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

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IB Chemistry on Limiting, Excess and Percentage Yield.

IB Chemistry on Limiting, Excess and Percentage Yield.

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

    • Tutorial on Limiting, Excess and Percentage Yield Prepared by Lawrence Kok http://lawrencekok.blogspot.com
    • Chemical Reaction Chemical equation Word equation Lead + Potassium → Nitrate iodide Lead + Potassium iodide nitrate Chemical formula 1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq)
    • Chemical Reaction Chemical equation Word equation Lead + Potassium → Nitrate iodide Lead + Potassium iodide nitrate Chemical formula 1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq) Reactants – Left side Products – Right side 1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq) Mole Ratio (stoichiometric ratio) Coefficient in front of reactants/products - moles 1:2→1:2 Conservation Mass Total Mass reactants = Total Mass products
    • Chemical Reaction Chemical equation Word equation Lead + Potassium → Nitrate iodide Chemical formula 1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq) Lead + Potassium iodide nitrate Reactants – Left side Products – Right side 1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq) Mole Ratio (stoichiometric ratio) Coefficient in front of reactants/products - moles 1:2→1:2 Mass of reactants (PbNO3 + KI) = 15.82 Before Conservation Mass Total Mass reactants = Total Mass products Video on conservation mass Mass of products (PbI3 + KNO3) = 15.82 After Chemical reaction • matter is neither created nor destroyed • Undergoes physical and chemical change. • LAW of conservation of mass.
    • Chemical Reaction Chemical equation Word equation Calcium + hydrochloric →Calcium + carbon + water carbonate acid chloride dioxide Reactants – Left side Chemical formula CaCO3 (s) + 2HCI → CaCI2(aq) + CO2(g) + H2O(l) Products – Right side 1CaCO3(s) + 2HCI(aq) → 11CaCI2(aq)+ 1CO2(g) + 1H2O(l) 1CaCO3(s) + 2HCI(aq) → CaCI2(aq) + 1CO2(g) + 1H2O(l) Physical states + symbols (s) – solid (I) - liqud (g) – gas (aq) – aqueous ∆ - heating ppt – precipitate/solid ↔ - reversible
    • Chemical Reaction Chemical equation Word equation Calcium + hydrochloric →Calcium + carbon + water carbonate acid chloride dioxide Reactants – Left side Chemical formula CaCO3 (s) + 2HCI → CaCI2(aq) + CO2(g) + H2O(l) Products – Right side 1CaCO3(s) + 2HCI(aq) → 11CaCI2(aq)+ 1CO2(g) + 1H2O(l) 1CaCO3(s) + 2HCI(aq) → CaCI2(aq) + 1CO2(g) + 1H2O(l) Mole Ratio (stoichiometric ratio) Coefficient in front of reactants/products - moles 1 : 2 → 1 :1: 2 Conservation Mass Total Mass reactants = Total Mass products 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 Physical states + symbols (s) – solid (I) - liqud (g) – gas (aq) – aqueous ∆ - heating ppt – precipitate/solid ↔ - reversible
    • Chemical Reaction Chemical equation Word equation Calcium + hydrochloric →Calcium + carbon + water carbonate acid chloride dioxide Reactants – Left side Chemical formula CaCO3 (s) + 2HCI → CaCI2(aq) + CO2(g) + H2O(l) Products – Right side 1CaCO3(s) + 2HCI(aq) → 11CaCI2(aq)+ 1CO2(g) + 1H2O(l) 1CaCO3(s) + 2HCI(aq) → CaCI2(aq) + 1CO2(g) + 1H2O(l) Mole Ratio (stoichiometric ratio) Coefficient in front of reactants/products - moles Physical states + symbols (s) – solid (I) - liqud (g) – gas (aq) – aqueous ∆ - heating ppt – precipitate/solid ↔ - reversible Video on conservation mass 1 : 2 → 1 :1: 2 Conservation Mass Total Mass reactants = Total Mass products 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 Before After
