Problem :
4.08 gm of a mixture of BaO and an unknown carbonate
M
C
O
3
M
C
O
3
is heated strongly. The residue weighs 3.64 gram. The residue is then dissolved in 100 ml of 1 N HCl. The excess acid requires 16 ml of 2.5 NaOH for complete neutralization. Identify the metal M, if atomic weight of Ba is 138.
Solution :
We need to determine atomic weight of M to identify it. From the question, we see that first part involves decomposition, which can be analyzed using mole concept. Note that difference of mass of mixture and residue is mass of
C
O
2
C
O
2
. On the other hand, analysis of back titration gives us meq of HCl required for neutralization of residue. To combine meq data with decomposition data, we shall convert required meq in milli-moles.
⇒
total meq of HCl
=
N
V
=
1
X
100
=
100
⇒
total meq of HCl
=
N
V
=
1
X
100
=
100
⇒
meq of excess HCl
=
N
V
=
2.5
X
16
=
40
⇒
meq of excess HCl
=
N
V
=
2.5
X
16
=
40
⇒
meq of HCl used for neutralization of residue
=
100
−
40
=
60
⇒
meq of HCl used for neutralization of residue
=
100
−
40
=
60
⇒
milli-moles of HCl for neutralization of residue
=
x
X
m
e
q
=
1
X
60
=
60
⇒
milli-moles of HCl for neutralization of residue
=
x
X
m
e
q
=
1
X
60
=
60
Now, we analyze the data which is given for decomposition of mixture. On heating, BaO is not decoposed. Only
M
C
O
3
M
C
O
3
is decomposed as :
M
C
O
3
=
M
O
+
C
O
2
M
C
O
3
=
M
O
+
C
O
2
Carbon dioxide is released and is not part of residue. The amount of
C
O
2
C
O
2
is equal to difference of mass of mixture and residue. Hence, moles of
C
O
2
C
O
2
released are :
⇒
n
C
O
2
=
4.08
−
3.64
12
+
2
X
16
=
0.44
44
=
0.01
m
o
l
e
⇒
n
C
O
2
=
4.08
−
3.64
12
+
2
X
16
=
0.44
44
=
0.01
m
o
l
e
Applying mole concept, moles of MO in the residue is 0.01 mole i.e. 10 milli-moles. Now, MO reacts with HCl as :
M
O
+
2
H
C
l
→
M
C
l
2
+
H
2
O
M
O
+
2
H
C
l
→
M
C
l
2
+
H
2
O
Moles of HCl required for MO is 2X0.01 moles = 20 milli-moles of HCl. But total milli-moles that reacted with residue is 60 milli-moles. Hence, milli-moles of HCl that reacted with BaO is 60-20 = 40 milli-moles of HCl. Applying mole concept,
B
a
O
+
2
H
C
l
→
B
a
C
l
2
+
H
2
O
B
a
O
+
2
H
C
l
→
B
a
C
l
2
+
H
2
O
The milli-moles of BaO is half of that of HCl. Hence, millimoles of BaO in the mixture is 20 milli-moles.
⇒
amount of BaO in the mixture
=
milli-moles
X
M
B
a
O
1000
=
20
X
138
+
16
1000
=
3.08
g
m
⇒
amount of BaO in the mixture
=
milli-moles
X
M
B
a
O
1000
=
20
X
138
+
16
1000
=
3.08
g
m
⇒
amount of MO in the residue
=
3.64
−
3.08
=
0.56
g
m
⇒
amount of MO in the residue
=
3.64
−
3.08
=
0.56
g
m
Using formula :
⇒
g
=
milli-moles
X
M
M
O
1000
=
10
X
M
M
O
1000
⇒
g
=
milli-moles
X
M
M
O
1000
=
10
X
M
M
O
1000
⇒
M
M
O
=
100
g
=
100
X
0.56
=
56
⇒
M
M
O
=
100
g
=
100
X
0.56
=
56
Clearly, atomic weight of M is 56-16 = 40. The element, therefore, is calcium.
