Number of moles of $\mathrm{KMnO}_{4}$ required to oxidize one mole of $\mathrm{Fe}\left(\mathrm{C}_{2}…

Number of moles of $\mathrm{KMnO}_{4}$ required to oxidize one mole of $\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}ight)$ in acidic medium is
  1. $0.167$
  2. $0.6$
  3. $0.2$
  4. $0.4$

Solution

The required equation is $2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow$
$\mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+3 \mathrm{H}_{2} \mathrm{O}+\quad 5[\mathrm{O}]$
$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ nascent oxygen
$2 \mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}ight)+3 \mathrm{H}_{2} \mathrm{SO}_{4}+3[\mathrm{O}] \longrightarrow$
$\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}ight)_{3}+2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}$
O required for 1 mol. of $\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}ight)$ is $1.5,5 \mathrm{O}$ are obtained from 2 moles of $\mathrm{KMnO}_{4}$
$\therefore 1.5[\mathrm{O}]$ will be obtained from
$=\frac{2}{5} \times 1.5=0.6$ moles of $\mathrm{KMnO}_{4}$. .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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