How many mL of 0.125M Cr³⁺ must be reacted with 12.00mL of 0.200M MnO₄^– if the redox products are Cr₂O₇^2– and Mn²⁺

${\mathrm{Cr}}^{3+}+{\mathrm{MnO}}_4^{-}⟶{\mathrm{Mn}}^{2+}+\frac{1}{2} {\mathrm{Cr}}_2{\mathrm{O}}_7^{-2}$

Equivalent of Cr³⁺=3× moles of Cr³⁺

Equivalents of ${\mathrm{MnO}}_4^{-}$​=5× moles of ${\mathrm{MnO}}_4^{-}$

Amount of Cr³⁺=0.125×V millimole =0.125×V×meq

Amount of ${\mathrm{MnO}}_4^{-}$ ​=0.200×12.00×5 milliequivalent

0.125×V×3=0.200×12.00×5

V=32 mL