what is the minimum mass of CaCo3,below which it decomposes completely,required to establish equilibrium in a 6.50 litre container for the reaction: CaCo3(s):::::CaO(s)+CO2(g)

CaCO3(s) --> CaO(s) + CO2(g) ...... Now at t=0 , moles are a (let) , 0 , 0 ,respectively .... At t= eqm. Moles are 0 , a ,a , respectively. ..... Now KC = [ CO2] ...as it is the only gas ...... So, ... 0.05 = a/6.5 ....., a = 0.325.......So moles of CaCO3 required....= 0.325 ....So mass of CaCO3 required. = Moles × Molar mass = 0.325 × 100 g = 32.5 g