284
5 – Applications
B – Sodium-sulfur battery
1. Examination of the formula Na 1.67 Mg 0.67 Al 10.33 O 17 shows that we can write
it in the form Na 1+x Mg x Al 11−x O 17 with x = 0.67.
Identification with the formula (Na 2 O) α (MgO) β (Al 2 O 3 ) γ leads to
2
1 0.67
0.835
0.670
2
11 0.67
5.165
α
β
γ
=
+
=
=
=
−
=
The formula is (
)
(
)
(
)
Na O
MgO
Al O
.
.
.
2
0835
0 670
2 3 5 165
In both forms, we verify that the number of oxygen atoms is 17.
2. We start by calculating the total number n t of moles of β ′′ alumina to prepare,
n
M
m
M
V
t
ρ
=
=
where M is the molar mass of the compound and V is the volume of the plate.
We find M = 605.6 g mol
−1
and V = 9.425 cm
3
, which gives n t = 5.011 # 10
−2
mol.
Based on the formula (Na 2 O) 0.835 (MgO) 0.670 (Al 2 O 3 ) 5.165 , we calculate for
a formula unit the molar fraction and the number of moles of each oxide.
The calculation for Na 2 O gives
.
.
.
.
.
x
0 835 0 670 5 165
0 835
0 1252
Na O
2
=
+
+
=
n
x
n
Na O
N a O
t
2
2
=
#
We obtain n Na 2 O = 6.274 # 10
−3
mol. By proceeding in the same manner,
we obtain n MgO = 5.034 # 10
−3
mol and n Al 2 O 3 = 3.88 # 10
−2
mol. From this
we deduce
m Na 2 O 3 = 2 n Na 2 O # M Na 2 O 3
m Na 2 O 3 = 2 # 6.274 # 10
−2 # 85
.
m
g
1 067
Na O
2 3
=
An identical calculation gives
.
m
g
0 747
(
)
Mg NO 3 2 =
.
m
g
16 529
(
)
Al NO 3 3 =
Note – To be absolutely rigorous, we should have accounted for shrinkage
during sintering.