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Electrolysis and Electrolytes

364.5 g of ...

Question

$364.5$ g of $K C l O_{3}$ will produce $x$ L of $O_{2}$ at NTP. The value of $x$ is :

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Solution

The reaction is as follows:
$2 K C l O_{3} \to 2 K C l + 3 O_{2}$
$122.5 \times 2$ $3 \times 22.4$ (in litres)
So, $364.5$ g will produce $= 3 \times 22.4 \times \frac{364.5}{2} \times 122.5 = 100$ L

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