1 4.6 Excess, Limiting Amounts, Percentage Yield, and Impurities 1.
IB Chemistry on Limiting, Excess and Percentage Yield.
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Transcript of IB Chemistry on Limiting, Excess and Percentage Yield.
![Page 1: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/1.jpg)
http://lawrencekok.blogspot.com
Prepared by Lawrence Kok
Tutorial on Limiting, Excess and Percentage Yield
![Page 2: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/2.jpg)
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)
![Page 3: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/3.jpg)
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
Lead + Potassium → Lead + Potassium Nitrate iodide iodide nitrate
1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq)
1 : 2 → 1 : 2
1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq)
![Page 4: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/4.jpg)
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
Lead + Potassium → Lead + Potassium Nitrate iodide iodide nitrate
1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq)
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.
1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq)
![Page 5: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/5.jpg)
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)
1CaCO3(s) + 2HCI(aq) → 1CaCI2(aq) + 1CO2(g)
+ 1H2O(l)
Reactants – Left side Products – Right side
![Page 6: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/6.jpg)
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
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
1CaCO3(s) + 2HCI(aq) → 1CaCI2(aq) + 1CO2(g)
+ 1H2O(l)
1CaCO3(s) + 2HCI(aq) → 1CaCI2(aq) + 1CO2(g)
+ 1H2O(l)
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 : 1 : 2
![Page 7: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/7.jpg)
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
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
1CaCO3(s) + 2HCI(aq) → 1CaCI2(aq) + 1CO2(g)
+ 1H2O(l)
1CaCO3(s) + 2HCI(aq) → 1CaCI2(aq) + 1CO2(g)
+ 1H2O(l)
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 : 1 : 2 Video on conservation mass
Before After
![Page 8: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/8.jpg)
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
![Page 9: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/9.jpg)
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
Balanced Chemical equation
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
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
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
![Page 10: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/10.jpg)
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
Balanced Chemical equation
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
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
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
![Page 11: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/11.jpg)
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
![Page 12: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/12.jpg)
Limiting and Excess
Limiting reactant – use up first, limits the products form- rxn stops if all used up
Excess reactant – left over, remains behind
Which is limiting and excess ?
How many hot dogs with 6 buns and 3 hot dogs?
Stoichiometric ratio /proportion 1 mol (bun) : 1 mol (hot dog) → 1 mol
+5 5 5
No Excess No limiting
Excess - BunsLimiting - Hot dogs are used up
Both hot dog and bun are used up
+ +
![Page 13: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/13.jpg)
Limiting and Excess
Limiting reactant – use up first, limits the products form- rxn stops if all used up
Excess reactant – left over, remains behind
Which is limiting and excess ?
How many hot dogs with 6 buns and 3 hot dogs?
Stoichiometric ratio /proportion 1 mol (bun) : 1 mol (hot dog) → 1 mol
+5 5 5
+
No Excess No limiting
Excess - BunsLimiting - Hot dogs are used up
Both hot dog and bun are used up
How many burgers with 12 buns and 6 patties?
+ +
Stoichiometric ratio/proportion 2 mol (bun) : 1 mol (burger) → 1 mol
No Excess No limiting
Simulation on limiting/excess
![Page 14: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/14.jpg)
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 ?
11
22
33
Which is limiting and excess ?
1Zn (s) + 2HCI(aq) → 1ZnCI2(aq) + 1H2(g)
1 mol Zn react 2 mol HCI
0.30 mol Zn + 0.52 mol HCl added
0.52 mol HCI0.30 mol Zn
![Page 15: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/15.jpg)
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
HCI is limiting
Moles reactants given, which is limiting and excess ?
11
22
33
Which is limiting and excess ?
1st method
1Zn (s) + 2HCI(aq) → 1ZnCI2(aq) + 1H2(g)
1 mol Zn react 2 mol HCI
0.30 mol Zn + 0.52 mol HCl added
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)
0.52 mol HCI0.30 mol Zn
![Page 16: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/16.jpg)
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
HCI is limiting
Moles reactants given, which is limiting and excess ?
11
22
33
Which is limiting and excess ?
