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JEE Main 24 Feb 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 36
Total: 90
A
l
2
O
3
was leached with alkali to get
X
. The solution of
X
on passing of gas
Y
, forms
Z
.
X
,
Y
and
Z
respectively are
[24 Feb 2021 Shift 1]
X
=
N
a
[
A
l
(
O
H
)
4
]
,
Y
=
S
O
2
,
Z
=
A
l
2
O
3
X
=
A
l
(
O
H
)
3
,
Y
=
S
O
2
,
Z
=
A
l
2
O
3
⋅
×
H
2
O
X
=
A
l
(
O
H
)
3
,
Y
=
C
O
,
Z
=
A
l
2
O
3
X
=
N
a
[
A
l
(
O
H
)
4
]
,
Y
=
C
O
2
,
Z
=
A
l
2
O
3
⋅
x
H
2
O
Validate
Solution:
A
l
2
O
3
(aluminium oxide) was leached with alkali to get
N
a
[
A
l
(
O
H
)
4
]
(
X
)
(sodium aluminate). The solution of sodium aluminate when passed through gas
Y
, i.e. carbon dioxide
(
C
O
2
)
forms
(
A
l
2
O
3
⋅
x
H
2
O
)
(
Z
)
.
Complete reactions are as follows
A
l
2
O
3
Aluminium
oxide
+
2
N
a
O
H
(
a
q
)
Sodium
hydroxide
+
3
H
2
O
(
l
)
⟶
2
N
a
[
A
l
(
X
)
(
O
H
)
4
]
(
a
q
)
+
C
O
2
(
Y
)
⟶
A
l
2
O
3
⋅
x
H
2
O
(
Z
)
+
2
N
a
H
C
O
3
Sodium
bicarbonate
Hence,
X
=
N
a
[
A
l
(
O
H
)
4
]
Y
=
C
O
2
Z
=
A
l
2
O
3
⋅
x
H
2
O
© examsnet.com
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