Solve the following

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:

(50)

M. eq. of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=\mathrm{M}$. eq. of $\mathrm{FeC}_{2} \mathrm{O}_{4}$

$\mathrm{FeC}_{2} \mathrm{O}_{4}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{CO}_{2}+\mathrm{Cr}^{3+}$

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

$\mathrm{V}=50 \mathrm{~mL}$

Leave a comment