Question

For the reaction
`CaCO_(3)(s) hArr CaO(s)` was put in to `10 L` container and heated to `800^(@)C`, what percentage of the `CaCO_(3)` would remain unreacted at equilibrium.

Answer 1

`CaCO(s) hArr CaO(s)+CO_(2)(g)`
`K_(p)=P_(CO_(2))=1.16 "atm"`
`n(CO_(2))=(PV)/(RT)=(1.16 "atm" xx10.0L)/((0.0821 L "atm mol"^(-1) K^(-1))(1073 K))`
`=0.132 "mol"`
moles of `CaCO_(3)` initial present `=20//100=0.2 "mol"`
`%` of dissociation of `CaCO_(3)=0.132/0.2xx100=66%`
`%` of `CaCO_(3)` left `=100-66=34%`