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Stoichiometric Calculations

A 5.0 cm 3 so...

Question

A ${5.0} {cm}^{3}$ solution of $H_{2} O_{2}$ liberates $0.508 g$ of iodine from acidified $K I$ solution. Calculate the strength of $H_{2} O_{2}$ solution in terms of volume strength at STP.

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Solution

$H_{2} O_{2} + H_{2} S O_{4} + 2 K I ⟶ K_{2} S O_{4} + I_{2} + 2 H_{2} O$
We know,
Molar mass of $H_{2} O_{2} = 34 g$
Molar mass of $I_{2} = 254 g$

So, $254 g$ of $I_{2}$ is obtained from $34 g$ $H_{2} O_{2}$
$0.508 g$ of $I_{2}$ is obtained from $\frac{34}{254} \times 0.508 g H_{2} O_{2} = 0.068 g H_{2} O_{2}$

$5 mL of the solution contains = 0.068 g H_{2} O_{2}$
$1 mL of the solution contains = \frac{0.068}{5} g H_{2} O_{2} = 0.0136 g H_{2} O_{2}$

$2 H_{2} O_{2} 68 g ⟶ 2 H_{2} O + O_{2} 22400 mL$
$68 g of H_{2} O_{2} produces 22400 mL O_{2}$
$0.0136 g of H_{2} O_{2} will produce = \frac{22400}{68} \times 0.0136 mL of O_{2} = 4.48 mL of O_{2}$

$1 mL$ of $H_{2} O_{2}$ solution gives $4.48 mL$ of oxygen
So, strength of $H_{2} O_{2}$ solution = $4.48 V$

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