Question

${\mathrm{FeC}}_2{\mathrm{O}}_4,{\mathrm{Fe}}_2{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3,{\mathrm{FeSO}}_4,{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$ 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

$2$

The explanation for the correct option:

The above question can be solved as:

Step 1:

For ${\mathrm{FeC}}_2{\mathrm{O}}_4$ :

${\mathrm{FeC}}_2{\mathrm{O}}_4+{\mathrm{KMnO}}_4\underset{}{\overset{}{\to }}{\mathrm{Fe}}^{+3}+{\mathrm{CO}}_2+{\mathrm{Mn}}^{+2}$

$\mathrm{v}.\mathrm{f}=3$ . $\mathrm{v}.\mathrm{f}=5$

$1\times 3=\mathrm{moles}\times 5$

$\mathrm{Mole}=3/5$

Step 2:

For ${\mathrm{Fe}}_2{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3$:

${\mathrm{Fe}}_2{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3+{\mathrm{KMnO}}_4\underset{}{\overset{}{\to }}{\mathrm{Fe}}^{+3}+{\mathrm{CO}}_2+{\mathrm{Mn}}^{+2}$

$\mathrm{v}.\mathrm{f}=6$ $\mathrm{v}.\mathrm{f}=5$

$1\times 3=\mathrm{moles}\times 5$$\mathrm{Mole}=6/5$

Step 3:

For ${\mathrm{FeSO}}_4$:

${\mathrm{FeSO}}_4+{\mathrm{KMnO}}_4\underset{}{\overset{}{\to }}{\mathrm{Fe}}^{+3}+{\mathrm{SO}}_4^{-2}+{\mathrm{Mn}}^{+2}$

$\mathrm{v}.\mathrm{f}=1$ $\mathrm{v}.\mathrm{f}=5$

$1\times 1=\mathrm{moles}\times 5$

$\mathrm{Mole}=1/5$

Step 4:
${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$ can't oxidize.

Total number of moles of ${\mathrm{KMnO}}_4$ $$\begin{gathered} =3/5+6/5+1/5 \\ =2 \end{gathered}$$

Therefore, option (B) is the correct answer.