A chemical reaction has happened when new substances are formed. All
chemical compounds can be represented by a chemical formula showing
the ratio of atoms or ions within the simplest formula unit. A chemical
equation shows the chemical change that happens in terms of the substances
reacting (the reactants) and the new substances produced (the products):
Reactants → Products
Syllabus reference R2.1.2
Reactivity 2.1.2 - 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.
- Calculate reacting masses and/or volumes and concentrations of reactants and products.
Guidance
- 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.
Tools and links
- Structure 1.5 - How does the molar volume of a gas vary with changes in temperature and pressure?
- Nature of science, Structure 1.4 - In what ways does Avogadro’s law help us to describe, but not explain, the behaviour of gases?
Stoichiometry
The stoichiometry is shown by the relative coefficients of the components
appearing in the chemical reaction.
Example
4NH3 + 5O2 → 4NO
+ 6H2O
The stoichoimetry of the reaction tells us that 4
molecules of ammonia react with 5 molecules
of oxygen and produce 4 molecules of nitrogen
monoxide and 6 molecules of water
1 mole is equivalent to 6.02 x 1023 molecules therefore the stoichiometry
also gives us the ratio of moles reacting.
Consider the reaction:
Fe + S → FeS
The stoichiometry of the equation shows us that one atom of iron is needed
to react with each atom of sulfur. Extending this idea we can see that the
same number of iron and sulfur atoms are always needed for a complete reaction.
Therefore the moles of iron are always equal to the moles of sulfur in this
reaction.
If we are told the mass of iron that we start with is 5.6g then we can calculate
the mass of sulfur needed. The calculation proceeds via the number of moles.
[Relative atomic mass of Fe=56, S=32]
Moles of iron = mass /RAM = 5.6/56 = 0.1 moles
Therefore moles of sulfur = 0.1 moles
RAM of sulfur = 32
Therefore mass of sulfur needed = moles x RAM = 0.1 x 32 = 3.2g
The procedure followed is:
- 1. Determine the moles of one of the components.
- 2. Use the stoichometric ratio to calculate the moles of the component that you require.
- 3. Change mol of required component to mass, volume, as desired.
Example: Calculate the mass of chlorine needed
to completely react with 2.24g of iron [relative atomic masses Fe=56,
Cl=35.5]
2Fe + 3Cl2 →
2FeCl3
Mass of iron = 2.24g, therefore moles of iron = 2.24/56 = 0.04 moles
from the equation stoichiometry it can be seen that 2 moles of iron
react with 3 moles of chlorine
therefore 0.04 moles iron react with 0.04 x 3/2 moles chlorine molecules
= 0.06 moles
relative mass of Cl2 = 2 x 35.5 = 71
therefore 0.06 moles of chlorine = 0.06 x 71 = 4.26g
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Worked examples
Q212-01 Calculate the mass
of Magnesium required to completely with 2.54g of iodine according to the equation:
Mg + I2 → MgI2
Answer
|
Relative atomic mass of Mg = 24, Relative mass of I2
= 254
Equation shows that 1 mole of magnesium reacts with 1 mole of iodine
Moles of iodine = 2.54/254 = 0.01 moles
Therefore moles of magnesium needed = 0.01 moles
1 mole Mg = 24g
Therefore mass of magnesium needed = 24 x 0.01 = 0.24
g
|
Q212-02 Calculate the mass
of sodium required to completely with 3.2g of oxygen, according to the equation:
4Na + O2 →
2Na2O
Answer
|
Relative atomic mass of Na = 23, Relative mass of O2
= 32
Equation shows that 4 moles of sodium react with 1 mole of oxygen
Moles of oxygen = 3.2/32 = 0.1 moles
Therefore moles of sodium needed = 0.1 x 4 = 0.4 moles
1 mole Na = 23g
Therefore mass of sodium needed = 23 x 0.4 = 9.2
g
|
Q212-03 Calculate the mass
of copper required to completely with 1.6g of oxygen, according to the equation:
2Cu + O2 →
2CuO
Answer
|
Relative atomic mass of copper = 63.5, Relative mass of
O2 = 32
Equation shows that 2 moles of copper react with 1 mole of oxygen
Moles of oxygen = 1.6/32 = 0.05 moles
Therefore moles of copper needed = 0.05 x 2 = 0.1 moles
1 mole Cu = 63.5g
Therefore mass of copper needed = 63.5 x 0.1 = 6.35
g
|
Q212-04 Calculate the mass
of sulfur required to react completely with 5.6g of iron according to the equation:
Fe + S →
FeS
Answer
|
Relative atomic mass of sulfur = 32, Relative atomic mass of Iron
= 56
Equation shows that 1 moles of sulfur reacts with 1 mole of iron
Moles of iron = 5.6/56 = 0.1 moles
Therefore moles of sulfur needed = 0.1 moles
1 mole sulfur = 32g
Therefore mass of sulfur needed = 32 x 0.1 = 3.2
g
|
Q212-05 In a thermit reaction
aluminium reacts with iron(III) oxide according to the equation:
2Al + Fe2O3 →
Al2O3 + 2Fe
Calculate the mass of aluminium needed to fully react with 80g of iron(III)
oxide.
