Given: $2\mathrm{C}_2\mathrm{H}_6+7\mathrm{O}_2\rightarrow4\mathrm{CO}_2+6\mathrm{H}_2\mathrm{O}$ $\because 2$ moles of ethane reacts with 7 moles of oxygen to produce -6 moles of $\mathrm{H}_2\mathrm{O}+4$ moles of carbondioxide. $2\mathrm{ml}$ . of ethane reacts with $7\mathrm{ml}$ . of oxygen = $4\mathrm{ml}$ . of carbon dioxide $+\mathrm{ml}$ of $\mathrm{H}_2\mathrm{O}$ . $\therefore 80\mathrm{ml}$ . of ethane reacts with $280\mathrm{ml}$ . of oxygen $=160\mathrm{ml}$ . of $\mathrm{CO}_2+240\mathrm{ml}$ . of $\mathrm{H}_2\mathrm{O}$In this reaction ethane in the limiting reagent and $20\mathrm{ml}$ . of oxygen will be left, unreacted.The composition of the resultant gaseous mixture when measured at room temperature will be:$20\mathrm{ml}$. of oxygen gas : $160\mathrm{ml}$ . of $\mathrm{CO}_2$ and $240\mathrm{ml}$ . of $\mathrm{H}_2\mathrm{O}$.