Given, $CH _{4}+2 O _{2} \longrightarrow CO _{2}+2 H _{2} O$
$\Delta H=-210 \,kcal / mol\,\,\, ...(i)$
$C _{2} H _{6}+\frac{7}{2} O _{2} \longrightarrow 2 CO _{2}+3 H _{2} O$
$\Delta H=-368 \,kcal / mol \,\,\,...(ii)$
On subtracting Eq (i) from Eq (ii), we get
$CH _{2}+\frac{3}{2} O _{2} \longrightarrow CO _{2}+ H _{2} O$
$\Delta H=-158 \,kcal / mol$
$\therefore $ Enthalpy of combustion of one $CH _{2}$ unit
$=-158\,kcal / mol$
$\Delta H_{\text {comb }}\left( C _{10} H _{22}\right)=\Delta H_{\text {comb }}\left( CH _{4}\right)+9 \times \Delta H_{\text {comb }}\left( CH _{2}\right)$
$=-210+(9 \times-158)$
$=-1632 \,kcal$