1st method 2nd method
1Zn (s) + 2HCI(aq) → 1ZnCI2(aq) + 1H2(g)
1 mol Zn react 2 mol HCI
0.30 mol Zn + 0.52 mol HCl added
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)
0.52 mol HCI0.30 mol Zn
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
Simulation on limiting/excess
![Page 17: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/17.jpg)
Balanced chemical eqn
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 ?
11
22
33
44
Which is limiting and excess ?
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
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
Mass → Moles Mass = 10.0
RMM 331.2
= 0.0302 mol
Mass = 10.0
RMM 149.9
= 0.0667 mol
![Page 18: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/18.jpg)
Balanced chemical eqn
Mass of reactants added
Mole ratio/stoichiometry ratio Mole ratio
1 : 2 → 1: 2
10.0g Pb(NO3)2 + 10.0g NaI added
Pb(NO3)2 is limiting
Mass reactants given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
1st method
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
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)
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
Mass → Moles Mass = 10.0
RMM 331.2
= 0.0302 mol
Mass = 10.0
RMM 149.9
= 0.0667 mol
NaI is excess
![Page 19: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/19.jpg)
Balanced chemical eqn
Mass of reactants added
Mole ratio/stoichiometry ratio Mole ratio
1 : 2 → 1: 2
10.0g Pb(NO3)2 + 10.0g NaI added
Pb(NO3)2 is limiting
Mass reactants given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
1st method 2nd method
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
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
Pb(NO3)2 is limiting
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
Mass → Moles Mass = 10.0
RMM 331.2
= 0.0302 mol
Mass = 10.0
RMM 149.9
= 0.0667 mol
Simulation on limiting/excess
NaI is excess
![Page 20: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/20.jpg)
Balanced chemical eqn
Mass of reactants added
Mole ratio/stoichiometry ratio Mole ratio
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 ?
11
22
33
44
Which is limiting and excess ?
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
![Page 21: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/21.jpg)
Balanced chemical eqn
Mass of reactants added
Mole ratio/stoichiometry ratio Mole ratio
1 : 2 → 1: 1
0.623g Mg + 27.3cm3, 1.25M HCI add
HCI is limiting
Mass (solid) and Vol/Conc (solution) given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
1st method
1 mol Mg react 2 mol HCI 0.623g Mg + 27.3cm3, 1.25M HCI
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)
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
![Page 22: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/22.jpg)
Balanced chemical eqn
Mass of reactants added
Mole ratio/stoichiometry ratio Mole ratio
1 : 2 → 1: 1
0.623g Mg + 27.3cm3, 1.25M HCI add
HCI is limiting
Mass (solid) and Vol/Conc (solution) given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
1st method 2nd method
1 mol Mg react 2 mol HCI 0.623g Mg + 27.3cm3, 1.25M HCI
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
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
Simulation on limiting/excess
![Page 23: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/23.jpg)
Balanced chemical eqn
Vol/Conc solution added
Mole ratio/stoichiometry ratio Mole ratio
2 : 1 → 1: 1
100ml, 0.2M, NaOH + 50.0ml, 0.5M H2SO4 add
Vol/Conc (solution) given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
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
![Page 24: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/24.jpg)
Balanced chemical eqn
Vol/Conc solution added
Mole ratio/stoichiometry ratio Mole ratio
2 : 1 → 1: 1
100ml, 0.2M, NaOH + 50.0ml, 0.5M H2SO4 add
NaOH is limiting
Vol/Conc (solution) given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
1st method
2 mol NaOH react 1 mol H2SO4
100ml, 0.2M NaOH + 50ml, 0.5M H2SO4
0.02 mol NaOH + 0.025 mol 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)
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
H2SO4 is excess
![Page 25: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/25.jpg)
Balanced chemical eqn
Vol/Conc solution added
Mole ratio/stoichiometry ratio Mole ratio
2 : 1 → 1: 1
100ml, 0.2M, NaOH + 50.0ml, 0.5M H2SO4 add
NaOH is limiting
Vol/Conc (solution) given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
1st method 2nd method
2 mol NaOH react 1 mol H2SO4
100ml, 0.2M NaOH + 50ml, 0.5M H2SO4
0.02 mol NaOH + 0.025 mol 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)
Reactants produce least amt of product → will be limiting
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
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
H2SO4 is excess
Click here for animation
![Page 26: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/26.jpg)
2 mol CO react 1 mol O2
45.42L CO + 11.36L O2
2 mol CO + 0.5 mol O2
Balanced chemical eqn
Vol gas added
Mole ratio/stoichiometry ratio Mole ratio
2: 1 → 2
45.42L CO + 11.36L O2 add
Vol (gas) given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
2CO(g) + 1O2(g) → 2CO2 (g)
Vol → Moles
Mole = Vol molar vol= 45.42 = 2.0 mol 22.4
Mole = Vol molar vol= 11.36 = 0.5 mol 22.4
![Page 27: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/27.jpg)
2 mol CO react 1 mol O2
45.42L CO + 11.36L O2
2 mol CO + 0.5 mol O2
Balanced chemical eqn
Vol gas added
Mole ratio/stoichiometry ratio Mole ratio
2: 1 → 2
45.42L CO + 11.36L O2 add
O2 is limiting
Vol (gas) given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
1st method
2 mol CO → 1 mol O2
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
Mole = Vol molar vol= 45.42 = 2.0 mol 22.4
Mole = Vol molar vol= 11.36 = 0.5 mol 22.4
![Page 28: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/28.jpg)
2 mol CO react 1 mol O2
45.42L CO + 11.36L O2
2 mol CO + 0.5 mol O2
Balanced chemical eqn
Vol gas added
Mole ratio/stoichiometry ratio Mole ratio
2: 1 → 2
45.42L CO + 11.36L O2 add
O2 is limiting
Vol (gas) given, which is limiting and excess ?
11
22
33
44
Which is limiting and excess ?
1st method 2nd method
2 mol CO → 1 mol O2
2 mol CO → 1 mol O2
= 1 mol O2 need = 0.5 mol O2 add, (limit) = 0.5 (add) < 1 (need)
Reactants produce least amt of product → will be limiting
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
Mole = Vol molar vol= 45.42 = 2.0 mol 22.4
Mole = Vol molar vol= 11.36 = 0.5 mol 22.4
Click here for animation
![Page 29: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/29.jpg)
Balanced chemical eqn
Mass of reactants added
Mole ratio/stoichiometry ratio Mole ratio
2 → 2: 1
1.00g HgO add
Theoretical, Actual and Percentage Yield
11
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33
44
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)
![Page 30: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/30.jpg)
Balanced chemical eqn
Mass of reactants added
Mole ratio/stoichiometry ratio Mole ratio
2 → 2: 1
1.00g HgO add
Theoretical, Actual and Percentage Yield
11
22
33
44
2HgO(s) → 2Hg(s) + O2(g)
Mass → Moles Mass = 1.00 = 4.6 x 10-3 mol RMM 216.6
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)
Calculate percentage yield O2 , when 1.00g HgO was added. (Actual yield from expt is 0.069g)
![Page 31: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/31.jpg)
Balanced chemical eqn
Mass of reactants added
Mole ratio/stoichiometry ratio Mole ratio
2 → 2: 1
1.00g HgO add
Theoretical, Actual and Percentage Yield
11
22
33
44
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)
Mass → Moles Mass = 1.00 = 4.6 x 10-3 mol RMM 216.6
Simulation on limiting/excess
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%
Calculate percentage yield O2 , when 1.00g HgO was added. (Actual yield from expt is 0.069g)
Actual yield given = 0.069g O2
![Page 32: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/32.jpg)
Pb(NO3)2 (s) + 2KI(aq) → PbI2(s) + 2KNO3 (aq)
Stoichiometry
Balanced Chemical equation
Coefficient
Mole proportion/ratio
Limiting
Reagent
Excess Reagen
t
Percentage yield
Concept Map
Actual/experimental yield
Theoretical yield
Chemical formula
![Page 33: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/33.jpg)
Pb(NO3)2 (s) + 2KI(aq) → PbI2(s) + 2KNO3 (aq)
Stoichiometry
Balanced Chemical equation
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
![Page 34: IB Chemistry on Limiting, Excess and Percentage Yield.](https://reader038.fdocuments.in/reader038/viewer/2022102616/55500d62b4c905af648b474a/html5/thumbnails/34.jpg)
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