Question

${\mathrm{FeC}}_2{\mathrm{O}}_4,{\mathrm{Fe}}_2{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3,{\mathrm{FeSO}}_4,{\mathrm{Fe}}_2{\left(\mathrm{SO}\right)}_4$ one mole each will react with how many moles of ${\mathrm{KMnO}}_4?$

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

The correct option is B

$2$

Explanation for the correct option:

(B)

$$\begin{gathered} \mathbf{Step}\mathbf{}\mathbf{1}\mathbf{:}{\mathrm{FeC}}_2{\mathrm{O}}_4+{\mathrm{KMnO}}_4\to {\mathrm{Fe}}^{3+}+{\mathrm{CO}}_2+{\mathrm{Mn}}^{2+} \\ \mathrm{v}.\mathrm{f}.=3\mathrm{v}.\mathrm{f}=5[\mathbf{v}\mathbf{.}\mathbf{f}\mathrm{stands}\mathrm{for}\mathrm{valency}\mathrm{factor}] \\ 1\times 3=\mathrm{mole}\times 5 \\ \mathrm{Mole}=\frac{\mathbf{5}}{\mathbf{3}} \\ \mathbf{Step}\mathbf{}\mathbf{2}\mathbf{:}{\mathrm{Fe}}_2{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3+{\mathrm{KMnO}}_4\to {\mathrm{Fe}}^{3+}+{\mathrm{CO}}_2+{\mathrm{Mn}}^{2+} \\ \mathrm{v}.\mathrm{f}=6\mathrm{v}.\mathrm{f}=5 \\ 1\times 6=\mathrm{mole}\times 5 \\ \mathrm{Mole}=\frac{\mathbf{6}}{\mathbf{5}} \\ \mathbf{Step}\mathbf{}\mathbf{3}\mathbf{:}\mathbf{}{\mathrm{FeSO}}_4+{\mathrm{KMnO}}_4\to {\mathrm{Fe}}^{3+}+{{\mathrm{SO}}_4}^{2-}+{\mathrm{Mn}}^{2+} \\ \mathrm{v}.\mathrm{f}=1\mathrm{v}.\mathrm{f}=5 \\ 1\times 1=\mathrm{moles}\times 5 \\ \mathrm{Mole}=\frac{\mathbf{1}}{\mathbf{5}} \\ \mathbf{Step}\mathbf{}\mathbf{4}\mathbf{}\mathbf{:}{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3\mathrm{doesn}\text{'}\mathrm{t}\mathrm{oxidise} \\ \mathbf{Step}\mathbf{}\mathbf{5}\mathbf{}\mathbf{}\mathbf{:}\mathbf{}\mathbf{Total}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{KMnO}}_{\mathbf{4}}\mathbf{}=\frac{5}{3}+\frac{6}{5}+\frac{1}{5}=\mathbf{2} \\ Hence,onemoleofeachgivenspecieswillreactwith\mathbf{2}\mathbf{}molesof{\mathrm{KMnO}}_4. \end{gathered}$$