What is the minimum mass of ${\mathrm{CaCO}}_3\left(s\right)$, below which it decomposes completely, required stablish equilibrium in a 6.50 litre container for the reaction : ${\mathrm{CaCO}}_3\left(s\right)\rightleftharpoons \mathrm{CaO}\left(s\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right); K_c=0.05$

For the given reaction,$CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$

here,

$K_c=\left[C{O}_2\right]=0.005 mo{l}^{\mathit{-}\mathit{1}}$

$\therefore \mathrm{for} 6.51 \mathrm{container}$

moles of $C{O}_2$ required $=6.5\times 0.005$ $=0.0325 \mathrm{mole}$

$\therefore$ moles of $CaC{O}_3$ required $=0.0325 \mathrm{mole}$

$mass\mathit{ }of\mathit{ }CaC{O}_3\left(min\right)=0.0325\times 100$

$=32.5 \mathrm{g}$