for the reaction XCO3(s)======>XO(s) + CO2(g) Kp=1.642 at 727 degree celsius. If 4 moles XCO3(s) was put into a 50 litre container and heated to 727 degree celsius. what mole percent of the XCO3 remains unreacted at equilibrium?

XCO₃  ==>   XO  +   CO₂

  4               0          0                                  (initial)

 4-x             x           x

partial pressure of CO₂ = p then

K_p = p                                                           (at equilibrium )

pv = nRT

p = nRT/V = xRT/V  so

K_p = XRT/V            

after plugging all the values we get

X = 1

moles of XCO₃ remaining are 3 ...mol % remain unreacted is 75%

                              XCO3(s)======>XO(s) + CO2(g)

GIVEN INITIALLY         4 MOLES             0            0  MOLES

AT EQ                       4-a                     a             a moles

k_p=1.642      =>    k_p=P_co2 =  1.642

therefore acc toPV=nRT         n =   1 mole

therefore unreacted xco₃=  4-a  =4-1  =  3 mole

so  mole percent of unreacted reactant =  3 x100 /4  =  75%