Question

How many $mL$ of $0.125\, M \,Cr ^{3+}$ must be reacted with $12.00 \,mL$ of $0.200\, M\, MnO _{4}^{-}$if the redox products are $CrO _{7}^{2-}$ and $Mn ^{2+}$ ?

• A. 8 mL
• B. 16 mL
• C. 24 mL
• D. 32 mL

Answer 1

The Correct Option is D

Equivalent of $Cr ^{3 +}=3 \times$ moles of $Cr ^{3+}$ Equivalents of $MnO _{4}^{-}=5 \times$ moles of $MnO _{4}^-$ Amount of $Cr ^{3+}=0.125 \times V$ millimol $=0125 \times V \times 3$ milliequivalent Amount of $MnO _{4}^{-} =0.200 \times 12.00 \times 5$ milli equivalent $0.125 \times V \times 3=0.200 \times 12.00 \times 5$ $V=32 \,mL$