Answer
|
Relative atomic mass of aluminium = 27, Relative formula mass of
iron(III) oxide = (56x2) + (16x3) = 160
Equation shows that 2 moles of aluminium reacts with 1 mole of iron(III)
oxide
Moles of iron(III) oxide = 80/160 = 0.5 moles
Therefore moles of aluminium needed = 0.5 x 2 = 1 moles
1 mole aluminium = 27g
Therefore mass of aluminium needed = 27
g
|
Q212-06 The reaction between
iodine and tin proceeds according to the equation:
Sn + 2I2 →
SnI4
Calculate the mass of iodine needed to fully react with 2.38g of tin [Ar
119].
Answer
|
Relative mass of iodine = 254, Relative atomic mass of
tin = 119
Equation shows that 2 moles of iodine reacts with 1 mole of tin
Moles of tin = 2.38/119 = 0.02 moles
Therefore moles of iodine needed = 0.02 x 2 = 0.04 moles
1 mole iodine = 254g
Therefore mass of iodine needed = 0.04 x 254 = 10.16
g
|
Q212-07 The reaction between
ammonium chloride and calcium hydroxide proceeds according to the equation:
2NH4Cl + Ca(OH)2 →
2NH3 + CaCl2 + 2H2O
Calculate the mass of ammonium chloride needed to produce 34g of ammonia.
Answer
|
Relative formula mass of ammonium chloride = (14 + 4 + 35.5) = 53.5,
Relative mass of ammonia = 17
Equation shows that 2 moles of ammonium chloride makes 2 moles of
ammonia
i.e. moles of ammonium chloride = moles of ammonia produced
Moles of ammonia to be produced = 34/17 = 2 moles
Therefore moles of ammonium chloride needed = 2 moles
1 mole ammonium chloride = 53.5g
Therefore mass of ammonium chloride needed = 2 x 53.5 = 107
g
|
Q212-08 Calculate the mass
of aluminium needed to produce 1.12kg of iron using the thermite reaction, assuming
that there is enough iron(III) oxide:
2Al + Fe2O3 →
Al2O3 + 2Fe
Answer
|
From the equation 2 moles of aluminium are needed to produce 2 moles
of iron
i.e. moles of aluminium = moles of iron
Mass of iron required = 1.12kg = 1120g
Moles of iron required = 1120/56 = 20 moles
Therefore moles of aluminium required = 20 moles
1 mole of aluminium has a mass = 27g
Therefore 20 moles of aluminium = 20 x 27 = 540g
needed
|
Q212-09 Calculate the mass
of carbon dioxide produced when 10g of calcium carbonate is completely dissolved
in excess hydrochloric acid. [Ca=40, O=16, C=12]
Answer
|
Equation for the reaction:
CaCO3 + 2HCl →
CaCl2 + CO2 + H2O
From the data given Mr (CaCO3) = 40 + 12 +
16 + 16 + 16 =100
Moles of CaCO3 = 10.0/100 = 0.1 moles
From the equation 1 mole CaCO3 produces 1 mole CO2
Therefore 0.1 moles CaCO3 produces 0.1 moles CO2
Mr of CO2 = 12 + 16 + 16 = 44
Mass = moles x Mr = 0.1 x 44 = 4.4g
of CO2 produced
|
Q212-10 Calculate the mass
of dibromocyclohexane formed when 20.5g of cyclohexene, C
6H
10,
reacts with excess bromine water, according to the equation:
C6H10 + Br2 →
C6H10Br2
Answer
| Relative mass (Mr) of C6H10 = (6 x
12) + (10 x 1) = 82
Moles of C6H10 = 20.5/82 = 0.25 moles
From the equation, this makes the same number of moles of C6H10Br2
Mr (C6H10Br2) = 72 + 10 + 80 + 80
= 242
therefore 0.25 moles of C6H10Br2
= 0.25 x 242 = 60.5g
|
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Now test yourself
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