What is the minimum mass of `CaCO_3(s)`, below which it decomposes completely, required to establish equilibrium in a 6.50 litre container for the reaction :
`CaCO_3(s)hArr CaO(s)+CO_2(g), K_c=0.05`

Options

A. `32.5g`
B. `24.6g`
C. `40.9g`
D. `8.0g`

Answer

A

Explanation

`K_(c)=[CO_(2)]=0.05` mole//litre
so moles of `CO_(2)=6.50xx0.05 "moles"=0.3250` moles
`CaCO_(2)hArrCaO+CO_(2)`
`1 "mole of" CO_(2)=1 "mole of" CaCO_(3)`
`0.3250 "moles of" CO_(2)=0.3250 "moles of" CaCO_(3)=0.3250xx100 g "of" CaCO_(3)=32.50 "gm of" CaCO_(3)`