Reactivity 2.
1
IB CHEMISTRY SL
Reactivity 2.1.1 and 2.1.2
Understandings:
• Chemical equations show the ratio of reactants and products in a reaction (2.1.1).
• The mole ratio of an equation can be used to determine:
the masses and/or volumes of reactants and products
the concentrations of reactants and products for reactions occurring in solution
(2.1.2).
Learning outcomes:
• Deduce chemical equations when reactants and products are specified (2.1.1).
• Calculate reacting masses and/or volumes and concentrations of reactants and
products (2.1.2).
Additional notes:
• Include the use of state symbols in chemical equations.
• Avogadro’s law and definitions of molar concentration are covered in Structure
1.4.
• The values for Ar given in the data booklet to two decimal places should be used
in calculations.
Linking questions:
• Reactivity 3.2 When is it useful to use half- equations?
• Structure 1.5 How does the molar volume of a gas vary with changes in
temperature and pressure?
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Balancing chemical equations
• The law of the conservation of mass states that mass is conserved in a chemical
reaction.
• Therefore, there must be the same number of each type of atom in the reactants
and products, as shown below.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
• To balance a chemical equation, we can only change the numbers in front of the
reactants or products which are called coefficients.
Example 1:
• Sodium reacts with chlorine to produce sodium chloride. The unbalanced equation
is shown.
…..Na(s) + .….Cl2(g) → …..NaCl(s)
There is one Na atom in the reactants and one in the products. However, there
are two Cl atoms in the reactants but only one in the products. Write the balanced
equation for the reaction.
Example 2:
• Calcium carbonate reacts with hydrochloric acid to produce calcium chloride,
water and carbon dioxide. The unbalanced equation is shown.
…..CaCO3(s) + …..HCl(aq) → …..CaCl2(aq) + …..H2O(l) + …..CO2(g)
Write the balanced equation for the reaction.
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Exercises: Balance the following chemical equations using whole number coefficients.
When each equation is balanced, determine the sum of coefficients in the equations.
1) …..CH4(g) + …..O2(g) → …..CO2(g) + …..H2O(l) sum of coefficients……..
2) …..C3H8(g) + …..O2(g) → …..CO2(g) + …..H2O(l) sum of coefficients……..
3) …..CH3OH(l) + …..O2(g) → …..CO2(g) + …..H2O(l) sum of coefficients……..
4) …..Mg(s) + …..HCl(aq) → …..MgCl2(aq) + …..H2(g) sum of coefficients……..
5) …..MgCO3(s) + …..HCl(aq) → …..MgCl2(aq) + …..H2O(l) + …..CO2(g)
sum of coefficients……..
6) …..NaCl(s) + …..CaO(aq) → …..CaCl2(aq) + …..Na2O(s) sum of coefficients……..
7) …..Al(s) + …..Fe3O4(s)→ …..Al2O3(s) + …..Fe(s) sum of coefficients……..
8) …..Mg3N2(s) + …..H2SO4(aq) → … ..MgSO4(aq) + ….. (NH4)2SO4(aq)
sum of coefficients……..
9) …..Fe2O3(s) + …..C(s) → …..Fe(s) + …..CO(g) sum of coefficients……..
10) …..Al(OH)3(s) + …..H2SO4(aq) → …..Al2(SO4)3(aq) + …..H2O(l)
sum of coefficients……..
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Mole ratios
• The coefficients in a balanced chemical equation tell us the mole ratios (or molar
ratio) of reactants and products.
• In the equation below we can determine that 2 mol of A react with 3 mol of B to
form 1 mol of C and 2 mol of D.
2A + 3B → C + 2D
Exercises:
1) State the mole ratios in the following chemical equations.
(a) 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l)
CH3OH : H2O
CH3OH : CO2
CH3OH : O2
(b) N2(g) + 3H2(g) → 2NH3(g)
N2 : H2
H2 : NH3
N2 : NH3
2) Determine the amount of the following.
(a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
The amount of CO2 produced from 0.10 mol of CH4
The amount of O2 required to react with 0.75 mol of CH4
(b) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
The amount of H2 produced from 1.50 mol of Mg
The amount of H2 produced from 0.80 mol of HCl
The amount of HCl required to react with 3.00 mol of Mg
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Reactivity 2.1.3
Understandings:
• The limiting reactant determines the theoretical yield.
Learning outcomes:
• Identify the limiting and excess reactants from given data.
Additional notes:
• Distinguish between the theoretical yield and the experimental yield.
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Limiting reactant and excess reactant
• The limiting reactant (reagent) is the reactant that limits the amount of product(s)
that can be formed.
• The excess reactant is the reactant that remains when the limiting reactant is
consumed.
