Question

The number of moles of $KMn{O}_4$ required, to react with 2 moles of ferrous oxalate is:

• A. $\frac{6}{5}$
• B. $\frac{2}{5}$
• C. $\frac{4}{5}$
• D. $1$

Answer 1

The correct option is A $\frac{6}{5}$

Here calculating the n factor,
$M{n}^{7+}+5e^{-}\to M{n}^{2+}]5$
$\begin{array}{c}F{e}^{2+}\to F{e}^{3+}+e^{-}\\ C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}\end{array}]3$


Using ,
Equivalents of $KMn{O}_4=$Equivalents of $Fe{C}_2{O}_4$

$\Rightarrow$ Moles $\times$ n factor of $KMn{O}_4$= Moles $\times$ n factor of $Fe{C}_2{O}_4$

putting values,

$\Rightarrow$ Moles of $KMn{O}_4$ $\times$ 5 = 2 $\times$ 3

$\Rightarrow$ Moles of $KMn{O}_4$ = $\frac{2\times 3}{5}$

$\Rightarrow$ Moles of $KMn{O}_4$ = $\frac{6}{5}$

Hence, (a) is the correct answer.