An excess of NaOH was added to 100 mL of a FeCl₃ solution which gives 2.14 of Fe(OH)₃. Calculate the normality of FeCl₃ solution.

$\begin{array}{l}3\mathrm{NaOH}+{\mathrm{FeCl}}_3 ⟶\mathrm{Fe}(\mathrm{OH}{)}_3+3\mathrm{NaCl}\\ \left(\mathrm{excess}\right) 100 \mathrm{mL} \\ \begin{array}{ll} 1\mathrm{mmol}\equiv 1\mathrm{mmol}& \\ & \\ {\mathrm{FeCl}}_3\equiv \mathrm{Fe}(\mathrm{OH}{)}_3& \\ \mathrm{M}\times 100\equiv \frac{2.14}{107}\times 1000& \\ {\mathrm{M}}_{{\mathrm{FeCl}}_3}=0.2& \\ {\mathrm{N}}_{{\mathrm{FeCl}}_3}=0.2\times 3(\text{ valency factor }=3)& \\ =0.6\mathrm{N}& \end{array}\end{array}$