$5 \ \text{moles}$ 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:

At $T \ K$,$K_{c}$ for the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$ is $16$. If initially one mole each of all the four gases are taken in a $1 \ L$ vessel,the equilibrium concentrations of $SO_{3(g)}$ and $SO_{2(g)}$ in $mol \ L^{-1}$ respectively are:

At $783 \, K$ in the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the molar concentrations $(mol \, L^{-1})$ of $H_2, I_2$ and $HI$ at some instant of time are $0.1, 0.2$ and $0.4$,respectively. If the equilibrium constant is $46$ at the same temperature,then as the reaction proceeds:

Two solids dissociate as follows:
$A_{(s)} \rightleftharpoons B_{(g)} + C_{(g)}$; $K_{p_1} = x \, atm^2$
$D_{(s)} \rightleftharpoons C_{(g)} + E_{(g)}$; $K_{p_2} = y \, atm^2$
The total pressure when both the solids dissociate simultaneously is:

For the reversible reaction $2NO_2 \leftrightarrow[K_2]{K_1} N_2O_4$,the rate of disappearance of $NO_2$ is given by:

$A$ solid $XY$ kept in an evacuated sealed container undergoes decomposition to form a mixture of gases $X$ and $Y$ at temperature $T$. The equilibrium pressure is $10 \, bar$ in the vessel. $K_p$ for this reaction is