Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
470 views
in Chemistry by (84.6k points)
closed by
The final product obtained for the following reaction is:
`KMnO_(4)("excess") +H_(2)SO_(4)("concentrated and cold") to`
A. `Mn_(2)O_(7)`
B. `MnO`
C. `Mn_(3)O_(4)`
D. `MnO_(3)^(+)`

1 Answer

0 votes
by (87.2k points)
selected by
 
Best answer
Correct Answer - A
`2KMnO_(4)+3H_(2)SO_(4)to2KHSO_(4)+(MnO_(3))_(2)SO_(4)+2H_(2)O`
`(MnO_(3))_(2)SO_(4)+H_(2)O toMn_(2)O_(7)+H_(2)SO_(4)`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...