In chemical processes the ratio of reacting particles is always fixed
for a specific reaction.
Once the number of particles of any component of a known reaction is
determined, it is possible to find the number of particles involved
of any other component by using the stoichoimetry of the equation.
Syllabus reference R2.1.1
Reactivity 2.1.1 - Chemical equations show the ratio of reactants and products in a reaction.
- Deduce chemical equations when reactants and products are specified.
Guidance
- Include the use of state symbols in chemical equations.
Word equations
Word equations simply show the names of the reacting chemical and products.
They are of limited use except for giving an overall description of the chemical
reaction. They give no indication of the relative amounts of the reactants
or products involved.
Sodium hydroxide + sulfuric
acid → sodium sulfate
+ water
To make equations useful, they must show the individual formulae of the reactants
and products and indicate the relative quantities in which they react. These
'formula equations' are dealt with below.
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Formula equations and coefficients
Formula equations show the formulae of the reactants and the products on
either side. Balancing numbers are used - called the coefficients of the reaction
- to ensure that the numbers of particles on both sides of the equation are
equal.
In this equation one nitrogen molecule is needed to react with every three
hydrogen molecules to produce 2 molecules of ammonia. The coefficient of nitrogen
is 1, that of hydrogen is 3, and that of ammonia is 2.
To balance a chemical equation it is important to remember that the formula
of the reactants and products cannot be changed and that coefficients may
only be placed before the formulae, multiplying them by whole numbers.
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Example: One stage in the manufacture of nitric acid is the oxidation
of ammonia, as shown below:
4NH3 + _O2 →
_NO + _H2O
What is the coefficient for O2 when the equation is balanced?
Solution:
Balance the nitrogen by counting up the nitrogen atoms on both sides:
4NH3 + _O2 →
4NO + _H2O
Balance the hydrogen atoms: 4NH3 + _O2 →
4NO + 6H2O
Now balance the oxygen atoms: 4NH3 + 5O2 →
4NO + 6H2O
Correct response : Coefficient = 5
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These show the actual numbers of reacting particles in a chemical reaction.
The reaction must be BALANCED to give the correct number of particles on each
side of the reaction arrow.
2NaOH + H2SO4 →
Na2SO4 + 2H2O
These equations are constructed by writing the formula of each of the compounds
in the reaction, and then by counting up the number of atoms on each side
to make sure they are equal. If they are not equal, balancing numbers (coefficients)
are added in front of each chemical formula (where needed), so that the numbers
of each type of particle on each side of the equation are the same
Step 1 - write the chemical equation
| ammonia + oxygen → nitrogen
monoxide + water |
Step 2 - write the formula of each of the reaction components
ammonia + oxygen →
nitrogen monoxide + water
NH3 + O2 →
NO + H2O
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Step 3 - add coefficients IN FRONT OF the formulae to balance the equation
Note Whenever an exam question asks for an equation,
it is the balanced formula equation that is required, unless specified
otherwise.
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Ionic equations
When ionic solutions react, the reaction only usually involves some of the
ions and not others. An ionic equation shows just the ions implicated in the
reaction. The other ions are often called "spectator ions".
2NaOH + H2SO4 → Na2SO4 + 2H2O
The ionic equation is written as:
H+ + OH- → H2O
Although the formula equation represents the overall process, the sodium
ions start off in solution as Na+(aq) and at the end of the reaction
they are still Na+(aq), nothing has changed, they are merely spectator ions.
The same applies to the sulfate ions, SO42-. The only
particles that actually react are the OH- ions from the sodium
hydroxide and the H+ ions from the sulfuric acid.
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State symbols
These are used to show the states of the various compounds that constitute
the equation for the reaction.
(s) means that the compound is in the solid state.
(l) means that the compound is in the liquid state.
(g) means that the compound is in the gaseous state.
(aq) means that the compound is dissolved in water, i.e. it is in solution.
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq)
+ 2H2O(l)
The sodium hydroxide and the sulfuric acid are in solution. These make
sodium sulfate in solution and water liquid.
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Worked examples
Q334-01 When the following
equation is correctly balanced what is the coefficient for O2?
