Question

$6\times {10}^{-3}$ mole $K_{2}C{r}_2{O}_7$ reacts completely with $9\times {10}^{-3}$ mole $X^{n+}$ to give $X{O}_3{}^{-}$ and $C{r}^{3+}$. The value of $n$ is:

Answer 1

$6\times {10}^{-3}$ moles of $K_{2}C{r}_2{O}_7$ reacts with $9\times {10}^{-3}$ moles of $X^{n+}$ to give ${X{O}_3}^{-}$ & $C{r}^{3+}$.

$K_{2}C{r}_2{O}_7+6e^{-}+14H^{+}⟶2Cr{O}_3^{+}+7H_{2}O$

$X^{n+}+3H_{2}O⟶X{O}_3^{-}+(5-n)e^{-}+6H^{+}$

Since, $K_{2}C{r}_2{O}_7:X^{n+}=6\times {10}^{-3}:9\times {10}^{-3}$ moles

From the equations, $K_{2}C{r}_2{O}_7:X^{n+}=(5-n):6$

$\Rightarrow \frac{5-n}{6}=\frac{6\times {10}^{-3}}{9\times {10}^{-3}}$

$\Rightarrow 5-n=4$

$\Rightarrow n=1$ .