IB Chemistry on Limiting, Excess and Percentage Yield.

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Prepared by Lawrence Kok

Tutorial on Limiting, Excess and Percentage Yield

Chemical Reaction

Word equation

Chemical equation

Chemical formula

Lead + Potassium → Lead + Potassium Nitrate iodide iodide nitrate

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

Chemical Reaction

Word equation

Chemical equation

Chemical formula

Reactants – Left side Products – Right side

Conservation MassTotal Mass reactants = Total Mass products

Mole Ratio (stoichiometric ratio)Coefficient in front of reactants/products - moles



1 : 2 → 1 : 2


Chemical Reaction

Word equation

Chemical equation

Chemical formula






Mass of reactants (PbNO3 + KI) = 15.82

Mass of products (PbI3 + KNO3) = 15.82

AfterBefore

Video on conservation mass1 : 2 → 1 : 2

Chemical reaction • matter is neither created nor destroyed • Undergoes physical and chemical change. • LAW of conservation of mass.


Chemical Reaction

Word equation

CaCO3 (s) + 2HCI → CaCI2(aq) + CO2(g) + H2O(l)

Physical states + symbols

(s) – solid(I) - liqud(g) – gas(aq) – aqueous ∆ - heatingppt –

precipitate/solid↔ - reversible

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

Chemical equation

Chemical formula

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

+ 1H2O(l)


+ 1H2O(l)


Chemical Reaction

Word equation






Reaction Stoichiometry• Quantitative relationship bet quantities of reactants/ products• Determine quantities/amt (mass, moles, volume) • Predicts how much reactants react and amt products formed• Chemical rxn react in definite ratios

Chemical equation

Chemical formula


+ 1H2O(l)


+ 1H2O(l)




1 : 2 → 1 : 1 : 2

Chemical Reaction

Word equation






Reaction Stoichiometry• Quantitative relationship bet quantities of reactants/ products• Determine quantities/amt (mass, moles, volume) • Predicts how much reactants react and amt products formed• Chemical rxn react in definite ratios

Chemical equation

Chemical formula


+ 1H2O(l)


+ 1H2O(l)




1 : 2 → 1 : 1 : 2 Video on conservation mass

Before After

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

Balanced Chemical equation

Concept Map

Chemical Reaction

Chemical Equation

Molecular

Equation

CompleteIonic

EquationNet Ionic

Equation

1Pb2+(aq) + 2NO3

-(aq) + 2Na+

(aq) + 2I-(aq) → 1PbI2(s) + 2Na+

(aq)

+ 2NO3-(aq)

1Pb2+(aq) + 2CI-

(aq) → 1PbCI2(s)

Chemical Change

leads to

represented by



Coefficient• Mole proportion/ratio •(reactant) → (product) 1 : 2 → 1 : 2

Concept Map

Chemical Reaction

Chemical Equation

Molecular

Equation

CompleteIonic

EquationNet Ionic

Equation

1Pb2+(aq) + 2NO3

-(aq) + 2Na+

(aq) + 2I-(aq) → 1PbI2(s) + 2Na+

(aq)

+ 2NO3-(aq)

1Pb2+(aq) + 2CI-

(aq) → 1PbCI2(s)

Chemical Change

leads to

represented by


Stoichiometry• Quantitative relationship bet quantities of reactants/products• Determine quantities/amt in (mass, moles, vol) • Predicts amt reactants react and amt products formed• Chemical rxn reacts in definite ratios

Limiting reactant –Use up first- Limit products form- Rxn stop if all used up

Excess reactant – left over- remains behind

Percentage Yield

mass of Actual Yield x 100%mass of Theoretical Yield - Moles /mass product can be used

Theoretical yield- Max amt product form if rxn completed- Stoichiometry ratio / ideal condition- Assume all limiting reagents used up

Actual yield- Amt of product formed experimentally- Less than theoretical yield due to experimental error



