For the reaction XCO_{3(s)} rightleftharpoons | vedclass

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(D) For the reaction: $XCO_{3(s)} \rightleftharpoons XO_{(s)} + CO_{2(g)}$At $t=0$,moles of $XCO_3 = 4$.At equilibrium,moles of $XCO_3 = 4-x$ and mole

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(D) For the reaction: $XCO_{3(s)} ightleftharpoons XO_{(s)} + CO_{2(g)}$At $t=0$,moles of $XCO_3 = 4$.At equilibrium,moles of $XCO_3 = 4-x$ and moles of $CO_2 = x$.$K_p = K_c(RT)^{\Delta n}$. Here $K_c = [CO_2] = \frac{x}{50}$,$T = 727+273 = 1000 	ext{ K}$,and $R = 0.0821 	ext{ L atm K}^{-1} 	ext{ mol}^{-1}$.$1.642 = \left(\frac{x}{50}ight) 	imes (0.0821 	imes 1000)^1$.$1.642 = \frac{x}{50} 	imes 82.1$.$x = \frac{1.642 	imes 50}{82.1} = \frac{82.1}{82.1} = 1 	ext{ mole}$.$\% 	ext{ reacted} = \frac{x}{4} 	imes 100 = \frac{1}{4} 	imes 100 = 25\%$.$\% 	ext{ unreacted} = 100\% - 25\% = 75\%$.

For the reaction $XCO_{3(s)} \rightleftharpoons XO_{(s)} + CO_{2(g)},$ $K_p = 1.642 \text{ atm}$ at $727^{\circ}C.$ If $4 \text{ moles}$ of $XCO_{3(s)}$ were placed into a $50 \text{ L}$ container and heated to $727^{\circ}C,$ what mole percent of the $XCO_3$ remains unreacted at equilibrium?

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