For the equilibrium:CaCO_{3(s)} rightleftharp | vedclass

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(D) The equilibrium reaction is: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$.Initial moles of $CaCO_{3} = \frac{50 \ g}{100 \ g/mol} = 0.5

Vedclass · Accepted Answer

$60$

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(D) The equilibrium reaction is: $CaCO_{3(s)} ightleftharpoons CaO_{(s)} + CO_{2(g)}$.Initial moles of $CaCO_{3} = \frac{50 \ g}{100 \ g/mol} = 0.5 \ mol$.For this reaction,$K_{p} = P_{CO_{2}} = 1.64 \ atm$.Using the ideal gas equation $PV = nRT$ for $CO_{2}$ gas:$1.64 \ atm 	imes 10 \ L = n_{CO_{2}} 	imes 0.082 \ L \ atm \ K^{-1} \ mol^{-1} 	imes 1000 \ K$.$16.4 = n_{CO_{2}} 	imes 82$.$n_{CO_{2}} = \frac{16.4}{82} = 0.2 \ mol$.Since $1 \ mol$ of $CaCO_{3}$ produces $1 \ mol$ of $CO_{2}$,the moles of $CaCO_{3}$ reacted $= 0.2 \ mol$.Moles of $CaCO_{3}$ remaining $= 0.5 - 0.2 = 0.3 \ mol$.Percentage of unreacted $CaCO_{3} = \frac{0.3}{0.5} 	imes 100 = 60 \%$.

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