    • Concept Map Chemical Reaction leads to Chemical Change Molecular Equation 1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) represented by Chemical Equation Balanced Chemical equation Complete Ionic Equation Net Ionic Equation 1Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2I-(aq) → 1PbI2(s) + 2Na+(aq) + 2NO3-(aq) 1Pb2+(aq) + 2CI-(aq) → 1PbCI2(s)
    • Concept Map Chemical Reaction leads to Molecular Equation Chemical Change 1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) represented by Complete Ionic Equation Chemical Equation Balanced Chemical equation Limiting reactant –Use up first - Limit products form - Rxn stop if all used up 1Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2I-(aq) → 1PbI2(s) + 2Na+(aq) + 2NO3-(aq) 1Pb2+(aq) + 2CI-(aq) → 1PbCI2(s) Net Ionic Equation Excess reactant – left over - remains behind Coefficient • Mole proportion/ratio •(reactant) → (product) 1 : 2 → 1 : 2 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 1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) 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 Percentage Yield mass of Actual Yield mass of Theoretical Yield x 100% - Moles /mass product can be used
    • Concept Map Chemical Reaction leads to Molecular Equation Chemical Change 1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) represented by Complete Ionic Equation Chemical Equation Balanced Chemical equation Limiting reactant –Use up first - Limit products form - Rxn stop if all used up 1Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2I-(aq) → 1PbI2(s) + 2Na+(aq) + 2NO3-(aq) 1Pb2+(aq) + 2CI-(aq) → 1PbCI2(s) Net Ionic Equation Excess reactant – left over - remains behind Coefficient • Mole proportion/ratio •(reactant) → (product) 1 : 2 → 1 : 2 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 1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) 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 Video on concept map above Percentage Yield mass of Actual Yield mass of Theoretical Yield x 100% - Moles /mass product can be used
    • Limiting and Excess Excess reactant Limiting reactant – use up first, limits the products form - rxn stops if all used up – left over, remains behind + 5 Stoichiometric ratio /proportion 1 mol (bun) : 1 mol (hot dog) → 1 mol 5 5 No Excess No limiting Both hot dog and bun are used up
    • Limiting and Excess Excess reactant Limiting reactant – use up first, limits the products form - rxn stops if all used up – left over, remains behind + 5 5 5 Stoichiometric ratio /proportion 1 mol (bun) : 1 mol (hot dog) → 1 mol No Excess No limiting Both hot dog and bun are used up How many hot dogs with 6 buns and 3 hot dogs? + + Which is limiting and excess ? Excess - Buns Limiting - Hot dogs are used up
    • Limiting and Excess Excess reactant Limiting reactant – use up first, limits the products form - rxn stops if all used up – left over, remains behind + 5 5 5 Stoichiometric ratio /proportion 1 mol (bun) : 1 mol (hot dog) → 1 mol No Excess No limiting Both hot dog and bun are used up How many hot dogs with 6 buns and 3 hot dogs? + + Excess - Buns Limiting - Hot dogs are used up Which is limiting and excess ? 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
    • Moles reactants given, which is limiting and excess ? 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mole of reactants added 1Zn (s) + 2HCI (aq) → 1ZnCI2(aq) + 1H2(g) Mole ratio 1 : 2 → 1: 1 0.30 mol Zn + 0.52 mol HCl added 1Zn (s) + 2HCI(aq) → 1ZnCI2(aq) + 1H2(g) 1 mol Zn react 2 mol HCI 0.30 mol Zn 0.52 mol HCI 0.30 mol Zn + 0.52 mol HCl added Which is limiting and excess ?