- 9
- 13
- 18
- 24
Answer
|
Balance the carbon - C4H10 + O2 → 4CO2
+ H2O
Balance the hydrogen - C4H10 + O2 → 4CO2
+ 5H2O
Now balace the oxygen - C4H10 + 13/2O2 → 4CO2
+ 5H2O
fractions are not allowed in the correctly balanced equation so multiply
through by 2 to remove the fraction.
2C4H10 + 13O2 → 8CO2
+ 10H2O
Response B: Coefficient = 13
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Q334-02 When the following
equation is balanced, what is the coefficient for oxygen?
|
_C2H2(g) + _O2(g) → _CO2(g) +
_H2O(g)
|
- 2
- 3
- 4
- 5
Answer
|
First balance the carbon
_C2H2(g) + _O2(g) →
2CO2(g) + _H2O(g)
Then the hydrogen
_C2H2(g) + _O2(g) →
2CO2(g) + 1H2O(g)
However there are now an odd number of oxygen atoms. This will not
make a whole number of O2 molecules. Therefore the whole
equation must be multiplied through by 2.
2C2H2(g) + _O2(g) →
4CO2(g) + 2H2O(g)
Now balance the oxygen atoms
2C2H2(g) + 5O2(g) → 4CO2(g)
+ 2H2O(g)
Response D: 5 x O2
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Q334-03 Hydrogen sulfide,
H
2S, reacts with oxygen to form sulfur dioxide and water as shown
below:
What is the whole number coefficient for oxygen when this equation is balanced?
- 1
- 2
- 3
- 6
Answer
|
Balance the sulfur
2H2S + _O2 → 2SO2
+_H2O
Then balance the hydrogen
2H2S + _O2 → 2SO2
+ 2H2O
Now work backwards and balance the oxygen
2H2S + 3O2 → 2SO2
+ 2H2O
Response C: 3 moles of oxygen
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Q334-04 When this equation:
| wC4H9OH + xO2 →
yCO2 + zH2O |
is balanced correctly the coefficient, x, for O2 is:
- 6
- 9
- 13/2
- 13
Answer
|
Assume w = 1, balance the carbon atoms:
C4H9OH + xO2 → 4CO2
+ zH2O
balance the hydrogen atoms
C4H9OH + xO2 → 4CO2
+ 5H2O
Balance the oxygen atoms, do not forget the oxygen atom in the butanol:
C4H9OH + 6O2 → 4CO2
+ 5H2O
Response A : 6
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Q334-05 The reaction of lead(II)
sulfide with oxygen at high temperature is represented by the unbalanced equation:
| PbS(s) + O2(g) → PbO(s)
+ SO2(g) |
What is the sum of the coefficents in the balanced equation?
- 5
- 9
- 8
- 13
Answer
|
Inspection of the equation shows that the Pb and the S are balanced,
but the oxygen atoms are unbalanced.
The right hand side must provide an even number of oxygen atoms while
maintaining the coefficients of PbO(s) and SO2(g) equal.
Double the right hand side:
PbS(s) + O2(g) →
2PbO(s) + 2SO2(g)
Now balance the PbS(s) and the O2(g)
2PbS(s) + 3O2(g) → 2PbO(s)
+ 2SO2(g)
Response B : sum of coefficients = 9
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Q334-06 The oxidation of nitrogen
monoxide can be represented by the word equation:
| nitrogen monoxide + oxygen →
nitrogen dioxide |
Write out the balanced equation.
Answer
|
First write down the formulae for each component of the reaction
NO + O2 →
NO2
Now inspect the number of atoms on either side. There are three oxygen
atoms on the left and only two on the right. Multiply the NO by 2
and the NO2 by 2
2NO + O2 →
2NO2
The equation is balanced!
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Q334-07 White phosphorus is
manufactured by heating phosphate ore with sand and
coke in an electric furnace at 1500ºC. The phosphorus(V)
oxide, initially formed in an
inert atmosphere of carbon monoxide, is then reduced by
the coketo phosphorus. The first stage in the process may
be represented by the
unbalanced equation.
|
Ca3(PO4)2 + SiO2
→ CaSiO3
+ P4O10
|
What is the coefficient for silicon dioxide when the equation is correctly
balanced?
