Question

How many mL of $0.125MC{r}^{3+}$ mist be reacted with 12.0 mL of 0.200 M $Mn{O}_4^{-}$ 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)

32 mL

Equivalents of $C{r}^{3+}=3\times molesofC{r}^{3+}$
Equivalents of $Mn{O}_4^{-}=5\times moleofMn{O}_4^{-}$
Amount of $C{r}^{3+}=0.125\times Vmillimol$
$=0.125\times V\times 3milliequiv$.
Amount of $Mn{O}_4^{-}=0.200\times 12.00\times 5milliequviv$
$\therefore 0.125\times V\times 3=0.200\times 12.00\times 5$
$V=32.0mL$