For the equilibrium : ( CaCO ₃(s) CaO (s)+ CO ₂(g) ; K _ p =1.64 ~atm ) at ( 1000 ~K ) ( 50 ~g ) of ( CaCO ₃ ) in a ( 10 ) litre closed vessel is heated to ( 1000 ~K ). Percentage of ( CaCO ₃ ) that remains unreacted at equilibrium is (Given ( R =0.082 ~L ~atm K⁻¹ ~mol ⁻¹ ) )

Step-by-step solution

CaCO ₃( ~S ) CaO ( S )+ CO ₂(g) ; K _ p =1.64 Number of moles of CaCO ₃= 50 100 =0.5 ~mol ~K _ p = P _ CO ₃ From the ideal gas equation pV = nRT 1.64 10=n 0.085 100n=0.2 Therefore, no. of moles of CO ₂ formed =0.2 ~mol No. of moles of unreacted CaCO ₃ left =0.5-0.2=0.3 ~mol Percentage of unreacted CaCO ₃= 0.3 0.5 100=60 %