Question

$Mn{O}_4^{-}$ in basic medium oxidises $N{O}_2^{-}$ to $N{O}_3^{-}$. How many moles of $N{O}_2^{-}$ will be oxidised by $1$ mole of $Mn{O}_4^{-}?$

• A. 0.25
• B. 1
• C. 1.5
• D. 2

Answer 1

The correct option is C 1.5

$\stackrel{+7}{M}n{O}_4^{-}+\stackrel{+3}{N}{O}_2^{-}\stackrel{O{H}^{-}}{⟶}\stackrel{+4}{M}n{O}_2+\stackrel{+5}{N}{O}_3^{-}$

$n_f=\left(|\text{Change in O.S.}|\right)\times \text{Number of atoms}$
$n_{f_{Mn{O}_4^{-}}}=|+4-(+7)|\times 1=3$
$n_{f_{N{O}_2^{-}}}=|+5-(+3)|\times 1=2$

From the law of equivalence,
$\text{Equivalents of}Mn{O}_4^{-}=\text{Equivalents of}N{O}_2^{-}$
$1\times 3=n_{N{O}_2^{-}}\times 2$
$\therefore n_{N{O}_2^{-}}=\frac{3}{2}=1.5mol$