Answer
|
From the equation: Ca3(PO4)2 + SiO2
→ CaSiO3
+ P4O10
It may be seen that all of the phosphorus atoms from the calcium
phosphate end up in the phosphorus pentoxide.
Therefore the coefficient for Ca3(PO4)2
must be 2 and working forwards the coefficient for CaSiO3
must be 6 to equalise the calcium atoms on both sides.
2Ca3(PO4)2
+ SiO2 →
6CaSiO3 + P4O10
Working backwards then the coefficient for the SiO2 must
also be = 6
2Ca3(PO4)2
+ 6SiO2 →
6CaSiO3 + P4O10
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Q334-08 When the following
equation is correctly balanced what is the
coefficient for oxygen?
Answer
|
From the equation: C6H6 + O2 →
CO2 + H2O
6 carbon atoms in the benzene must make 6 carbon dioxide molecules
C6H6 + O2 →
6CO2 + H2O
And 6 hydrogen atoms from the benzene must produce 3 water molecules
C6H6 + O2 →
6CO2 + 3H2O
Counting back the oxygens on the right hand side = 15. This would
make an odd number of O2 molecules so the whole eqation
must be mutiplied by 2
2C6H6 + O2 → 12CO2
+ 6H2O
Now there are 30 oxygen atoms on the right hand side = 15 x O2
molecules, giving
2C6H6 + 15O2 → 12CO2
+ 6H2O
Therefore the coefficient for oxygen = 15
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Q334-09 When the following
equation is correctly balanced what is the
coefficient for potassium chlorate(V)?
Answer
|
By inspection of the equation: KClO3 →
KClO4 + KCl
It may be seen that the oxygen atoms must be equalised. They only
occur in two compounds therefore mutiplying each by its opposite number
of oxygens makes them equal, i.e mutiply KClO3 by 4, and
KClO4 by 3.
4KClO3 →
3KClO4 + KCl
The equation is now balanced so the coefficient for potassium chlorate
(V) = 4
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Q334-10 In the manufacture
of antimony, the sulfide ore is
roasted in oxygen to produce antimony oxide according
to the unbalanced equation:
What is the coefficient for oxygen when the equation is correctly
balanced?
Answer
|
Inspection of the equation: Sb2S3 + O2 → SO2 + Sb2O3
Shows us that all of the sulfur in the antimony sulfide Sb2S3
turns to sulfur dioxide SO2
Sb2S3 + O2 →
3SO2 + Sb2O3
There are now 9 x oxgen atoms on the right hand side - an odd number
so the equation must be multiplied by 2
2Sb2S3 + O2 → 6SO2
+ 2Sb2O3
now there are 18 oxygen atoms on the right hand side = 9 x O2
molecules
2Sb2S3 +
9O2 →
6SO2 + 2Sb2O3
Therefore the coefficient for oxygen = 9
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Q334-11 Write an equation
to represent the following reaction:
| sodium + oxygen → sodium
oxide |
Answer
|
Step 2: write down the symbols for each component of the equation
Na + O2 →
Na2O
Step 3: balance the numbers of atoms on either side of the arrow
4Na + O2 →
2Na2O
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Q334-12 Write a formula equation
to represent the following reaction:
|
sodium carbonate + hydrochloric acid →
sodium chloride + carbon dioxide + water
|
Answer
|
Step 2: write down the symbols for each component of the equation
Na2CO3 + HCl →
NaCl + CO2 + H2O
Step 3: balance the numbers of atoms on either side of the arrow
Na2CO3 + 2HCl → 2NaCl
+ CO2 + H2O
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Q334-13 Write an equation
to represent the following decomposition:
| Calcium nitrate → calcium
oxide + nitrogen dioxide + oxygen |
Answer
|
Step 2: write down the symbols for each component of the equation
Ca(NO3)2 →
CaO + NO2 + O2
Step 3: balance the numbers of atoms on either side of the arrow
2Ca(NO3)2 →
2CaO + 4NO2
+ O2
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Q334-14 Write an equation
to represent the following neutralisation:
| Calcium hydroxide + hydrochloric acid →
calcium chloride + water |
Answer
|
Step 2: write down the symbols for each component of the equation
Ca(OH)2 + HCl →
CaCl2 + H2O
Step 3: balance the numbers of atoms on either side of the arrow
Ca(OH)2 + 2HCl → CaCl2 +
2H2O
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Q334-15 Write an equation
to represent the following synthesis reaction:
|
Iron + chlorine → iron(III) chloride
|
Answer
|
Step 2: write down the symbols for each component of the equation
Fe + Cl2 → FeCl3
Step 3: balance the numbers of atoms on either side of the arrow
2Fe + 3Cl2 → 2FeCl3
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Q334-16 Write the ionic equation
for the reaction betweeen sulfuric acid and sodium hydroxide.