Coefficient• Mole proportion/ratio •(reactant) → (product) 1 : 2 → 1 : 2

Concept Map

Chemical Reaction

Chemical Equation

Molecular

Equation

CompleteIonic

EquationNet Ionic

Equation

1Pb2+(aq) + 2NO3

-(aq) + 2Na+

(aq) + 2I-(aq) → 1PbI2(s) + 2Na+

(aq)

+ 2NO3-(aq)

1Pb2+(aq) + 2CI-

(aq) → 1PbCI2(s)

Chemical Change

leads to

represented by


Stoichiometry• Quantitative relationship bet quantities of reactants/products• Determine quantities/amt in (mass, moles, vol) • Predicts amt reactants react and amt products formed• Chemical rxn reacts in definite ratios

Video on concept map above

Limiting reactant –Use up first- Limit products form- Rxn stop if all used up

Excess reactant – left over- remains behind

Percentage Yield

mass of Actual Yield x 100%mass of Theoretical Yield - Moles /mass product can be used

Theoretical yield- Max amt product form if rxn completed- Stoichiometry ratio / ideal condition- Assume all limiting reagents used up

Actual yield- Amt of product formed experimentally- Less than theoretical yield due to experimental error

Limiting and Excess

Limiting reactant – use up first, limits the products form- rxn stops if all used up

Excess reactant – left over, remains behind

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

+5 5 5

No Excess No limiting

Both hot dog and bun are used up

Limiting and Excess



Which is limiting and excess ?

How many hot dogs with 6 buns and 3 hot dogs?


+5 5 5


Excess - BunsLimiting - Hot dogs are used up


+ +

Limiting and Excess




How many hot dogs with 6 buns and 3 hot dogs?


+5 5 5

+


Excess - BunsLimiting - Hot dogs are used up


How many burgers with 12 buns and 6 patties?

+ +

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


Simulation on limiting/excess

Balanced chemical eqn

Mole of reactants added

Mole ratio/stoichiometry ratio

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

Mole ratio

1 : 2 → 1: 1

0.30 mol Zn + 0.52 mol HCl added

Moles reactants given, which is limiting and excess ?

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

1 mol Zn react 2 mol HCI


0.52 mol HCI0.30 mol Zn





Mole ratio

1 : 2 → 1: 1


HCI is limiting


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1st method




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 NOT enough (limiting) = 0.52 (added) < 0.60 (needed)






Mole ratio

1 : 2 → 1: 1


HCI is limiting


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1st method 2nd method




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 NOT enough (limiting) = 0.52 (added) < 0.60 (needed)


Reactants that produce least amt of 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 2 = 0.26 mol H2

HCI is limiting



Mass of reactants added

Mole ratio/stoichiometry ratio Mole ratio

1 : 2 → 1: 2

10.0g Pb(NO3)2 + 10.0g NaI added

Mass reactants given, which is limiting and excess ?

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44


1 mol Pb(NO3)2 react 2 mol NaI 10.0g Pb(NO3)2 + 10.0g NaI added

0.0302 mol Pb(NO3)2 + 0.0667 mol NaI


Mass → Moles Mass = 10.0

RMM 331.2

= 0.0302 mol

Mass = 10.0

RMM 149.9

= 0.0667 mol




1 : 2 → 1: 2


Pb(NO3)2 is limiting


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1st method


0.0302 mol Pb(NO3)2 + 0.0667 mol NaI

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, (excess) = 0.0667 (added) > 0.0604 (needed)



RMM 331.2

= 0.0302 mol

Mass = 10.0

RMM 149.9

= 0.0667 mol

NaI is excess




1 : 2 → 1: 2




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0.0302 mol Pb(NO3)2 + 0.0667 mol NaI

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, (excess) = 0.0667 (added) > 0.0604 (needed)

Reactants that produce least amt of product → will be limiting

Assume Pb(NO3)2 limiting 1 mol Pb(NO3)2→ 1 mol PbI2

0.0302 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 2 = 0.0334 mol PbI2




RMM 331.2

= 0.0302 mol

Mass = 10.0

RMM 149.9

= 0.0667 mol


NaI is excess




1 : 2 → 1: 1

0.623g Mg + 27.3cm3, 1.25M HCI add

Mass (solid) and Vol/Conc (solution) given, which is limiting and excess ?

