The standard heat of formation (Δ_f H⁰₂₉₈) of ethane (in kJ/mo1), if the heat of combustion of ethane, hydrogen and graphite are – 1560, -393.5 and -286 kJ/mol, respectively is(X)kJ/mol. Find the value of $\frac{\mathrm{X}-111.5}{9}$

To find out $∆$Hf of C₂H₆
$2\mathrm{C}\left(\text{ graphite }\right)+3{\mathrm{H}}_2\to {\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{\Delta H}}_{\mathrm{f}}=?$
Given : 
$\text{ - (a) }{\mathrm{H}}_2\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}{\mathrm{\Delta H}}_{\mathrm{c}}=-393.5\mathrm{kJ}/\mathrm{moI}$
$\text{ (b) }{\mathrm{C}}_2{\mathrm{H}}_6+\frac{7}{2}{\mathrm{O}}_2\to 2{\mathrm{co}}_2+3{\mathrm{H}}_2\mathrm{O}{\mathrm{\Delta H}}_{\mathrm{c}}=1560\mathrm{kJ}/\mathrm{mol}$
$\text{ (c) }\mathrm{C}\text{ (graphite) }+{\mathrm{O}}_2\to {\mathrm{CO}}_2{\mathrm{\Delta H}}_{\mathrm{c}}=-286\mathrm{kJ}/\mathrm{mol}$
2 x eqn. (c) +3 eqn (a) + reverse equation (b) = Aim
=> $∆$H_f = 3 x – 393.5 + 2 x – 286 + 1560 = - 192.5 kJ/mo1=X
$\frac{192.5-111.5}{9}=\frac{81}{9}=9$