Question

What volume of $O_2$ (g) measured at 1atm and 273 K will be formed by action of 100mL of 0.5 N KMnO${}_4$ on hydrogen peroxide in an acid solution? The skeleton equation for the reaction is
KMnO${}_4$ + H${}_2$SO${}_4$ + H${}_2$O${}_2$ $\to$ K${}_2$SO${}_4$ + MnSO${}_4$ + O${}_2$ + H${}_2$O

• A. 0.12 litre
• B. 0.028 litre
• C. 0.56 litre
• D. 1.12 litre

Answer 1

The correct option is C 0.56 litre

$2KMn{O}_4+3H_{2}S{o}_4+5H_2{O}_2⟶K_{2}S{O}_4+2MnS{O}_4+5O_2+8H_{2}O⟶\left(1\right)$

Normality of $KMn{O}_4$ solution=$0.5$

$\Rightarrow$ Molarity of $KMn{O}_4$ solution=$\frac{0.5}{5}=0.1M$

$[\because$ Normality=$Molarity\times nfactor$

Here, for $KMn{O}_4$ $nfactor$ is $5$ as $e^{-}$ change is $5]$

Now, Molarity =$0.1M$, which means

$0.1$ moles of $KMn{O}_4$ dissolved in $1000ml$ solution

$\therefore$ 1n $100ml$ solution, moles of $KMn{O}_4$ is $\frac{100\times 0.1}{1000}=0.01$ moles

From $\left(1\right)$ , it is clear that

$2$ moles $KMn{O}_4$ produce $5$ moles $O_2$

$\Rightarrow 0.01$ moles $KMn{O}_4$ will give $0.025$ moles of $O_2$

$\therefore$ Molar volume of $0.025$ moles of $O_2$ gas is as $1$ mole=$22.4L$ of $O_2$

$\Rightarrow 0.025$ moles=$0.025\times 22.4L$

$=0.56L$