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For the reaction ${XCO}_{3} (s) ⇌ XO (s) + {CO}_{2} (g), K_{p} = 1 .642 atm at 727^{\circ} C . If 4$ Mole of ${XCO}_{3} (s)$ was put into a 50 liter container and heated to 727^OC What mole percent of the XCO₃, remains unreacted at equilibrium?

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Detailed Solution

Moles ofCO₂ present at equilibrium

$= \frac{1.642 \times 50}{0.0821 \times 1000} = 1 (∵ K_{p} = P_{{CO}_{2}} = \frac{nRT}{V}) Mole % of {XCO}_{3} decomposed = \frac{1}{4} \times 100 = 25 %$

Hence, 75% remains undecomposed

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