Download (1) Identify all the species

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Learning aims:
 How to work out the products of a chemical
reaction.
 To complete new material – reacting masses
 To prepare for the theory exam next week
 Carry out practise of practical for practical exam in
two weeks.
News
!
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There are lots of different types of reactions:
We will look at some briefly here.
The only one you will need to know for the
exam is the acid-base reaction
We will come back to others later in the
course.
Periodic table
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
Direct Combination or Synthesis Reaction
In a synthesis reaction two or more chemical species combine
to form a more complex product.
A + B → AB
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8 Fe + S8 → 8 FeS
Chemical Decomposition
In a decomposition reaction a compound is broken into smaller
chemical species.
AB → A + B
The electrolysis of water into oxygen and hydrogen gas is an
example of a decomposition reaction:
2 H2O → 2 H2 + O2
Periodic table

Single Displacement or Substitution Reaction
A substitution or single displacement reaction is characterized by
one element being displaced from a compound by another element.
A + BC → AC + B
e.g. zinc combines with hydrochloric acid, the zinc replaces the
hydrogen:
Zn + 2 HCl → ZnCl2 + H2
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Double Displacement Reaction
In a double displacement or metathesis reaction two compounds
exchange bonds or ions in order to form different compounds.
AB + CD → AD + CB
E.g. sodium chloride and silver nitrate to form sodium nitrate and
silver chloride.
NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
Periodic table
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Oxidation-Reduction or Redox Reaction
Oxidation Is Loss Reduction Is Gain
Redox reactions involve the transfer of electrons between chemical
species.
The reaction that occurs when In which Cl2 is reduced to Cl- and
KBr (bromide anion) is oxidized to bromine provides an example of
a redox reaction:
KBr(aq) + Cl2(aq) → Br2(aq) + 2KCl(aq)
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
A combustion reaction - oxygen combines with another compound
to form carbon dioxide and water.
E.g. the burning of naphthalene:
C10H8 + 12 O2 → 10 CO2 + 4 H2O
Periodic table

An acid-base reaction is type of double
displacement reaction that occurs between an
acid and a base. The H+ ion in the acid reacts
with the OH- ion in the base to form water and an
ionic salt:
HA + BOH → H2O + BA
The reaction between hydrochloric acid (HCl) and
sodium hydroxide is an example of an acid-base
reaction:
HCl + NaOH → NaCl + H2O
Periodic table
Acid + base gives salt and water
HA + BOH  BA + H2O
Let’s try a few:
Periodic table
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Periodic table
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We have learnt how to work out the Mr of a
compound.
We have learnt how to calculate the mass of a
compound by using m = Mr n
And to calculate the number of moles n =
m/Mr
We have used the titration formula to
calculate an unknown concentration or
volume. MAVA = x
MBVB y
Periodic table
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Chemists often need to calculate the amount
of a particular product that will be obtained
from a certain amount of reactant.
Or the amount of reactant required in order
to obtain a certain amount of product.
We always start this calculation off with a
balance equation.
Periodic table
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12 g of magnesium burns in air. Calculate
the mass of magnesium oxide formed
2Mg
+
O2
2MgO
12 g
?g
STAGE 3
convert moles of
MgO into mass of
MgO
STAGE 1
convert mass of Mg
into moles of Mg
moles of Mg
n=m/Mr
m =Mr n
STAGE 2
moles of MgO
use the equation to convert moles
of Mg into moles of MgO
Periodic table
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STAGE 1
n(Mg) =
STAGE 2
n=m/Mr
Mr(Mg) = 24
12
= 0.5 mol
24
from the equation n(Mg) = n(MgO)
n(MgO) = 0.5 mol
STAGE 3
Mr(MgO) = 40
m(MgO) = 0.5 x 40
m =Mr n
= 20 g
Periodic table
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What mass of barium sulphate would be produced from
10 g of barium chloride in the following reaction?
BaCl2 + H2SO4
10 g
BaSO4 + 2HCl

