Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
XCl2( textexcess)+YCl2 arrow XCl4+Y downarrow YO xrightarrow [> 400o C]Δ (1/2)O2+Y , Ore of Y would be
Q.
XC
l
2
(
excess
)
+
Y
C
l
2
→
XC
l
4
+
Y
↓
Y
O
Δ
>
40
0
o
C
2
1
O
2
+
Y
, Ore of
Y
would be
170
161
NTA Abhyas
NTA Abhyas 2022
Report Error
A
Siderite
B
Cinnabar
C
Malachite
D
Hornsilver
Solution:
The reaction sequence is as follows
(
XC
l
2
)
S
n
C
l
2
+
(
Y
C
l
2
)
H
g
C
l
2
→
(
XC
l
4
)
S
n
C
l
4
+
(
Y
)
H
g
H
g
O
Δ
>
40
0
∘
C
H
g
+
2
1
O
2
H
g
S
→
Cinnabar