**Answer: $\Delta E_{(combustion)} = - \frac{44Y}{X} J mol^{-1}$**

$CH_3CHO + \frac{5}{2} O_2 + \longrightarrow 2CO_2 + 2H_2O$

Heat of combustion when one burns to give desirable products is

X gm ethanal give heat = Y joules

$\therefore$ 44 gm ethanal gives heat = $\frac{44Y}{X} J mol^{-1}$

As reaction is exothermic,

$\therefore \Delta E_{(combustion)} = - \frac{44Y}{X} J mol^{-1}$