Acetylene burns in air according to the following equation:
$$\begin{array}{r}\mathrm{C}_{2} \mathrm{H}_{2}(g)+{ }_{2}^{5} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\
\Delta H_{\mathrm{ru}}^{\circ}=-1255.8 \mathrm{~kJ}\end{array}$$Given $\Delta H_{\mathrm{f}}^{\circ}$ of $\mathrm{CO}_{2}(g)=-393.5 \mathrm{~kJ} / \mathrm{mol}$ and $\Delta H_{\mathrm{f}}^{\circ}$ of $\mathrm{H}_{2} \mathrm{O}(g)=$$-241.8 \mathrm{~kJ} / \mathrm{mol},$ find $\Delta H_{i}$ of $\mathrm{C}_{2} \mathrm{H}_{2}(g)$