A gaseous mixture of methane and ethane (total 0.0200 mol) is burned completely in excess oxygen in a bomb calorimeter whose heat capacity (including…

A bomb calorimeter measures the heat at constant volume, so the energy change $\Delta U$ must be used rather than the enthalpy change. Convert using $\Delta U=\Delta H-\Delta n_gRT$. For $\mathrm{CH_4+2O_2\to CO_2+2H_2O(l)}$, $\Delta n_g=1-3=-2$, so $\Delta U=-890+2(8.314\times10^{-3})(298)=-885.04\ \mathrm{kJ\,mol^{-1}}$. For $\mathrm{C_2H_6+\tfrac72O_2\to2CO_2+3H_2O(l)}$, $\Delta n_g=2-4.5=-2.5$, so $\Delta U=-1560+2.5(2.4777)=-1553.81\ \mathrm{kJ\,mol^{-1}}$. Heat released $=5.00\times4.48=22.4\ \mathrm{kJ}$, which is $1120\ \mathrm{kJ}$ per mole of mixture. With $x$ as the mole fraction of ethane, $885.04(1-x)+1553.81x=1120$, so $x=(1120-885.04)/(668.77)=0.351$. The mole fraction of ethane is $0.35$. Using the $\Delta H$ values directly in the heat balance would ignore the $\Delta n_gRT$ term and give a slightly different, incorrect composition.