$25mL$ of $0.50M{H}_2{O}_2$ solution is added to $50mL$ of $0.20MKMn{O}_4$ in acid solution. Which of the following statements are true?

The correct option is C $0.005$ mole of $KMn{O}_4$ is left
Finding the limiting reagent

The reaction involved is:

$KMn{O}_4+H_2{O}_2\to M{n}^{2+}+H_{2}O+O_2$

The balanced reaction is:

$2KMn{O}_4+5H_2{O}_2\downarrow 5O_2+2MnO+2KOH+4H_{2}O$

Moles of $KMn{O}_4$ taken

= Molarity $\times$ Volume (in L)

$=0.20\times \frac{50}{1000}=0.01mol$

Moles of $H_2{O}_2$ taken

$=Molarity\times Volume\left(inL\right)$

$\Rightarrow 0.50\times \frac{25}{1000}$

$\Rightarrow 0.0125$

From the balanced chemical equation, $2mol$ of $KMn{O}_4$ reacts with $5mol$ of $H_2{O}_2.$

$1mol$ of $KMn{O}_4$ reacts with $2.5mol$ of $H_2{O}_2.0.01mol$ of $KMn{O}_4$ will react with $0.025mol$ of $H_2{O}_2$.

But the total moles of $H_2{O}_2$ present is $\mathrm{0.0125.}$

Hence, $H_2{O}_2$ is the limiting reagent.

Analysing the option

Option (A):

$5mol$ of $H_2{O}_2$ produces $5mol$ of $O_2$
Therefore, $0.0125mol$ of $H_2{O}_2$ will produce $0.0125mol$ of $O_2$.

Option (B):

$5mol$ of $H_2{O}_2$ react with $2mol$ of $KMn{O}_4$

$1mol$ of $H_2{O}_2$ react with $\frac{2}{5}mol$ of $KMn{O}_4$

$0.0125moles$ of $H_2{O}_2$ will react with $0.005moles$ of $KMn{0}_4$.

So, the moles of $KMn{O}_4$ left

$=0.01-0.005=0.005mol$

Option (C):

Moles of oxygen evolved $=0.0125mol$
Mass of oxygen evolved = Number of moles $\times$ Molar mass

$=0.0125mol\times 32gmo{l}^{-1}=0.4g$

Option (D):

Since $H_2{O}_2$ is the limiting reagent, so all of $H_2{O}_2$ reacts.

So, option (B) is the correct answer.