Question

In redox reaction, $H_{2}O_{2}$ oxidises $K_{4}\left[\right.Fe\left(\right.CN\left(\left.\right)_{6}\left]\right.$ into $K^{+},Fe^{+ 3},CO_{3}^{ - 2}$ and $NO_{3}^{-}$ ions in acidic medium, than how many moles of $H_{2}O_{2}$ will react with $1mole$ of $K_{4}\left[\right.Fe\left(\right.CN\left(\left.\right)_{6}\left]\right.$

• A. $5 \, moles$
• B. $9 \, moles$
• C. $8 \, moles$
• D. $30.5 \, moles$

Answer 1

Answer: (D)

$n-factor \, of \, H_{2}O_{2}=2$
$n-factor \, of \, K_{4}\left[\right.Fe\left(\right.CN\left(\left.\right)_{6}\left]\right.=61$
$\because \frac{m o l e \, o f \, H_{2} O_{2}}{1 \, m o l e \, o f \, K_{4} F e \left(\right. C N \left(\left.\right)_{6}}=\frac{61}{2}$
$\therefore \, \, mole \, of \, H_{2}O_{2}=\frac{61}{2}\times 1=30.5$