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Ion Electron Method

What volume o...

Question

What volume of $O_{2} (g)$ measured at 1 atm and 273 K will be formed by the action of 100 ml of 0.5 N of $K M n O_{4}$ on hydrogen peroxide in an acidic solution?
$K M n O_{4} + H_{2} S O_{4} + H_{2} O_{2} \to K_{2} S O_{4} + M n S O_{4} + O_{2} + H_{2} O$

0.12 L

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0.03 L

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0.56 L

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1.12 L

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Solution

The correct option is C 0.56 L
$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 5 O_{2} + 8 H_{2} O$
Normality of $K M n O_{4}$ solution= 0.5 N
Since Normality = Mortality* n-factor
And n-factor for $K M n O_{4}$ is 5
Molarity of $K M n O_{4}$ solution= $\frac{5}{0.5} = 0.1 M$
Now 0.1 M means 0.1 moles of $K M n O_{4}$ are dissolved in 1000ml
so in 100 ml moles dissolved are 0.01 moles
Now according to the reaction, 2 moles of $K M n O_{4}$ produce 5 moles of $O_{2}$
hence 0.01 moles will produce 0.025 moles of $O_{2}$
1 mole of oxygen gas occupies 22.4 L of gas
so 0.025 moles of oxygen gas will occupy
= 0.025*22.4 L = 0.56 L of $O_{2}$

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