Question

How many mL of $0.3$M $K_{2}C{r}_2{O}_7$(acidic) is required for complete oxidation of $5$ mL of $0.2$ M $Sn{C}_2{O}_4$ solution.

(Round off to 1 decimal place)

Answer 1

R.E.F image

$2K_{2}C{r}_2{O}_7+35n{C}_2{O}_4+8H_{2}S{O}_4\to 2k_{2}S{O}_4+2c{r}_2(S_4{)}_3+8H_{2}O+6C{O}_2$

ion electron method

$2$ mole of $K_{2}C{r}_2{O}_7$ react with $3$ mole of $S_n{C}_2{O}_4$

mole of $K_{2}C{r}_2{O}_7=M\times V$

$=3\times x\times x\times {10}^{-3}$

mole of $S_n{C}_2{O}_n=2\times 5\times {10}^{-3}$

$\frac{2}{3}=\frac{moleof{K}_{2}C{r}_2{O}_7}{moleofSn{C}_2{O}_4}$

$\frac{2}{3}=\frac{3x\times {10}^{-3}}{2\times 5\times {10}^{-3}}$

$9x=2$

$x=\frac{20}{9}=2.22ml$