Question

An excess $KI$ solution is mixed in a solution of $K_{2}C{r}_2{O}_7$ and liberated iodine required $72mL$ of $0.05NN{a}_2{S}_2{O}_3$ for complete reaction. How many grams of $K_{2}C{r}_2{O}_7$ were present in the solution of $K_{2}C{r}_2{O}_7$? The reaction occurs as:
${C{r}_2{O}_7}^{2-}+6I^{-}+14H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$

Answer 1

The reaction involved may be given as:
$C{r}_2{O}_7^{2-}+6I^{-}+14H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$
$3[I_2+2N{a}_2{S}_2{O}_3\to 2NaI+N{a}_2{S}_4{O}_6]$
$1mole{K}_{2}C{r}_2{O}_7=6moleN{a}_2{S}_2{O}_3....\left(i\right)$
No. of moles of hypo $=\frac{Mass}{M.w.\left(158\right)}=\frac{E\times N\times V}{1000\times 158}$
$N_{N{a}_2{S}_2{O}_3}=\frac{158\times 0.05\times 72}{1000\times 158}=3.6\times {10}^{-3}$.
No. of moles of $K_{2}C{r}_2{O}_7=\frac{1}{6}$ [No. of moles of $N{a}_2{S}_2{O}_3]$
$=\frac{1}{6}[3.6\times {10}^{-3}]=6\times {10}^{-4}mole$
Mass of $K_{2}C{r}_2{O}_7$ in the given solution
$=$ No. of moles $\times$ Molecular weight
$=6\times {10}^{-4}\times 294=0.1764$.