Consider the reversible reaction, $\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{g}\right)$ . Initially two moles of ‘A’ is taken in a two litre vessel. Total moles at equilibrium were found to be three. Percentage of decomposition of A(g) is 

$\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{g}\right)$

 Initial moles 

$\begin{array}{cccccc} & {\mathrm{A}}_{\left(\mathrm{g}\right)}& \rightleftharpoons & {\mathrm{B}}_{\left(\mathrm{g}\right)}& +& {\mathrm{C}}_{\left(\mathrm{g}\right)}\\ \mathrm{initial}\mathrm{moles}& 2& & -& & -\\ \mathrm{Moles}\mathrm{at}\mathrm{equilibrum}& 2-\mathrm{x}& & \mathrm{x}& & \mathrm{x}\end{array}$

           Total moles at equilibrium = 2 - X + X + X = 3

                                                                 X = 1

          % of decomposition $=\frac{1}{2}\times 100=50\mathrm{\%}$