wiz-icon
MyQuestionIcon
MyQuestionIcon
7
You visited us 7 times! Enjoying our articles? Unlock Full Access!
Question

A mixture of FeO and Fe2O3 is completely reacted with 100 mL of 0.25 M acidified KMnO4 solution. The resultant solution was then titrated with Zn dust which converted the Fe3+ ions to Fe2+ ions. The Fe2+ ions required 1000 mL of 0.10 M K2Cr2O7 solution. Find out the weight percentage of Fe2O3 in the mixture.

A
80.85
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
19.15
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
50
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
89.41
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 80.85
Milli-equivalents of FeOis equal to
Milli-equivalents of KMnO4
=0.25×5×100

Milli-moles ofFeO(n=1)=0.25×100×5=125

Total m-eq or m-mol of Fe2+
=1000×0.1×6=600

(from FeO and Fe2O3 after reaction with Zn dust)

Milli-moles of Fe2+ from Fe2O3
=600125=475
Milli-moles of Fe2O3=4752
Weight of FeO=0.125×(5616)g=9g
Weight of Fe2O3=4752×1601000=38g
% of Fe2O3=389+38×100=80.85

flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Balancing of Redox Reactions
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon