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Tardigrade
Question
Chemistry
How many mL of 0.125 text M text Cr3+ must be reacted with 12.0 mL of 0.200 text M text MnO4- if the redox products are Cr2O72- and Mn2+ ?
Q. How many mL of
0.125
M
C
r
3
+
must be reacted with 12.0 mL of
0.200
M
M
n
O
4
−
if the redox products are
C
r
2
O
7
2
−
and
M
n
2
+
?
2339
178
Manipal
Manipal 2013
Report Error
A
32 mL
76%
B
24 mL
7%
C
16 mL
14%
D
8 mL
3%
Solution:
Equivalents of
C
r
3
+
=
3
×
moles of
C
r
3
+
Equivalents of
M
n
O
4
−
=
5
×
moles of
M
n
O
4
−
Amount of
C
r
3
+
=
0.125
×
V
milli mol
=
0.125
×
V
×
3
miliequiv. Amount of
M
n
O
4
−
=
0.200
×
12.00
×
5
milliequiv
∴
0.125
×
V
×
3
=
0.200
×
12.00
×
5
V
=
32.0
m
L