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. 32.5 g
B. 24.6 g
C. 40.6 g
D. 8.0 gm

Answer

A

Explanation

`K_C=[CO_2]=0.05` mole/litre
so moles of `CO_2=6.50xx0.05` moles=0.3250 moles
`CaCO_3 hArr CaO+CO_3`
1 mole of `CO_2` =1 mole of `CaCO_3`
0.3250 moles of `CO_2`=0.3250 moles of `CaCO_3=0.3250xx100` gm of `CaCO_3` =32.5 gm of `CaCO_3`