    • Moles reactants given, which is limiting and excess ? 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mole of reactants added 1Zn (s) + 2HCI (aq) → 1ZnCI2(aq) + 1H2(g) Mole ratio 1 : 2 → 1: 1 0.30 mol Zn + 0.52 mol HCl added 1Zn (s) + 2HCI(aq) → 1ZnCI2(aq) + 1H2(g) 1 mol Zn react 2 mol HCI 0.30 mol Zn 0.52 mol HCI 0.30 mol Zn + 0.52 mol HCl added Which is limiting and excess ? 1st method 1 mol Zn → 2 mol HCI 0.30 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) HCI is limiting
    • Moles reactants given, which is limiting and excess ? 11 Balanced chemical eqn 22 1Zn (s) + 2HCI (aq) → 1ZnCI2(aq) + 1H2(g) Mole ratio/stoichiometry ratio 33 Mole of reactants added Mole ratio 1 : 2 → 1: 1 Simulation on limiting/excess 0.30 mol Zn + 0.52 mol HCl added 1Zn (s) + 2HCI(aq) → 1ZnCI2(aq) + 1H2(g) 1 mol Zn react 2 mol HCI 0.30 mol Zn 0.52 mol HCI 0.30 mol Zn + 0.52 mol HCl added Which is limiting and excess ? 1st method 1 mol Zn → 2 mol HCI 0.30 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) HCI is limiting 2nd method Reactants that produce least amt of product → will be limiting Assume Zn limiting 1 mol Zn → 1 mol H2 gas 0.3 mol Zn → 1 x 0.3 = 0.3 mol H2 Assume HCI limiting 2 mol HCI → 1 mol H2 gas 0.52 mol HCI → 1 x 0.52 2 = 0.26 mol H2 HCI is limiting
    • Mass reactants given, which is limiting and excess ? 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mass of reactants added 1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) 44 Mass → Moles Mole ratio 1 : 2 → 1: 2 10.0g Pb(NO3)2 + 10.0g NaI added Mass = 10.0 RMM 331.2 = 0.0302 mol Mass = 10.0 RMM 149.9 = 0.0667 mol 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 Which is limiting and excess ?
    • Mass reactants given, which is limiting and excess ? 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) Mass of reactants added 44 Mass → Moles Mole ratio 1 : 2 → 1: 2 10.0g Pb(NO3)2 + 10.0g NaI added Mass = 10.0 RMM 331.2 = 0.0302 mol Mass = 10.0 RMM 149.9 = 0.0667 mol 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 Which is limiting and excess ? 1st method 1 mol Pb(NO3)2 → 2 mol NaI 0.0302 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) NaI is excess Pb(NO3)2 is limiting
    • Mass reactants given, which is limiting and excess ? 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq) Mass of reactants added 44 Mass → Moles Mole ratio 1 : 2 → 1: 2 10.0g Pb(NO3)2 + 10.0g NaI added Mass = 10.0 RMM 331.2 = 0.0302 mol Mass = 10.0 RMM 149.9 = 0.0667 mol Simulation on limiting/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 Which is limiting and excess ? 1st method 1 mol Pb(NO3)2 → 2 mol NaI 0.0302 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) NaI is excess Pb(NO3)2 is limiting 2nd method 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 Pb(NO3)2 is limiting Assume NaI limiting 2 mol NaI → 1 mol PbI2 0.0667 mol NaI → 1 x 0.0667 2 = 0.0334 mol PbI2
    • Mass (solid) and Vol/Conc (solution) given, which is limiting and excess ? Mg(s) + 2HCI(aq) → 1MgCI2(aq) + H2 (g) 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mass of reactants added 44 Mass /Conc → Moles Mole ratio 1 : 2 → 1: 1 0.623g Mg + 27.3cm3, 1.25M HCI add 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 mol Mg react 2 mol HCI 0.623g Mg + 27.3cm3, 1.25M HCI 0.0256 mol Mg + 0.0341 mol HCI Which is limiting and excess ?
