Solutions to exercises
27
b. 2 The reactions are
xSrO + LaMnO 3 $ LaSr x MnO 3+x
(1)
xSrO + (1 −x) LaMnO 3 $ La 1−x Sr x Mn 1−x O 3−2x
(2)
xSrO + La 1−x MnO 3−1.5x $ La 1−x Sr x MnO 3−0.5x
(3)
2 The normal structure elements and the point defects in the solid solution are given below:
xSrO + LaMnO 3 $ La
#
La + x Sr ′
La + (3 + x)O
#
O
+ Mn
#
Mn + x V
3
′
Mn + 2 x V
••
O
(1)
x SrO + (1− x)LaMnO 3 $ (1 − x)La
#
La + x Sr ′
La + (1 − x)Mn
#
Mn
+ (3 − 2 x)O
#
O + xV
3
′
Mn + 2 xV
••
O
(2)
x SrO + La 1−x MnO 3−1.5x $ (1 − x)La
#
La + x Sr ′
La + Mn
#
Mn
+ (3 − 0.5 x)O
#
O + 0.5 xV
••
O
(3)
2 Preparation of the solid solution La 0.84 Sr 0.16 MnO 2.92 :
Calculation of the molar mass of the solid solution to prepare
M = (0.84 # M La ) + (0.16 # M Sr ) + M Mn + (2.92 # M O )
M = 232.38 g mol
–1
10 g of this solid solution corresponds to an amount of substance n of
.
n
232 38
10
=
n = 0.043 mol
We are left to calculate the molar masses of the reagents and the masses to weigh given the stoichiometric coefficients of each element in
the solid solution. The molar masses and the quantity of reagent and
product are listed in table 3.
Table 3 – Molar masses and quantity of reagents and product
Parameters
La(NO 3 ) 3 SrCO 3 Mn(NO 3 ) 3 La 0.84 Sr 0.16 MnO 2.92
Molar mass [g mol
–1
] 324.92
147.63
240.95
232.38
n [mol per 10 g]
0.0361 0.00688
0.043
0.043
m [g per 10 g]
11.73
1.016
10.36
10.00