Enthalpy of combustion of methane and ethane are - 210kcal / mol...

Given,&

${CH}_{4} + 2{O}_{2} ⟶ {CO}_{2} + 2{H}_{2}O; \Delta H = - 210kcal / mol... ( i )$

${C}_{2}{H}_{6} + \dfrac{7}{2}{O}_{2} ⟶ 2{CO}_{2} + 3{H}_{2}O; \Delta H = - 368kcal / mol... ( ii )$

On subtracting Eq. $( i )$ from Eq $( ii )$, we get

${CH}_{2} + \dfrac{3}{2}{O}_{2} ⟶ {CO}_{2} + {H}_{2}O; \Delta H = - 158kcal / mol$

$∴$ Enthalpy of combustion of one ${CH}_{2}unit$

$= - 158kcal / mol$

${ΔH}_{comb}\left({C}_{10}{H}_{22}\right) = {ΔH}_{comb} \left({CH}_{4}\right) +9 \times {ΔH}_{comb} \left({CH}_{2}\right)$

$= - 210 + \left(9 \times -158\right)$

$= - 1632kcal$