Question

The volume, in $mL ,$ of $0.02 M K _{2} Cr _{2} O _{7}$ solution required to react with $0.288 g$ of ferrous oxalate in acidic medium is (Molar mass of $\left. Fe =56 g mol ^{-1}\right)$

Answer 1

$\underset{n =6 }{ K_2Cr_2O_7} + \underset{n = 3}{FeC_2O_4} \to Cr^{+3}+ Fe^{+3} + CO_2$

$ \frac{0.02 \times 6 \times V(mL)}{1000} = \frac{0.288}{144} \times 3$

$\Rightarrow V = 50 mL$