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1 mol Mg react 2 mol HCI 0.623g Mg + 27.3cm3, 1.25M HCI

0.0256 mol Mg + 0.0341 mol HCI

Mg(s) + 2HCI(aq) → 1MgCI2(aq) + H2 (g)

Mass /Conc → Moles

Mole = Mass RMM= 0.623 = 0.0256 mol 24.31

Mole = M x V 1000= 1.25 x 27.3 = 0.0341 mol 1000




1 : 2 → 1: 1

0.623g Mg + 27.3cm3, 1.25M HCI add

HCI is limiting


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1st method


0.0256 mol Mg + 0.0341 mol HCI

1 mol Mg → 2 mol HCI0.0256 mol Mg → 2 x 0.0512 mol HCI = 0.0512 mol HCI need = 0.0341 mol HCI add, (limit) = 0.0341 (add) < 0.0512 (need)



Mole = Mass RMM= 0.623 = 0.0256 mol 24.31

Mole = M x V 1000= 1.25 x 27.3 = 0.0341 mol 1000




1 : 2 → 1: 1

0.623g Mg + 27.3cm3, 1.25M HCI add

HCI is limiting


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44




0.0256 mol Mg + 0.0341 mol HCI

1 mol Mg → 2 mol HCI0.0256 mol Mg → 2 x 0.0512 mol HCI = 0.0512 mol HCI need = 0.0341 mol HCI add, (limit) = 0.0341 (add) < 0.0512 (need)

Reactants produce least amt of product → will be limiting

Assume Mg limiting 1 mol Mg→ 1 mol H2

0.0256 mol Mg→ 0.0256mol H2

= 0.0256 mol H2

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

HCI is limiting



Mole = Mass RMM= 0.623 = 0.0256 mol 24.31

Mole = M x V 1000= 1.25 x 27.3 = 0.0341 mol 1000



Vol/Conc solution added


2 : 1 → 1: 1

100ml, 0.2M, NaOH + 50.0ml, 0.5M H2SO4 add

Vol/Conc (solution) given, which is limiting and excess ?

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2 mol NaOH react 1 mol H2SO4

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

0.02 mol NaOH + 0.025 mol H2SO4

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

Vol/Conc → Moles

Mole = M x V 1000= 0.5 x 50 = 0.025 mol 1000

Mole = M x V 1000= 0.2 x 100 = 0.02 mol 1000




2 : 1 → 1: 1


NaOH is limiting


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1st method


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


2 mol NaOH → 1 mol H2SO4

0.02 mol NaOH → 1 x 0.02 mol H2SO4

2 = 0.01 mol H2SO4need = 0.025 mol H2SO4add, (excess) = 0.025 (add) > 0.01 (need)


Vol/Conc → Moles

Mole = M x V 1000= 0.5 x 50 = 0.025 mol 1000

Mole = M x V 1000= 0.2 x 100 = 0.02 mol 1000

H2SO4 is excess




2 : 1 → 1: 1


NaOH is limiting


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44




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


2 mol NaOH → 1 mol H2SO4

0.02 mol NaOH → 1 x 0.02 mol H2SO4

2 = 0.01 mol H2SO4need = 0.025 mol H2SO4add, (excess) = 0.025 (add) > 0.01 (need)


Assume NaOH limiting 2 mol NaOH→ 1 mol H2O0.02 mol NaOH→ 1 x 0.02 mol H2O 2 = 0.01 mol H2O

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

NaOH is limiting


Vol/Conc → Moles

Mole = M x V 1000= 0.5 x 50 = 0.025 mol 1000

Mole = M x V 1000= 0.2 x 100 = 0.02 mol 1000

H2SO4 is excess

Click here for animation

2 mol CO react 1 mol O2

45.42L CO + 11.36L O2

2 mol CO + 0.5 mol O2


Vol gas added


2: 1 → 2

45.42L CO + 11.36L O2 add

Vol (gas) given, which is limiting and excess ?

