Question:
The volume, in $\mathrm{mL}$, of $0.02 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution required to react with $0.288 \mathrm{~g}$ of ferrous oxalate in acidic medium is
(Molar mass of $\mathrm{Fe}=56 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Solution:
$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{FeC}_{2} \mathrm{O}_{4} \longrightarrow \mathrm{Cr}^{+3}+\mathrm{Fe}^{+3}+\mathrm{CO}_{2}$
$\mathrm{n}=6 \quad \mathrm{n}=3$
$\frac{0.02 \times 6 \times V(\mathrm{~mL})}{1000}=\frac{0.288}{144} \times 3$
$\Rightarrow \mathrm{V}=50 \mathrm{~mL}$