Question

Determine the volume of $M/8KMn{O}_4$ solution (in $mL$) that required to react completely with $25.0c{m}^3$ of $M/4FeS{O}_4$ solution in acidic medium.

Answer 1

The balanced ionic equation for the reaction is,
$Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$
From the balanced ionic equation, it is evident that 1 mole of $KMn{O}_4$ require 5 moles of $FeS{O}_4$, applying law of chemical equivalence, we get
$M_1{V}_1{n}_1\left(KMn{O}_4\right)=M_2{V}_2{n}_2\left(FeS{O}_4\right)$
(where $n$ is n-factor)
so, $\frac{1}{8}\times V_1\times 5=\frac{1}{4}\times 25\times 1$
$V_1=10c{m}^3$
or $V_1=10mL$