What is the percentage strength of the $11.2 L$ volume solution of $H_{2} O_{2}$ ?

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Solution

Step 1: Given 11.2 L volume solution of $H_{2} O_{2}$
It means that $1 L$ of $H_{2} O_{2}$ will give $11.2 L$ of oxygen $(O_{2})$ at N.T.P
Step 2: Balanced chemical equation:
$2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)$
Step 3: Calculation:
$22.4 L$ of O₂ at N.T.P is produced from $68 g H_{2} O_{2}$
so, $11.2 L$ of O₂ at N.T.P will be produced by $H_{2} O_{2}$ = $\frac{68}{22.4} \times 11.2 = 34 g L^{- 1}$
Step 4: Percentage strength ( % w/w )
$\frac{34}{1000} \times 100 = 3.4 %$
Thus, the percentage volume strength of solution is $3.4 %$ .

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$1) I_{2} + H_{2} O_{2} + 2 {OH}^{-} \to 2 I^{-} + 2 H_{2} O + O_{2}$

$2) H_{2} O_{2} + HOCl \to {Cl}^{-} + H_{3} O^{+} + O_{2}$

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