Question

What volume of $0.1MKMn{O}_4$ is needed to oxidize $100mg$ of $Fe{C}_2{O}_4$ in acidic solution?

• A. $4.1mL$
• B. $8.2mL$
• C. $10.2mL$
• D. $4.6mL$

Answer 1

The correct option is A $4.1mL$

Total number of electrons involved per molecule oxidation of $Fe{C}_2{O}_4$ to $F{e}^{3+}$ and $C{O}_2$ is 3.
The oxidation of ferrous ions to ferric ions involves 1 electron.
$F{e}^{2+}\to F{e}^{3+}+e^{-}$
The oxidation of oxalate ions to carbon dioxide involves 2 electrons.
$C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$
In acidic solution, 1 mole of $KMn{O}_4$ involves 5 moles of electrons.
Thus, 3 moles of $KMn{O}_4$ will oxidize 5 moles of ferrous oxalate.
The molar mass of ferrous oxalate is 143.91 g/mol.
100 mg of ferrous oxalate correspond to $\frac{100}{1000\times 143.91}=6.95\times {10}^{-4}mol$.
They will require $6.95\times {10}^{-4}mol\times \frac{3}{5}=4.1\times {10}^{-4}mol$ of $KMn{O}_4$.
The volume of 0.1 M $KMn{O}_4$ required will be $\frac{4.1\times {10}^{-4}mol}{0.1}=4.1\times {10}^{-3}L=4.1mL$.