$5 \ \text{moles}$ of $SO_2$ and $5 \ \text{moles}$ of $O_2$ are allowed to react to form $SO_3$ in a closed vessel. At the equilibrium stage,$60\%$ of $SO_2$ is used up. The total number of moles of $SO_2$,$O_2$,and $SO_3$ in the vessel now is:

The values of $K_p$ for the reactions,
$X \rightleftharpoons Y + Z$ $...(i)$
$A \rightleftharpoons 2B$ $...(ii)$
are in the ratio $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the ratio of total pressure at equilibrium for $(i)$ and $(ii)$ is:

$K_{c}$ 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 $mol \ L^{-1}$?

In the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,one mole of $PCl_5$ is started in a $5 \ L$ vessel. If $0.3 \ mol$ of $PCl_5$ is present at equilibrium,find the concentration of $PCl_3$,total moles,and the value of $K_c$.

One mole of $O_{2(g)}$ and two moles of $SO_{2(g)}$ were heated in a closed vessel of $1 \, L$ capacity at $1098 \, K$. At equilibrium,$1.6 \, moles$ of $SO_{3(g)}$ were found. The equilibrium constant $K_c$ of the reaction would be

When $5 \ mol$ of $SO_2$ and $5 \ mol$ of $O_2$ are reacted,$60\%$ of $SO_2$ is consumed at equilibrium. If the total pressure of the equilibrium mixture is $1 \ atm$,then the partial pressure of $O_2$ will be ...... $atm$.