Transcript Understanding Chemical Reactions

Understanding
Chemical
Reactions
Lesson: Calculations in
Chemistry 2
Page 175 Q 4
a) One mole of ZnO has a RAM of
65 + 16 = 81g
0.2 moles will have a mass of
0.2 X 81 = 16.2 g
Page 175 Q 4
b) 1 mole of H2S has a mass of
(2x1) + 32 = 34g
2.5 moles will have a mass of
2.5 X 34 = 85 g
Page 175 Q 4
c) 1 mole of CuSO4 has a mass of
63.5 + 32 + (16 x 4) = 259.5g
0.45 moles will have a mass of
0.45 X 259.5 = 116.8 g
Calculating
formulae
Page 178 of Chemistry
text
Working out the formula of
magnesium oxide
• A student heated
some Mg as
shown.
• When Mg burns in
air it combines
with oxygen to
make magnesium
oxide.
Working out the formula of
magnesium oxide
Step 1 - Results
• Mass of crucible +
lid + Mg before
heating = 25.24g
• Mass of crucible
and lid = 25.00g
• Therefore, mass of
Mg = 0.24g
Working out the formula of
magnesium oxide
Step 2 - Results
• Mass of crucible + lid
+ magnesium oxide
after heating =
25.40g
• Mass of crucible + lid
+ Mg before heating=
25.24g
• Therefore, mass of
oxygen in magnesium
oxide = 0.16g
Working out the formula of
magnesium oxide
Step 3 – Change the masses into moles
Magnesium = 0.24 / 24 = 0.01 mole
Oxygen = 0.16 / 16 = 0.01 mole
Working out the formula of
magnesium oxide
Step 4 – Work out the ratio of moles
Mg
0.01
1
:
:
:
O
0.01
1
Therefore the formula of magnesium oxide
is MgO
One for you!
• A compound of nitrogen and
hydrogen was broken down into its
elements.
• It was found that 1.4g of nitrogen
had combined with 0.3g of hydrogen
in the compound.
• What was the formula of the
compound?
• (R.A.M.s N = 14, H = 1)
One for you!
Step 1 – work out the number of
moles
Moles of N = 1.4 / 14 = 0.1
Moles of H = 0.3 / 1 = 0.3
One for you!
Step 2 – work out the ratio of the number of
moles to the lowest whole numbers
N
:
H
0.1
:
0.3
1
:
3
Therefore there is 3 times as many H atoms
as N atoms.
Its formula must be NH3
Reacting iron with copper
sulphate
• A more reactive metal will displace a
less reactive metal from its solution:
• Iron
• Fe(s)
+ copper sulphate
+ CuSO4(aq)
iron sulphate
FeSO4(aq)
+ copper
+ Cu(s)
• How much copper will 0.5g of iron
produce?
Reacting iron with copper
sulphate
• Fill your beaker with copper sulphate
solution.
• Add precisely 5g of iron to the
beaker and stir gently for 4 minutes.
• Filter the mixture – making sure all
the solid is removed from the beaker
– rinse beaker with water if needed.
Reacting iron with copper
sulphate
• Rinse the filter paper with propanone
to dry the copper.
• When dry – re-weigh the solid.
• 1 mole of Fe should produce 1 mole
of Cu
• 5 g of Fe is 5 / 56 = 0.09 moles
• 0.09 moles of Fe should produce
0.09 moles of Cu
• 0.09 moles of Cu will have a mass of
0.09 x 64 = 5.7 g