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

We know that,

$K_{2}C{r}_2{O}_7+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

$X^{nt}+3H_{2}O\to X{O}^{3-}+(5-n)e^{-}+6H^{+}$

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

From the egn, $K_{2}C{r}_2{O}_7:X^{nt}$

$=(5-n):6$

Therefore,

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

or $5-n=\frac{36}{9}=4$

$5-n=4$

$n=1$