$15 g$ sample of $B a O_{2}$ is heated to $794^{o} C$ in a close evacuated vessel of $5$ litre capacity. How many $g$ of peroxide are converted to $B a O (s)$ .
$B a O_{2} (s) ⇌ 2 B a O (s) + O_{2} (g), K_{P} = 0.5 a t m$ .

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Solution

$K_{P} = (P_{O_{2}}) = 0.5 a t m$
$∵ P V = \frac{m}{M} R T$ or $P V = n R T$
$∴$ no. of moles of $O_{2} = n = \frac{P V}{R T} = \frac{0.5 \times 5}{0.0821 \times 1067} = 0.028$
1 mole of $B a O_{2}$ gives 1 mole of $O_{2}$
$∴$ 0.028 moles of $B a O_{2}$ gives 0.028 moles of $O_{2}$ .
$∴$ Weight of $B a O_{2}$ converted to $B a O = 0.028 \times M_{B a O_{2}}$
$= 0.028 \times 169 = 4.732 g$

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