If $8 \, \text{mol}$ of $PCl_5$ is heated in a closed vessel of $10 \, L$ capacity and $25\%$ of it dissociates into $PCl_3$ and $Cl_2$ at equilibrium,then the value of $K_p$ is .....

For the equilibrium:
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$; $K_{p} = 1.64 \ atm$ at $1000 \ K$.
$50 \ g$ of $CaCO_{3}$ in a $10 \ L$ closed vessel is heated to $1000 \ K$. The percentage of $CaCO_{3}$ that remains unreacted at equilibrium is:
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

The amount of $PCl_5$ (in moles) that needs to be added to a $1\,L$ vessel at $250\,^oC$ in order to obtain $0.1\,mol$ of $Cl_2$ for the given reaction is:
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$; $K_C = 0.0414\,mol\,L^{-1}$

$K_{c}$ for the following reaction is $99.0$: $A_{2(g)} \rightleftharpoons B_{2(g)}$. In a $1 \ L$ flask,$2 \ moles$ of $A_{2}$ were heated to $T(K)$ and equilibrium was reached. The concentrations at equilibrium of $A_{2}$ and $B_{2}$ are $C_{1}(A_{2})$ and $C_{2}(B_{2})$ respectively. Now,$1 \ mole$ of $A_{2}$ was added to the flask and heated to $T(K)$ to establish equilibrium again. The concentrations of $A_{2}$ and $B_{2}$ are $C_{3}(A_{2})$ and $C_{4}(B_{2})$ respectively. What is the value of $C_{3}(A_{2})$ in $mol \ L^{-1}$?

$A$ mixture of $SO_2$ and $O_2$ at $5 \, atm$ pressure reacts $30\%$ until equilibrium is reached. Determine the total pressure of the equilibrium mixture in $atm$.
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$

The reaction $2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$ is at equilibrium in a closed $15 \ L$ vessel at $300 \ K$. The total weight of the mixture of $NO_2$ and $N_2O_4$ in the vessel is $64.4 \ g$. The equilibrium constant for the reaction is $K_p = 6.67$. Assuming ideal gas behavior,the total pressure in the vessel (in $atm$) is: [Given: Gas constant $R = 0.082 \ atm \ L \ K^{-1} \ mol^{-1}$]