(aq) +

CaCO3 (s)

Ca2+(aq)+ 2Cl-

(aq) + 2Na+(aq)+ CO3

2-(aq) → 2Na+

(aq) + 2Cl-(aq) +

CaCO3 (s)

OH-(aq) + H+

(aq) → H2O(l)

2Na+(aq) + 2OH-

(aq) + 2H+(aq) + SO4

2-(aq) → 2Na+

(aq) + SO42-

(aq) + 2H2O(l)

2Na+(aq) + 2OH-

(aq) + 2H+(aq) + SO4

2-(aq) → 2Na+

(aq) + SO42-

(aq) + 2H2O(l)

Cancel out

Chemical Eqn

Cancel out


Molecular eqn

Complete ionic eqn

Net ionic eqn

Net ionic eqn

Complete ionic eqn

Molecular eqn

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

CaCl2 (aq) + Na2CO3(aq) → 2 NaCl(aq) + CaCO3(s)

4 g H2 508 g I2

512 g HI 150 g NO2

300 g N2O4

Simulation conservation mass/balancing eqn

Click to view animation

+2 moles H2 2 moles I2 4 moles HI

150 g NO2

3 moles NO23 moles NO2 3 moles N2O4+ +

3NO2 + 3NO2 → 3N2O4 2H2 + 2I2 → 4HI

= +

Video on stoichiometry

=

Concept Map

Chemical Rxn Chemical change Chemical eqn Balancing chemical eqn

Molecular Eqn

1Pb(NO3)2(s) + 2KI(aq) → 1PbI2(s) + 2KNO3 (aq)

1Pb2+(aq) + 2NO3

-(aq) + 2K+

(aq) + 2I-(aq) → 1PbI2(s) + 2K+

(aq)

+ 2NO3-(aq)

1Pb2+(aq) + 2CI-

(aq) → 1PbCI2(s)

Coefficient Mole ratio (reactant) →

(product) 1 : 2 → 1

: 2

Complete ionic eqn

Net ionic eqn

Reaction StoichiometryQuantitative relationship bet quantities react/

prodFind quantities/amt (mass, mole, vol)

Predicts how much react and amt prod formChemical rxn react in definite ratio

http://glencoe.mcgraw-hill.com/olcweb/cgi/pluginpop.cgi?it=swf::800::600::/sites/dl/free/0035715985/117354/Stoch_Nav.swf::Stoichiometry%20Simulation

Exercise

Write balanced eqn for following rxn1. Rxn of sulphur dioxide and oxygen to form sulphur trioxide2SO2(g) + O2(g) → 2SO3(g)

2. Neutralization bet potassium hydroxide and sulphuric acid to form potassium sulphate and water2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(I)

3. Combustion of ethane (C2H6) to form carbon dioxide and water 2C2H6(g) + 7O2(g ) → 4CO2(g) + 6H2O(I)

4. Displacement rxn bet zinc metal and copper(II) sulphate sol to form copper metal and zinc sulphate Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)

5. Decomposition of zinc carbonate to form zinc oxide and carbon dioxide when heated ZnCO3(s) → ZnO(s) + CO2(g)

6. Ammonia react with oxygen to form nitrogen monoxide and water. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(I)

• Manganese(IV) oxide react with hydrochloric acid to form manganese(II) chloride sol, chlorine and water MnO2(s) + 4HCI(aq) → MnCI2(aq) + CI2(g) + 2H2O(I)

•Neutralization bet aq ammonia with hydrochloric acid to form ammonium chloride and water. NH4OH(aq) + HCI(aq) → NH4CI(aq) + H2O(I)

Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/

Prepared by Lawrence Kok

Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com

ib chemistry limiting, excess, theoretical and percentage yield

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