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Byju's Answer
Standard VIII
Chemistry
Mixture
Cl2 required ...
Question
C
l
2
required in this reaction is: (in mol)
Open in App
Solution
2
T
i
O
2
+
3
C
+
4
C
l
2
70
%
−
−
→
2
T
i
C
l
4
+
C
O
2
+
2
C
O
7500gm
7500
80
=
93.75
mole
So,
C
l
2
requires
=
93.75
×
2
=
187.5
×
0.7
×
0.93
=
122.1
mole
=
8.66
k
g
.
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Similar questions
Q.
C
l
2
required in this reaction is
:
Q.
C
l
2
required in this reaction is (in kg) :
Q.
For the following reaction:
P
C
l
5
(
g
)
⇌
P
C
l
3
(
g
)
+
C
l
2
(
g
)
0.4
m
o
l
of
P
C
l
5
(
g
)
,
0.2
m
o
l
of
P
C
l
3
(
g
)
and
0.6
m
o
l
of
C
l
2
(
g
)
are taken in a
1
L
flask.
If
K
c
=
0.2
, then the direction in which reaction proceeds is :
Q.
Which reaction is this?
2
K
I
+
C
l
2
→
2
K
C
l
+
I
2
Q.
The following series of reactions show the formation of
N
a
C
l
O
3
.
i
)
K
M
n
O
4
+
H
C
l
→
K
C
l
+
M
n
C
l
2
+
H
2
O
+
C
l
2
i
i
)
C
l
2
+
B
a
(
O
H
)
2
→
B
a
(
C
l
O
3
)
2
+
B
a
C
l
2
+
H
2
O
i
i
i
)
B
a
(
C
l
O
3
)
2
+
N
a
2
S
O
4
→
B
a
S
O
4
+
N
a
C
l
O
3
Pick the incorrect statement(s) about this process:
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