The stoichiometric numbers appearing from left to right in the reaction…
$\mathrm{MnO}_{4}^{2-}+\mathrm{H}^{+} ightarrow \mathrm{MnO}_{4}^{-}+\mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}$
are
- $3,2,2,1,2$
- $3,4,2,1,2$
- $2,4,1,1,2$
- $2,3,1,1,2$
Solution
$3 \mathrm{MnO}_{4}^{2-}+4 \mathrm{H}^{+} ightarrow 2 \mathrm{MNO}_{4}^{-}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
Asked in: JEE-TOPICTESTS-CHEMISTRY