Chemical Reaction

Word equation

Chemical equation

Chemical formula

Reactants – Left side Products – Right side

Conservation Mass Total 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

After Before

Video on conservation mass 1 : 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)

Chemical Reaction

Word equation

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

Physical states + symbols (s) – solid (I) - liqud (g) – gas (aq) – aqueous ∆ - heating ppt – precipitate/solid ↔ - reversible

Calcium + hydrochloric → Calcium + carbon + water carbonate 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)

Reactants – Left side Products – Right side

Conservation Mass Total 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

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

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)

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

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 - Buns Limiting - 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

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 ?

1

2

3

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 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)

0.52 mol HCI 0.30 mol Zn

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

Simulation on limiting/excess

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 ?

1

2

3

4

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 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)

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

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 ?

1

2

3

4

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 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)

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

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 ?

1

2

3

4

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 H2O 0.02 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

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 ?

1

2

3

4

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

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

1

2

3

4

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% 0.074g Percentage 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

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

Balanced Chemical equation Coefficient Mole proportion/ratio Limiting

Reagent Excess

Reagent

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