$5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(aq\right)+2{\mathrm{MnO}}_4\left(aq\right)+6{\mathrm{H}}^{+}\left(aq\right)$$⟶2{\mathrm{Mn}}^{2+}\left(aq\right)+10{\mathrm{CO}}_2\left(g\right)+8{\mathrm{H}}_2\mathrm{O}\left(l\right)$  Oxalic acid, $H_2{C}_2{O}_4$, reacts with permanganate ion according to the balanced equation above. How many mL of  $0.0154 MKMn{O}_4$  solution are required to react with $25.0 \mathrm{mL}$ of $0.0208M{H}_2{C}_2{O}_4$ solution?

In the reaction

${\mathrm{MnO}}_4^{-}\to {\mathrm{Mn}}^{+2}$

thus nf = 5

$\begin{array}{r}{\mathrm{C}}_2^{+3}{\mathrm{O}}_4^{2-}\to {\mathrm{CO}}_2^{+4}\\ 2\cdot {\mathrm{C}}^{+3}\to {\mathrm{C}}^{+4}\\ \mathrm{nf}=2\times 1=2\end{array}$

thus by equation

$\begin{array}{r}\left({\mathrm{KMnO}}_4\right){\mathrm{M}}_1{\mathrm{V}}_1{\mathrm{nf}}_1={\mathrm{M}}_2{\mathrm{V}}_2{\mathrm{nf}}_2\left({\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\right)\\ 0.0162\mathrm{M}\times {\mathrm{V}}_1\times 5=0.022\mathrm{M}\times 25.0\mathrm{mL}\times 2\\ {\mathrm{V}}_1=13.58\mathrm{mL}\simeq 13.6\mathrm{mL}\end{array}$

As a result, in order to react with 25.0 mL of 0.022M ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ solution, 13.58 mL of ${\mathrm{KMnO}}_4$ solution is needed.