    • Mass (solid) and Vol/Conc (solution) given, which is limiting and excess ? Mg(s) + 2HCI(aq) → 1MgCI2(aq) + H2 (g) 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mass of reactants added 44 Mass /Conc → Moles Mole ratio 1 : 2 → 1: 1 0.623g Mg + 27.3cm3, 1.25M HCI add 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 mol Mg react 2 mol HCI 0.623g Mg + 27.3cm3, 1.25M HCI 0.0256 mol Mg + 0.0341 mol HCI Which is limiting and excess ? 1st method 1 mol Mg → 2 mol HCI 0.0256 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) HCI is limiting
    • Mass (solid) and Vol/Conc (solution) given, which is limiting and excess ? Mg(s) + 2HCI(aq) → 1MgCI2(aq) + H2 (g) 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mass of reactants added 44 Mass /Conc → Moles Mole ratio 1 : 2 → 1: 1 0.623g Mg + 27.3cm , 1.25M HCI add Simulation on limiting/excess 3 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 mol Mg react 2 mol HCI 0.623g Mg + 27.3cm3, 1.25M HCI 0.0256 mol Mg + 0.0341 mol HCI Which is limiting and excess ? 2nd method 1st method 1 mol Mg → 2 mol HCI 0.0256 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) HCI is limiting 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
    • Vol/Conc (solution) given, which is limiting and excess ? 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Vol/Conc solution added 44 Vol/Conc → Moles 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) Mole ratio 2 : 1 → 1: 1 100ml, 0.2M, NaOH + 50.0ml, 0.5M H2SO4 add Mole = M x V 1000 = 0.2 x 100 = 0.02 mol Mole = M x V 1000 = 0.5 x 50 = 0.025 mol 1000 1000 2 mol NaOH react 1 mol H2SO4 100ml, 0.2M NaOH + 50ml, 0.5M H2SO4 0.02 mol NaOH + 0.025 mol H2SO4 Which is limiting and excess ?
    • Vol/Conc (solution) given, which is limiting and excess ? 11 22 Mole ratio/stoichiometry ratio 33 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) Balanced chemical eqn Vol/Conc solution added Mole ratio 2 : 1 → 1: 1 100ml, 0.2M, NaOH + 50.0ml, 0.5M H2SO4 add Mole = M x V 1000 = 0.2 x 100 = 0.02 mol 44 Vol/Conc → Moles Mole = M x V 1000 = 0.5 x 50 = 0.025 mol 1000 1000 2 mol NaOH react 1 mol H2SO4 100ml, 0.2M NaOH + 50ml, 0.5M H2SO4 0.02 mol NaOH + 0.025 mol H2SO4 Which is limiting and excess ? 1st method 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) H2SO4 is excess NaOH is limiting
    • Vol/Conc (solution) given, which is limiting and excess ? 11 22 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) Balanced chemical eqn Mole ratio/stoichiometry ratio 33 Mole ratio 2 : 1 → 1: 1 100ml, 0.2M, NaOH + 50.0ml, 0.5M H2SO4 add Vol/Conc solution added Mole = M x V 1000 = 0.2 x 100 = 0.02 mol 44 Vol/Conc → Moles Click here for animation Mole = M x V 1000 = 0.5 x 50 = 0.025 mol 1000 1000 2 mol NaOH react 1 mol H2SO4 100ml, 0.2M NaOH + 50ml, 0.5M H2SO4 0.02 mol NaOH + 0.025 mol H2SO4 Which is limiting and excess ? 1st method 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) H2SO4 is excess NaOH is limiting 2nd method Reactants produce least amt of product → will be limiting Assume NaOH limiting 2 mol NaOH→ 1 mol H2O 0.02 mol NaOH→ 1 x 0.02 mol H2O 2 = 0.01 mol H2O NaOH is limiting Assume H2SO4 limiting 1 mol H2SO4 → 1 mol H2O 0.025 mol H2SO4 → 0.025 mol H2O = 0.025 mol H2O
    • Vol (gas) given, which is limiting and excess ? 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Vol gas added 2CO(g) + 1O2(g) → 2CO2 (g) 44 Vol → Moles Mole ratio 2: 1 → 2 45.42L CO + 11.36L O2 add Mole = Vol molar vol = 45.42 = 2.0 mol 22.4 Mole = Vol molar vol = 11.36 = 0.5 mol 22.4 2 mol CO react 1 mol O2 45.42L CO + 11.36L O2 2 mol CO + 0.5 mol O2 Which is limiting and excess ?
