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Calculations
on the mole
Francine
Taylor-Campbell, Contributor
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LET'S
CONSIDER the following question: 1.
34 cm³
of hydrochloric acid of unknown concentration
required 25 cm³
NaOH of concentration 2.00 mol/dm³
to be completely neutralised.
Write
the equation for the reaction between
sodium hydroxide and hydrochloric
acid. (include/state symbols) Calculate
the number of moles of sodium hydroxide
in 25 cm³
of the solution used.
Calculate
the number of moles of hydrochloric
acid in the volume of hydrochloric
acid used.
Calculate
the number of moles of hydrochloric
acid in 1 dm³
of solution.
ANSWERS
(a)
HCl (aq) + NaOH (aq) = NaCl (aq) +
H2O (l)
(b)
Concentration of NaOH is 2.0 mol/dm³
Thus
2 mol NaOH are present in 1000 cm³
X
mol NaOH is present in 25 cm³
X
mol = (25 x 2)/1000 = 0.050 mol
(c)
Since NaOH and HCl react in a 1:1
ratio, the number of moles of HCl
that reacted is also 0.050 mol which
is present in 34 cm³.
(d)
0.050 mol HCl is in 34 cm³
Thus
X mol are in 1000 cm³
(1 dm³)
X
mol = (0.050 x 1000)/34 = 1.47 mol
Concentration
of HCl = 1.47 mol/dm³
QUESTION
2:
28.50
cm³
of 0.050 mol/dm³
H2SO4
exactly neutralised 25.00cm³
X2CO3
of concentration 6.00g/dm³.
Calculate (a) Mr for X2CO3
(b) Ar for X
(i)
Equation: X2CO3
+ H2SO4
= X2SO4
+ H2O + CO2
28.50cm³
H2SO4
contains the same number of moles
as 25.00cm³
X2CO3
(ii)
moles of H2SO4
in 28.50cm³
of 0.050 mol/dm³
= (28.50x0.050)/1000 = 0.001425 mols
(iii)
moles X2CO3
in 25.00cm³
= 0.001425 mols
(iv)
Moles X2CO3
in 1000cm³
= (0.001425x1000)/25 = 0.057 mol/dm³
(v)
But 1dm³
X2CO3(aq)
contains 6.00g
6.00g
has 0.057 mol
ie
6.00/0.057 = 1 mol = 105g
Mr
= 105.0
(vi)
Mr of X2CO3
= (Ar of X x 2) + 12 + (3 x 16) =
105
Ar
of X = (105 - (12 + 48))/2 = 22.5
ie Ar = 22.5
QUESTION
3:
A
small piece of lithium of mass 0.35g
is added to cold water. The resulting
solution is titrated with 2.00 mol/dm³
hydrocholoric acid.
(a)
Write the equation between lithium
and water and the lithium solution
and hydrochloric acid.
(b)
What volume of hydrochloric acid is
needed to neutralise the lithium solution?
ANSWERS
(a)
2Li (s) + 2H2O
(l) = 2LiOH (aq) + H2 (g)
Remember
that the alkali metals dissolve in
water to form a metal hydroxide and
give off hydrogen.
LiOH
(aq) + HCl (aq) = LiCl (aq) + H2O
(l)
(b)
Molar mass of lithium is 7
#
mol Li = 0.35/7 = 0.05 mol
Based
on the equation Li reacts in a 1:1
ratio to form LiOH
Thus
the # mol LiOH = 0.05 mol
LiOH
and HCl also react in a 1:1 ratio
#
mol of HCl is also 0.05 mol
Since
1000 cm³
contain 2.00 mol
X
cm³
contain 0.05 mol thus X cm³
= (1000 x 0.05)/2 = 25 cm³
Q3.
Now attempt these questions:
What
is the concentration of a solution
of sodium hydroxide if 25cm³
of it requires 20cm³
of hydrochloric acid of concentration
0.100 mol/dm³
for neutralisation?
Q4.
37.50cm³
of HCl containing 0.100 mol/dm³
neutralised 25.00cm³
XHCO3 of concentration
15.00g/dm³.
Calculate Mr for XHCO3
and Ar for X
Q5.
In this titration 25.0 cm³
of 1.0 mol/dm³
NaOH was used. Calculate the volume
of 2.0 mol/dm³
HCl needed to neutralise the alkali.
Calculate the mass of sodium chloride
formed.
*
Francine Taylor-Campbell is
an independent contributor.
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