Question

Calculate the mass of $K_{2}C{r}_2{O}_7$ in grams required to produce $254g{I}_2$ from $KI$ solution.
$K_{2}C{r}_2{O}_7+KI\to C{r}^{3+}+I_2$

(Molar mass of $K_{2}C{r}_2{O}_7$ is $294gmo{l}^{-1}$)

Answer 1

$K_2\stackrel{+6}{C}{r}_2{O}_7+K\stackrel{-1}{I}\to \stackrel{+3}{C}{r}^{3+}+{\stackrel{0}{I}}_2$

$n_f=\left(|\text{Change in O.S.}|\right)\times \text{Number of atoms}$
$n_{f_{K_{2}C{r}_2{O}_7}}=|+3-(+6)|\times 2=6$
$n_{f_{I_2}}=|0-(-1)|\times 2=2$

From the law of equivalence,
$\text{Equivalents of}{K}_{2}C{r}_2{O}_7=\text{Equivalents of}{I}_2$
$\frac{W_{K_{2}C{r}_2{O}_7}}{294}\times 6=\frac{254}{254}\times 2$
$\therefore W_{K_{2}C{r}_2{O}_7}\approx 98g$