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2CO(g) + 1O2(g) → 2CO2 (g)

Vol → Moles

Mole = Vol molar vol= 45.42 = 2.0 mol 22.4



45.42L CO + 11.36L O2



Vol gas added


2: 1 → 2

45.42L CO + 11.36L O2 add

O2 is limiting


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1st method

2 mol CO → 1 mol O2


= 1 mol O2 need = 0.5 mol O2 add, (limit) = 0.5 (add) < 1 (need)

2CO(g) + 1O2(g) → 2CO2 (g)

Vol → Moles




45.42L CO + 11.36L O2



Vol gas added


2: 1 → 2

45.42L CO + 11.36L O2 add

O2 is limiting


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44





= 1 mol O2 need = 0.5 mol O2 add, (limit) = 0.5 (add) < 1 (need)


Assume CO limiting 2 mol CO→ 2 mol CO2

2 mol CO→ 2 mol CO2

= 2 mol CO2

Assume O2 limiting 1 mol O2 → 2 mol CO2 0.5 mol O2 → 2 x 0.5 mol CO2 = 1 mol CO2 O2 is limiting

2CO(g) + 1O2(g) → 2CO2 (g)

Vol → Moles



Click here for animation




2 → 2: 1

1.00g HgO add

Theoretical, Actual and Percentage Yield

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2HgO(s) → 2Hg(s) + O2(g)

Mass → Moles Mass = 1.00 = 4.6 x 10-3 mol RMM 216.6

Calculate percentage yield O2 , when 1.00g HgO was added. (Actual yield from expt is 0.069g)




2 → 2: 1

1.00g HgO add


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2HgO(s) → 2Hg(s) + O2(g)


Theoretical yield- Max amt product form if rxn complete- Stoichiometry ratio/ideal condition - Assume all limiting reagents used up

Actual yield- Amt of product form experimentally- Less than theoretical yield due to experimental error

Percentage Yield

mass of Actual Yield x 100%mass of Theoretical Yield - Moles/mass product can be used

2HgO(s) → 2Hg(s) + O2(g)





2 → 2: 1

1.00g HgO add


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Theoretical yield O2 = 0.074g

Actual yield of O2 = 0.069g

Percentage yield = 93.2%

2 mol HgO→ 1 mol O2

4.6 x 10-3 mol HgO→ 4.6 x 10-3 mol O2

2 Mole = 2.23 x 10-3 mol O2

x RMM O2(32)

Mass = 2.23 x 10-3 x 32 Theoretical yield = 0.074g O2

2HgO(s) → 2Hg(s) + O2(g)



Theoretical yield- Max amt product form if rxn complete- Stoichiometry ratio/ideal condition - Assume all limiting reagents used up

Actual yield- Amt of product form experimentally- Less than theoretical yield due to experimental error

Percentage Yield

mass of Actual Yield x 100%mass of Theoretical Yield - Moles/mass product can be used

2HgO(s) → 2Hg(s) + O2(g)

Percentage = Mass of Actual Yield x 100% Yield Mass of Theoretical Yield = 0.069g x 100% 0.074gPercentage Yield = 93.2%


Actual yield given = 0.069g O2

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

Stoichiometry


Coefficient

Mole proportion/ratio

Limiting

Reagent

Excess Reagen

t

Percentage yield

Concept Map

Actual/experimental yield

Theoretical yield

Chemical formula

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

Stoichiometry


Coefficient

Mole proportion/ratio

Limiting

Reagent

Excess Reagen

t

Percentage yield

Click here for limiting excess notes

Click here for online tutorial

Click here notes

Click here tutorial on austute

Concept Map

Actual/experimental yield

Theoretical yield

Chemical formula

Online tutorial limiting/excess

Click here tutorial on chemwiki Click here tutorial on chemtamu

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 on Limiting, Excess and Percentage Yield.

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Transcript of IB Chemistry on Limiting, Excess and Percentage Yield.