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For the reaction CaCO(3)(s) hArr CaO(s) ...

For the reaction `CaCO_(3)(s) hArr CaO(s)` `K_p=1.16" atm"` at `800^(@)C`.
If `20 g` of `CaCO_3` was put in to `10 L` container and heated to `800^(@)C`, what percentage of the `CaCO_(3)` would remain unreacted at equilibrium.

Text Solution

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`CaCO(s) hArr CaO(s)+CO_(2)(g)`
`K_(p)=P_(CO_(2))=1.16 "atm"`
`T=800^(@)C=800+273=1073 K`
`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%`
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