Question

How many mL of $0.125MC{r}^{3+}$ must be reacted with 12.0 mL of $0.200MMn{O}_4^{-}$ if the redox products are $C{r}_2{O}_7^{2-}$ and $M{n}^{2+}$ ?

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

Answer 1

Answer: (A)

Equivalents of $C{r}^{3+}=3\times$ moles of $C{r}^{3+}$ Equivalents of $Mn{O}_4^{-}=5\times$ moles of $Mn{O}_4^{-}$ Amount of $C{r}^{3+}=0.125\times V$ milli mol $=0.125\times V\times 3$ miliequiv. Amount of $Mn{O}_4^{-}=0.200\times 12.00\times 5$ milliequiv $\therefore$ $0.125\times V\times 3=0.200\times 12.00\times 5$ $V=32.0mL$