Question

Number of moles of $K_{2}C{r}_2{O}_7$ reduced by one mole of iodide ions is :

• A. $3$
• B. $\frac{1}{3}$
• C. $6$
• D. $\frac{1}{6}$

Answer 1

The correct option is D $\frac{1}{6}$

Number of moles of $K_{2}C{r}_2{O}_7$ reduced by one mole of iodide ions is $\frac{1}{6}$.
The blanced reaction is as follows:
$K_{2}C{r}_2{O}_7+6KI+7H_{2}S{O}_4\to 4K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+7H_{2}O+3I_2$
We can see, $6$ moles of $I^{-}$ is required to reduce $1$ mole of $K_{2}C{r}_2{O}_7$.