Question

One mole of acidified $K_{2}C{r}_2{O}_7$ on reaction with excess of KI will liberate moles of $l_2$.

• A. 6
• B. 1
• C. 7
• D. 3

Answer 1

The correct option is D 3

When one mole of acidified potassium dicromate $\left(K_{2}C{r}_2{O}_7\right)$ react with excess of $KI$ solution the balanced chemical reaction is as :

$C{r}_2{O}_7^{2-}+14H^{+}+6I^{-}\to 2C{r}^{3+}+3I_2+7H-2O$

Thus, $3$ moles of $I_2$ is evolved , when excess of $KI$ react with one mole of acidified potassium dicromate $\left(K_{2}C{r}_2{O}_7\right)$.