What is the weight of available oxygen from a solution of $H_{2} O_{2}$ , if $20 m l$ of this solution needs $25 m l$ , $\frac{N}{20} K M n O_{4}$ for complete oxidation?

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Solution

Step 1: Write down the chemical formula
Hydrogen peroxide reacts with potassium permanganate which acts as an oxidizing agent giving oxygen as a product and reducing itself to a manganese ion as given below:
$5 H_{2} O_{2} (l) + 2 {KMnO}_{4} (s) + 3 H_{2} {SO}_{4} (l) \to 5 O_{2} (g) + K_{2} {SO}_{4} (s) + 2 {MnSO}_{4} (s) + 8 H_{2} O (l)$
This reaction occurs in an acidic medium hence the presence of sulphuric acid.
Step 2: Given information
Normality of $H_{2} O_{2}$ $N_{1}$ =?
Volume of $H_{2} O_{2}$ $V_{1} = 20 m l$
Normality of ${KMnO}_{4}$ $N_{2} = \frac{1}{20}$
Volume of ${KMnO}_{4}$ $V_{2} = 25 ml$
Step 3: Formulae to apply
$N_{1} V_{1} = N_{2} V_{2}$
$Weight of$ $O_{2}$ = $N_{1}$ $\times$ $M o l e c u l a r w e i g h t o f O_{2}$ ( $32 g$ )
Step 4: Calculations
Putting the values in the formula
$N_{1} V_{1} = N_{2} V_{2}$
$N_{1}$ $20 = \frac{1}{20} \times 25$
$N_{1}$ $= 0.0625$
molarity of hydrogen peroxide = molarity of oxygen therefore, the normality will be the same as well.
Putting the values in the second formula
$Weight of$ $O_{2}$ = $N_{1}$ $\times$ $M o l e c u l a r w e i g h t o f O_{2}$
$Weight of$ $O_{2}$ = $0.0625 \times 32 = 2 g$
Therefore, $2 g of O_{2}$ per Litre.

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