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What is the minimum mass of CaCO3(s), be...

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 : `[K^@=0.05` mole/litre]
`CaCO_3(s)hArrCaO(s)+CO_2(g)`

A

`32.5g`

B

`24.6g`

C

`40.9g`

D

`8.0g`

Text Solution

Verified by Experts

The correct Answer is:
A
`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)`
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