An electric discharge is passed through a mixture containing 50 c.c. of ${\mathrm{O}}_2$ and 50 c.c. of ${\mathrm{H}}_2.$ The volume of the gases formed (i) at room temperature and (ii) at ${110}^0\mathrm{C}$ will be 

 $$\begin{gathered} \left(\mathrm{i}\right)\mathrm{At}\mathrm{room}\mathrm{temperature}2{\mathrm{H}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)} \\ 2\mathrm{moles} 1\mathrm{mole} 2\mathrm{moles} \\ 2\times 22.4\mathrm{l} 22.4\mathrm{l} \\ 50\mathrm{ml} ?(=25\mathrm{ml}) \\ \mathrm{given} \mathrm{volume} \mathrm{of} {\mathrm{O}}_2=50 \mathrm{ml} \\ \mathrm{volume} \mathrm{of} {\mathrm{O}}_2 \mathrm{released}(\mathrm{which} \mathrm{is} \mathrm{unreacted})=50-25 =25\mathrm{ml} \\ \mathrm{volume} \mathrm{of} \mathrm{gases} \mathrm{formed} \mathrm{at} \mathrm{room} \mathrm{temperature} = 25\mathrm{cc}(\mathrm{since} {\mathrm{H}}_2\mathrm{O} \mathrm{formed} \mathrm{is} \mathrm{liquid}) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{At} 110°\mathrm{c} : \\ 2{\mathrm{H}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)} \\ 50\mathrm{ml} 25\mathrm{ml} 50\mathrm{ml} \\ \mathrm{volume} \mathrm{of} \mathrm{gases} \mathrm{formed} =\mathrm{volume} \mathrm{of} {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \mathrm{formed}+\mathrm{volume} \mathrm{of} {\mathrm{O}}_2 \mathrm{unreacted} \\ = 50 + 25 =75\mathrm{ml} \end{gathered}$$