The molar ratio of $F{e}^{2+}$to $F{e}^{3+}$ in a mixture of $FeS{O}_4$ and $F{e}_2{\left(S{O}_4\right)}_3$ having equal number of sulphate ions in both ferrous and ferric sulphates is

$FeS{O}_4\underset{ }{\overset{ }{\to }}F{e}^{2+}+S{O}_4^{2-}$

$F{e}_2{\left(S{O}_4\right)}_3\underset{ }{\overset{ }{\to }}2F{e}^{3+}+3S{O}_4^{2-}$

Suppose  $x$  mol  $S{O}_4^{2-}$  ions are furnished by both  $FeS{O}_4$  and  $F{e}_2{\left(S{O}_4\right)}_3$

Number of moles of  $F{e}^{2+}=x$
Number of moles of  $F{e}^{3+}=\frac{2}{3}x$
  $F{e}^{2+}:F{e}^{3+}::x:\frac{2}{3}x=3:2$