?
n=m/Mr
n(BaCl2)
=
10/208
n(BaSO4)
=
m(BaSO4)
=
=
Mr(BaCl2) = 208
Mr(BaSO4) = 233
0.048 mol
0.048 mol
m =Mr n
0.048 x 233
= 11.184 g
Periodic table
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Note! Even though
we have 2 : 2 ratio
we can still deal with
1 amount to 1
amount because
they are in the same
proportions
What mass of calcium carbonate is needed to remove
10.0 g of SO2 in the following reaction?
2CaCO3(s) + 2SO2(g) + O2  2CaSO4(s) + 2CO2(g)
?
10 g
Mr(SO2) = 64
n=m/Mr
n(SO2)
=
10/64
Mr(CaCO3) = 100
=
n(CaCO3)
=
0.156 mol
m(CaCO3)
=
0.156 x 100
0.156 mol
m =Mr n
=
15.6 g
Periodic table
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What mass of KIO3 is required to give 10 g
the following reaction?
This time we have
a 1:3 ratio, so for
every 10g of
iodine produced
we need 1/3 of
amount
of the
iodine
inof
KIO3
KIO3(aq) + KI(aq) + 6H+(aq)  3I2(aq) + 6K+(aq) + 3H2O(l)
?
10 g
n=m/Mr
n(I2)
=
10/254
Mr(I2) = 254
Mr(KIO3) = 214
=
0.0394 mol
n(KIO3)
=
0.0394/3
m(KIO3)
=
0.0131 x 214
0.0131mol
=
m =Mr n
=
2.81 g
Periodic table
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here we have 2:1 ratio
so we will need ½ the
amount of sodium
carbonate.
What mass of sodium carbonate can be obtained by
heating 100 g of sodium hydrogencarbonate?
2NaHCO3(s)
100 g

Na2CO3(s)
+
?
CO2(g)
+
H2O(g)
Mr(NaHCO3) = 84
Mr(Na2CO3) = 106
n(NaHCO3)
= 100/84
=
n(Na2CO3)
= 1.19/2
m(Na2CO3)
= 0.595 x 106
1.19 mol
0.595 mol
=
=
63.07 g
Periodic table
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What mass of sodium chloride is required to produce 1
kg of sodium hydrogencarbonate in the following
reaction?
NaCl(aq) + NH3(g) + H2O(l) + CO2(g)  NaHCO3(s) + NH4Cl(aq)
?
1000 g
Mr(NaHCO3) = 84
Mr(NaCl) = 58.5
n(NaHCO3)
= 1000/84
=
n(NaCl)
=
11.90 mol
m(NaCl)
=
11.90 x 58.5
11.90 mol
=
696.15 g
Periodic table
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What mass of ethyne can be prepared from 10.0 g of
calcium carbide by the following reaction?
CaC2(s)
+
2H2O(l)

Ca(OH)2(aq)
+
C2H2(g)?
10 g
?
Mr(CaC2) = 64
Mr(C2 H2) = 26
n(CaC2)
=
10/64
=
n(C2H2)
=
0.156 mol
m(C2H2)
=
0.156 x 26
0.156 mol
=
4.056 g
Periodic table
18
Calculate the mass of ammonium chloride required to
produce 1 kg of ammonia
2NH4Cl(s) + Ca(OH)2  2NH3(g) + CaCl2(s) + 2H2O(g)
?
1 kg
Mr(NH4Cl) = 53.5
Mr(NH3) = 17
n(NH3)
=
1000/17
=
n(NH4Cl)
=
58.82 mol
m(NH4Cl)
=
58.82 x 53.5
58.82 mol
=
3.15 kg
Periodic table
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What mass of potassium chloride would be produced
from the reaction of 20 g of potassium carbonate with
excess hydrochloric acid?
K2CO3(s) + 2HCl(aq)  2KCl(aq) + CO2(g) + H2O(l)
20 g
?
Mr(K2CO3) = 138
Mr(KCl) = 74.5
n(K2CO3)
=
20/138
n(KCl)
=
m(KCl)
= 0.290 x 74.5
=
0.145 mol
2 x 0.145 =
=
0.290 mol
21.61 g
Periodic table
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Periodic table
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This is a solution of known concentration.
 This will be 250cm3 0.1M solution of sodium
hydroxide
 We need to know how many grams to weigh
out.
 Use the equation n = MV
1000
moles = 0.1 x 250 / 1000 = 0.025
Convert moles to mass …….m = Mr n
mass = (23+16+1) x 0.025 = 1g

Periodic table
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Periodic Table
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