Answer
|
Word equation:
sulfuric acid and sodium hydroxide →
sodium sulfate + water
formulae:
NaOH(aq) + H2SO4(aq) →
Na2SO4(aq) + H2O(l)
Balanced:
2NaOH(aq) + H2SO4(aq) →
Na2SO4(aq) + 2H2O(l)
Here we can see that the sodium ions (Na+) in the sodium
hydroxide solution remain as sodium ions in the sodium sulfate solution
- they are SPECTATOR IONS.
The same applies to the sulfate ions SO42-.
They start off as sulfate ions in the sulfuric acid solution and end
up as sulfate ions in the sodium sulfate solution - they are also
spectator ions. The only ions that actually react together are the
hydrogen ions from the sulfuric acid and the hydroxide ions from the
sodium hydroxide:
H+(aq) + OH-(aq) →
H2O(l)
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Q334-17 Write an ionic equation
for the reaction between silver nitrate and sodium chloride forming a white
silver chloride precipitate.
Answer
|
Word equation:
silver nitrate + sodium chloride →
sodium nitrate + silver chloride
formulae:
AgNO3(aq) + NaCl(aq) →
NaNO3(aq) + AgCl(s)
Already balanced:
Here we can see that the silver chloride is formed as a solid precipitate.
These ions must then be removed from the solution. All of the other
ions remain behind as spectator ions in solution
Ag+(aq) + Cl-(aq) →
AgCl(s)
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Q334-18 Write an ionic equation
to represent the reaction between barium chloride and magnesium sulfate
Answer
|
Barium chloride and sodium sulfate react together to form a precipitate
of barium sulfate - this is used to identify the sulfate group.
The unbalanced equation
BaCl2(aq) + Na2SO4(aq) → BaSO4(s)
+ NaCl(aq)
balanced
BaCl2(aq) + Na2SO4(aq) → BaSO4(s)
+ 2NaCl(aq)
BUT, the sodium chloride is still in solution so its ions have not
reacted.
The only ions involved in reaction are those that come together to
make the precipitate:
Ba2+(aq) + SO34-(aq) → BaSO4(s)
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Q334-19 Write an ionic equation
to represent the reaction between iron(III) sulfate and sodium hydroxide
Answer
|
iron(III) sulfate and sodium hydroxide react together to form a precipitate
of iron(III) hydroxide - this is used to identify the iron(III) ion.
The unbalanced equation
NaOH(aq) + Fe2(SO4)3(aq) → Na2SO4(aq)
+ Fe(OH)3(s)
balanced
6NaOH(aq) + Fe2(SO4)3(aq) → 3Na2SO4(aq)
+ 2Fe(OH)3(s)
BUT, the sodium ions have not reacted and neither have the sulfate
ions - the sodium sulfate 'produced' is in solution.
The only ions involved in reaction are those that come together to
make the precipitate:
Fe3+(aq) + 3OH-(aq) →
Fe(OH)3(s)
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Q334-20 Write an ionic equation
for the reaction between manganese(IV) oxide and hydrochloric acid.
Answer
|
This is the reaction used to prepare chlorine in the laboratory:
manganese(IV) oxide and hydrochloric acid →
manganese(II) chloride + chlorine + water
The balanced equation:
MnO2(s) + 4HCl(aq) →
MnCl2(aq) + Cl2(g) + 2H2O(l)
Two of the chloride ions from the left hand side are spectator ions
whereas all of the other ions are involved in reaction:
MnO2(s) + 4H+(aq) + 2Cl-(aq) → Mn2+(aq)
+ Cl2(g) + 2H2O(l)
If you check out the charge totals on both sides it may
be seen that they are equal.
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Now test yourself
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