Question

Volume of 0.1 M $K_{2}C{r}_2{O}_7$ required to oxidise 35 mL of 0.5 M $FeS{O}_4$ solution is

• A. 29.2 mL
• B. 17.5 mL
• C. 175 mL
• D. 145 mL

Answer 1

Answer: (A)

$K_{2}C{r}_2{O}_7+4H_{2}S{O}_4\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+4H_{2}O+3\left[O\right]$
$2FeS{O}_4+H_{2}S{O}_4+\left[O\right]\to F{e}_2{\left(S{O}_4\right)}_3+H_{2}O\times 3$
$\frac{K_{2}C{r}_2{O}_7+7H_{2}S{O}_4+6FeS{O}_4\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+7H_{2}O+3F{e}_2{\left(S{O}_4\right)}_3}{\frac{M_1{V}_1}{n_1}=\frac{M_2{V}_2}{n_2}}$
$(K_{2}C{r}_2{O}_7)(FeS{O}_4)$
$\frac{0.1\times V_1}{1}=\frac{0.5\times 35}{6\times 0.1}$
$V_1=\frac{0.5\times 35}{6\times 0.1}=29.2mL$