• How many sandwiches can be made with 12 pieces of bread and 7 slices of ham?
• Which is the limiting reactant?
• Which is the excess reactant?
How to determine the limiting and excess reactant.
1. Determine the amount (in mol) of each reactant.
2. Divide the amount of each reactant by its coefficient in the balanced equation.
3. The lowest value is the limiting reactant and the highest is the excess reactant.
Exercises:
1) 50.0 g of N2H4 is reacted with 75.0 g of N2O4 according to the following equation.
2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g)
(a) Determine the limiting and excess reactants.
(b) Determine the amount of excess reactant that remains at the end of the reaction.
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2) 3.00 g of Zn is reacted with 50.0 cm3 of 1.00 mol dm-3 HCl according to the
following equation.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
(a) Determine the limiting and excess reactants.
(b) Determine the amount of excess reactant that remains at the end of the reaction.
(c) 30.0 g of ammonium nitrate (NH4NO3) and 50.0 g of sodium phosphate (Na3PO4)
are reacted together. Determine the limiting and excess reactants.
3NH4NO3 + Na3PO4 → (NH4)3PO4 + 3NaNO3
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Reactivity 2.1.4
Understandings:
• The percentage yield is calculated from the ratio of experimental yield to
theoretical yield.
Learning outcomes:
• Solve problems involving reacting quantities, limiting and excess reactants,
theoretical, experimental and percentage yields.
Theoretical yield and percentage yield
• The theoretical yield is the maximum amount of product that can be formed in a
chemical reaction (based on the stoichiometry of the reaction and amount of the
limiting reactant).
• The actual yield is the actual amount of product that is formed in a chemical
reaction.
• The percentage yield is the actual yield divided by the theoretical yield.
experimental yield
Percentage yield (%) = × 100
theoretical yield
Exercises:
1) A 15.0 g sample of pure K2O produces 7.62 g of K2CO3. Determine the
percentage yield of K2CO3 in the reaction.
4K2O(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g)
2) Aluminium reacts with excess oxygen according to the following equation.
Determine the percentage yield if 20.0 g of Al reacts with oxygen to produce 32.7
g of Al2O3.
4Al(s) + 3O2(g) → 2Al2O3(s)
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3) A 20.0 g sample of pure Fe3O4 produces 5.98 g of Fe. Determine the percentage
yield of Fe in the reaction.
Fe3O4(s) + 4H2(g) → 3Fe(s) + 4H2O(l)
4) 100.0 g of iron(III) oxide is reacted with 100.0 g of carbon. 46.73 g of iron is
produced. Calculate the % yield of Fe.
2Fe2O3(s) + 3C(s) → 4Fe(s) + CO2(g)
5) 15.0 g of CaCO3 is reacted with 50.0 cm3 of 2.00 mol dm-3 HCl. 1.85 g of CO2 is
produced. Calculate the % yield of CO2.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
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Reactivity 2.1.5
Understandings:
• The atom economy is a measure of efficiency in green chemistry.
Learning outcomes:
• Calculate the atom economy from the stoichiometry of a reaction.
Additional notes:
• Include discussion of the inverse relationship between atom economy and
wastage in industrial processes.
• The equation for calculation of the atom economy is given in the data booklet.
Linking questions:
• Structure 2.4, Reactivity 2.2 The atom economy and the percentage yield both
give important information about the “efficiency” of a chemical process. What
other factors should be considered in this assessment?
Atom economy
• The atom economy of a chemical reaction is a measure of the amount of starting
materials that become useful products.
• A high % atom economy means that the reaction has a high efficiency which
results in less waste being produced.
How to calculate the atom economy for a reaction:
1) Calculate the sum of the molar masses of the reactants.
2) Calculate the molar mass of the desired product and multiply by the coefficient.
3) Calculate the % atom economy.
Example: Calculate the atom economy for the following reaction in which ethanol
(C2H4O) is the desired product.
2C2H5OCl + Ca(OH)2 → 2C2H4O + CaCl2 + 2H2O
1) Calculate the sum of the molar masses of the reactants.
2) Calculate the molar mass of the desired product and multiply by the coefficient.
3) Calculate the % atom economy.
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Exercises:
1) Calculate the atom economy for the following reaction for making hydrogen (H2)
by reacting coal with steam. Comment on the efficiency of the reaction.
C(s) + 2H2O(g) → CO2(g) + 2H2(g)
2) Titanium is manufactured by reacting titanium(IV) chloride with magnesium.
Calculate the atom economy for this method of producing titanium. Comment on
the efficiency of the reaction.
TiCl4(s) + 2Mg(s) → Ti(s) + 2MgCl2(s)
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