What volume of 0.2 M Ba(MnO4)2 solution is required for complete oxidation of 25 gm of 89.6% pure Fe C2O4 in acidic medium according to the reaction MnO4−+FeCr2O4→Fe+3+Cr2O7−2+Mn+2

No of milli equivalents of Ba(MnO4 )2 =No.of milli equivalents of FeCr2 O4n - factor of Ba(MnO4 )2 = 10n - factor of FeCr2 O4 = 7∴V×0.2×10=25×89.6×103100×2247 ∴V=350ml

What volume of 0.2 M Ba(MnO4)2 solution is required for complete oxidation of 25 gm of 89.6% pure Fe C2O4 in acidic medium according to the reaction MnO4−+FeCr2O4→Fe+3+Cr2O7−2+Mn+2

No of milli equivalents of Ba(MnO4 )2 =No.of milli equivalents of FeCr2 O4n - factor of Ba(MnO4 )2 = 10n - factor of FeCr2 O4 = 7∴V×0.2×10=25×89.6×103100×2247 ∴V=350ml

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

What volume of 0.2 M Ba(MnO₄)₂ solution is required for complete oxidation of 25 gm of 89.6% pure Fe C₂O₄ in acidic medium according to the reaction ${MnO}_{4}^{-} + {FeCr}_{2} O_{4} \to {Fe}^{+ 3} + {Cr}_{2} O_{7}^{- 2} + {Mn}^{+ 2}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

200 ml

350 ml

700 ml

175 ml

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

No of milli equivalents of Ba(MnO4 )2 =No.
of milli equivalents of FeCr2 O4
n - factor of Ba(MnO₄ )₂ = 10
n - factor of FeCr₂ O₄ = 7
$∴ V \times 0.2 \times 10 = \frac{25 \times 89.6 \times 10^{3}}{100 \times (\frac{224}{7})} ∴ V = 350 ml$