    • Vol (gas) given, which is limiting and excess ? 11 22 Mole ratio/stoichiometry ratio 33 2CO(g) + 1O2(g) → 2CO2 (g) Balanced chemical eqn Vol gas added 44 Vol → Moles Mole ratio 2: 1 → 2 45.42L CO + 11.36L O2 add Mole = Vol molar vol = 45.42 = 2.0 mol 22.4 Mole = Vol molar vol = 11.36 = 0.5 mol 22.4 2 mol CO react 1 mol O2 45.42L CO + 11.36L O2 2 mol CO + 0.5 mol O2 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) O2 is limiting
    • Vol (gas) given, which is limiting and excess ? 11 22 2CO(g) + 1O2(g) → 2CO2 (g) Balanced chemical eqn Mole ratio/stoichiometry ratio 33 Mole ratio 2: 1 → 2 45.42L CO + 11.36L O2 add Vol gas added 44 Vol → Moles Click here for animation Mole = Vol molar vol = 45.42 = 2.0 mol 22.4 Mole = Vol molar vol = 11.36 = 0.5 mol 22.4 2 mol CO react 1 mol O2 45.42L CO + 11.36L O2 2 mol CO + 0.5 mol O2 Which is limiting and excess ? 2nd method 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) O2 is limiting 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
    • Theoretical, Actual and Percentage Yield 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mass of reactants added 44 Mass → Moles 2HgO(s) → 2Hg(s) + O2(g) Mole ratio 2 → 2: 1 1.00g HgO add 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)
    • Theoretical, Actual and Percentage Yield 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mass of reactants added 44 Mass → Moles 2HgO(s) → 2Hg(s) + O2(g) Mole ratio 2 → 2: 1 1.00g HgO add 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) 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
    • Theoretical, Actual and Percentage Yield 11 Balanced chemical eqn 22 Mole ratio/stoichiometry ratio 33 Mass of reactants added 44 Mass → Moles 2HgO(s) → 2Hg(s) + O2(g) Simulation on limiting/excess Mole ratio 2 → 2: 1 1.00g HgO add 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) 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 2 mol HgO→ 1 mol O2 4.6 x 10-3 mol HgO→ 4.6 x 10-3 mol O2 2 3 Mole = 2.23 x 10- mol O2 x RMM O2(32) Mass = 2.23 x 10-3 x 32 Theoretical yield = 0.074g O2 Actual yield - Amt of product form experimentally - Less than theoretical yield due to experimental error Actual yield given = 0.069g O2 Percentage Yield mass of Actual Yield x 100% mass of Theoretical Yield - Moles/mass product can be used Percentage = Mass of Actual Yield x 100% Yield Mass of Theoretical Yield = 0.069g x 100% 0.074g Percentage Yield = 93.2% Theoretical yield O2 = 0.074g Actual yield of O2 = 0.069g Percentage yield = 93.2%
    • Concept Map Theoretical yield Actual/experimental yield Chemical formula Mole proportion/ratio Percentage yield Pb(NO3)2 (s) + 2KI(aq) → PbI2(s) + 2KNO3 (aq)  Coefficient Limiting Reagent Excess Reagent Stoichiometry Balanced Chemical equation
    • Concept Map Theoretical yield Actual/experimental yield Chemical formula Mole proportion/ratio Percentage yield Pb(NO3)2 (s) + 2KI(aq) → PbI2(s) + 2KNO3 (aq)  Coefficient Limiting Reagent Balanced Chemical equation Excess Reagent Click here notes Stoichiometry Click here for limiting excess notes Click here tutorial on austute Online tutorial limiting/excess Click here tutorial on chemwiki Click here tutorial on chemtamu Click here for online tutorial
    • Acknowledgements Thanks to source of pictures and video used in this presentation Thanks to Creative Commons for excellent contribution on licenses http://creativecommons.org/licenses/ Prepared by Lawrence Kok Check out more video tutorials from my site and hope you enjoy this tutorial http://lawrencekok.blogspot.com